Sample Chapter
INSTANT DOWNLOAD COMPLETE TEST BANK WITH ANSWERS
Test Bank For Business Statistics A Decision Making Approach 9th Edition by David F. Groebner
SAMPLE QUESTIONS
Business Statistics, 9e (Groebner/Shannon/Fry)
Chapter 3 Describing Data Using Numerical Measures
1) If after graphing the data for a quantitative variable of interest, you notice that the distribution is highly skewed in the positive direction, the measure of central location that would likely provide the best assessment of the center would be the median.
Answer: TRUE
Diff: 2
Keywords: skew, median, center
Section: 3-1 Measures of Center and Location
Outcome: 1
2) A statistic is just another name for a parameter.
Answer: FALSE
Diff: 1
Keywords: statistic, parameter
Section: 3-1 Measures of Center and Location
Outcome: 1
3) The owner of a local gasoline station has kept track of the number of gallons of regular unleaded sold at his station every day since he purchased the station. This morning, he computed the mean number of gallons. This value would be considered a statistic.
Answer: FALSE
Diff: 2
Keywords: statistic, parameter
Section: 3-1 Measures of Center and Location
Outcome: 1
4) The Parks and Recreation manager for the city of Detroit recently submitted a report to the city council in which he indicated that a random sample of 500 park users indicated that the average number of visits per month was 4.56. This value should be viewed as a statistic by the city council.
Answer: TRUE
Diff: 2
Keywords: statistic, parameter, average
Section: 3-1 Measures of Center and Location
Outcome: 1
5) A statistic is a value that describes a population characteristic while a parameter is computed from a sample.
Answer: FALSE
Diff: 2
Keywords: statistic, parameter
Section: 3-1 Measures of Center and Location
Outcome: 1
6) The symbol μ is used to represent the sample mean.
Answer: FALSE
Diff: 1
Keywords: measure, spread, mean
Section: 3-1 Measures of Center and Location
Outcome: 1
7) The marketing manager for Voice-talk, a cell phone company, has taken a sample of 300 customers from the list of 4,356 total customers. The mean monthly bill for the last October based on the sample data is $45.62. The manager should realize that the mean bill for all 4,356 customers will actually be higher than $45.62.
Answer: FALSE
Diff: 2
Keywords: sample, mean
Section: 3-1 Measures of Center and Location
Outcome: 1
8) You are given the following data:
23 | 34 | 11 | 40 | 25 |
If these data were considered to be a population and you computed the mean, you would get the same answer as if these data were considered to be a sample from a larger population.
Answer: TRUE
Diff: 2
Keywords: population, mean, sample
Section: 3-1 Measures of Center and Location
Outcome: 1
9) You are given the following data:
23 | 34 | 11 | 40 | 25 | 47 |
Assuming that the data reflect a sample from a larger population, the sample mean is 30.00.
Answer: TRUE
Diff: 1
Keywords: sample, mean, population
Section: 3-1 Measures of Center and Location
Outcome: 1
10) You are given the following data:
23 | 34 | 11 | 40 | 25 | 47 |
Assuming that the data reflect the population of interest, the mean of the population is 36.00.
Answer: FALSE
Diff: 1
Keywords: mean, population
Section: 3-1 Measures of Center and Location
Outcome: 1
11) Data are considered to be right-skewed when the mean lies to the right of the median.
Answer: TRUE
Diff: 1
Keywords: skewed, skew, mean, median
Section: 3-1 Measures of Center and Location
Outcome: 1
12) The sample mean is an estimate of μ and may be either higher or lower than μ depending on the sample.
Answer: TRUE
Diff: 2
Keywords: mean, sample, population
Section: 3-1 Measures of Center and Location
Outcome: 1
13) When news articles report on household income level they usually report the median income, rather than the mean income. This would be because income is usually a right-skewed distribution.
Answer: TRUE
Diff: 2
Keywords: mean, median, skewed
Section: 3-1 Measures of Center and Location
Outcome: 1
14) You are given the following data:
9 | 11 | 14 | 22 | 31 |
Assuming that these data reflect the population of interest, these data can be considered symmetric.
Answer: FALSE
Diff: 2
Keywords: mean, median, symmetric
Section: 3-1 Measures of Center and Location
Outcome: 1
15) You are given the following data:
23 | 34 | 11 | 40 | 25 | 47 |
Assuming that these data are a sample selected from a larger population, the median value for these sample data is 25.5.
Answer: FALSE
Diff: 2
Keywords: median, sample
Section: 3-1 Measures of Center and Location
Outcome: 1
16) A distribution is said to be symmetric when the sample mean and the population mean are equal.
Answer: FALSE
Diff: 2
Keywords: mean, sample, population, symmetric
Section: 3-1 Measures of Center and Location
Outcome: 1
17) In a recent study of the sales prices of houses in a Midwestern city, the mean sales price has been reported to be $167,811 while the median sales price was $155,600. From this information, you can determine that the data involved in the study are left-skewed.
Answer: FALSE
Diff: 2
Keywords: mean, median, skew, skewed
Section: 3-1 Measures of Center and Location
Outcome: 1
18) One of the primary advantages of using the median as a measure of the center for a set of data is that the median is not affected by extreme values in the data.
Answer: TRUE
Diff: 1
Keywords: median, center, extreme
Section: 3-1 Measures of Center and Location
Outcome: 1
19) Suppose a study of houses that have sold recently in your community showed the following frequency distribution for the number of bedrooms:
Bedrooms | Frequency |
1 | 1 |
2 | 18 |
3 | 140 |
4 | 57 |
5 | 11 |
Based on this information, the mode for the data is 140.
Answer: FALSE
Diff: 2
Keywords: mode, frequency
Section: 3-1 Measures of Center and Location
Outcome: 1
20) Suppose a study of houses that have sold recently in your community showed the following frequency distribution for the number of bedrooms:
Bedrooms | Frequency |
1 | 1 |
2 | 18 |
3 | 140 |
4 | 57 |
5 | 11 |
Based on this information the mean number of bedrooms in houses that sold is approximately 3.26.
Answer: TRUE
Diff: 2
Keywords: mean, weighted
Section: 3-1 Measures of Center and Location
Outcome: 1
21) Suppose a study of houses that have sold recently in your community showed the following frequency distribution for the number of bedrooms:
Bedrooms | Frequency |
1 | 1 |
2 | 18 |
3 | 140 |
4 | 57 |
5 | 11 |
Based on this information, the median number of bedrooms in houses sold is 3.20.
Answer: FALSE
Diff: 2
Keywords: median
Section: 3-1 Measures of Center and Location
Outcome: 1
22) Suppose a study of houses that have sold recently in your community showed the following frequency distribution for the number of bedrooms:
Bedrooms | Frequency |
1 | 1 |
2 | 18 |
3 | 140 |
4 | 57 |
5 | 11 |
Based on this information, it is possible to determine that the distribution of bedrooms in homes sold is right-skewed.
Answer: TRUE
Diff: 3
Keywords: mean, median, skew, skewed
Section: 3-1 Measures of Center and Location
Outcome: 1
23) A data set in which the mean, median, and mode are all equal is said to be a skewed distribution.
Answer: FALSE
Diff: 1
Keywords: mean, median, mode, symmetric, skewed
Section: 3-1 Measures of Center and Location
Outcome: 1
24) First Pacific Bank has determined that the mean checking account balance for all its customers is currently $743.50. Based on this, it is fair to say that about half the customers have balances exceeding $743.50.
Answer: FALSE
Diff: 2
Keywords: mean, median
Section: 3-1 Measures of Center and Location
Outcome: 1
25) When analyzing annual incomes of adults in a market area, the marketing manager’s report indicated that the 90th percentile is $123,400. That means that 90 percent of the adult incomes in the market area fall at or below $123,400.
Answer: TRUE
Diff: 1
Keywords: percentile
Section: 3-1 Measures of Center and Location
Outcome: 1
26) When the median of a data set is 110 and the mean is 127, the percentile associated with the mean must be higher than 50 percent.
Answer: TRUE
Diff: 2
Keywords: mean, median, percentile
Section: 3-1 Measures of Center and Location
Outcome: 1
27) The second quartile for a set of data will have the same value as the 50th percentile only when the data are symmetric.
Answer: FALSE
Diff: 2
Keywords: quartile, percentile, symmetric
Section: 3-1 Measures of Center and Location
Outcome: 1
28) If a set of data has 1,500 values, the 30th percentile value will correspond to the 450th value in the data when the data have been arranged in numerical order.
Answer: TRUE
Diff: 3
Keywords: percentile, location
Section: 3-1 Measures of Center and Location
Outcome: 1
29) If a set of data has 540 values, the 3rd quartile corresponds to approximately the 135th value when the data have been arranged in numerical order.
Answer: FALSE
Diff: 3
Keywords: quartile, percentile, location
Section: 3-1 Measures of Center and Location
Outcome: 1
30) A set of data is considered to be symmetric if the 3rd quartile is three times larger than the 1st quartile.
Answer: FALSE
Diff: 2
Keywords: quartile, percentile, symmetric
Section: 3-1 Measures of Center and Location
Outcome: 1
31) If the mean value of a variable is 200 and the median is 150, the third quartile must be at least 200.
Answer: FALSE
Diff: 2
Keywords: mean, median, quartile
Section: 3-1 Measures of Center and Location
Outcome: 1
32) Recently an article in a newspaper stated that 75 percent of the households in the state had incomes of $20,200 or below. Given this input, it is certain that mean household income is less than $20,200.
Answer: FALSE
Diff: 3
Keywords: mean, percentile
Section: 3-1 Measures of Center and Location
Outcome: 1
33) It is possible for a set of data to have multiple modes as well as multiple medians, but there can be only one mean.
Answer: FALSE
Diff: 2
Keywords: mean, median, mode
Section: 3-1 Measures of Center and Location
Outcome: 1
34) A box and whisker plot shows where the mean value falls relative to the median for a variable.
Answer: FALSE
Diff: 1
Keywords: box, whisker, mean, median
Section: 3-1 Measures of Center and Location
Outcome: 2
35) The right and left edges of the box in a box and whisker plot represent the 3rd and 1st quartiles, respectively.
Answer: TRUE
Diff: 1
Keywords: quartile, box, whisker, edge
Section: 3-1 Measures of Center and Location
Outcome: 2
36) A recent study involving a sample of 3,000 vehicles in California showed the following statistics related to the number of miles driven per day: Q1 = 12, Q2 = 45, and Q3 = 56. Based on these data, we know that the distribution is skewed.
Answer: TRUE
Diff: 3
Keywords: quartile, skew, skewed, median
Section: 3-1 Measures of Center and Location
Outcome: 1
37) A recent study involving a sample of 3,000 vehicles in California showed the following statistics related to the number of miles driven per day: Q1 = 12, Q2 = 45, and Q3 = 56. Based on these data, if a box and whisker plot is developed, a value of 110 is an outlier.
Answer: FALSE
Diff: 3
Keywords: box, whisker, limit
Section: 3-1 Measures of Center and Location
Outcome: 2
38) A recent study involving a sample of 3,000 vehicles in California showed the following statistics related to the number of miles driven per day: Q1 = 12, Q2 = 45, and Q3 = 56. Based on these data, if a box and whisker plot is developed, the upper limit value is 122 miles.
Answer: TRUE
Diff: 3
Keywords: box, whisker, limit
Section: 3-1 Measures of Center and Location
Outcome: 2
39) In drawing a box and whisker plot the upper limit length of the whiskers is 1.5(Q3-Q1).
Answer: TRUE
Diff: 2
Keywords: box, whisker, limit
Section: 3-1 Measures of Center and Location
Outcome: 2
40) When surveyed, a sample of 1,250 patients at a regional hospital provided interviewers with the following summary statistics pertaining to the hospital charges:
Minimum = $278.00 Q1 = $1,245 Q2 = $3,567 Q3= $4,702.
Based on these data, if you were to construct a box and whisker plot, the value corresponding to the right-hand edge of the box would be $4,702.
Answer: TRUE
Diff: 2
Keywords: box, whisker, edge
Section: 3-1 Measures of Center and Location
Outcome: 2
41) When surveyed, a sample of 1,250 patients at a regional hospital provided interviewers with the following summary statistics pertaining to the hospital charges:
Minimum = $278.00 Q1 = $1,245 Q2 = $3,567 Q3= $4,702.
Based on these data, if you were to construct a box and whisker plot, the value $278 would be considered an outlier.
Answer: FALSE
Diff: 3
Keywords: box, whisker, outlier
Section: 3-1 Measures of Center and Location
Outcome: 2
42) When surveyed, a sample of 1,250 patients at a regional hospital provided interviewers with the following summary statistics pertaining to the hospital charges:
Minimum = $278.00 Q1 = $1,245 Q2 = $3,567 Q3= $4,702.
Based on these data, the distribution is seen to be symmetric.
Answer: FALSE
Diff: 2
Keywords: median, quartile, symmetric
Section: 3-1 Measures of Center and Location
Outcome: 1
43) A dairy farm in Wisconsin bottles milk in one gallon containers. At a recent meeting, the production manager asked top management for a new filling machine that he argued would assure that all containers had exactly one gallon of milk. Based on sound statistical principles, the top management group should conclude that the production manager could have merit to his argument.
Answer: FALSE
Diff: 2
Keywords: variation
Section: 3-2 Measures of Variation
Outcome: 3
44) The range is an ideal measure of variation since it is not sensitive to extreme values in the data.
Answer: FALSE
Diff: 1
Keywords: range, variation, sensitive
Section: 3-2 Measures of Variation
Outcome: 3
45) When a variance is calculated for a data set, the resulting value is the same regardless of whether the data set is treated as a population or a sample.
Answer: FALSE
Diff: 1
Keywords: variance, population, sample
Section: 3-2 Measures of Variation
Outcome: 3
46) The Good-Guys Car Dealership has tracked the number of used cars sold at its downtown dealership. Consider the following data as representing the population of cars sold in each of the 8 weeks that the dealership has been open.
3 | 5 | 2 | 7 | 7 | 7 | 9 | 0 |
The population range is 9.
Answer: TRUE
Diff: 1
Keywords: range, population
Section: 3-2 Measures of Variation
Outcome: 3
47) The Good-Guys Car Dealership has tracked the number of used cars sold at its downtown dealership. Consider the following data as representing the population of cars sold in each of the 8 weeks that the dealership has been open.
3 | 5 | 2 | 7 | 7 | 7 | 9 | 0 |
The population variance is approximately 9.43.
Answer: FALSE
Diff: 2
Keywords: population, variance
Section: 3-2 Measures of Variation
Outcome: 3
48) The Good-Guys Car Dealership has tracked the number of used cars sold at its downtown dealership. Consider the following data as representing the population of cars sold in each of the 8 weeks that the dealership has been open.
3 | 5 | 2 | 7 | 7 | 7 | 9 | 0 |
The population standard deviation is approximately 2.87 cars.
Answer: TRUE
Diff: 2
Keywords: population, standard, deviation
Section: 3-2 Measures of Variation
Outcome: 3
49) One of the reasons that the standard deviation is preferred as a measure of variation over the variance is that the standard deviation is measured in the original units.
Answer: TRUE
Diff: 1
Keywords: standard, deviation, variation, units
Section: 3-2 Measures of Variation
Outcome: 3
50) The interquartile range is the difference between the mean and the median.
Answer: FALSE
Diff: 1
Keywords: interquartile range, median, mean
Section: 3-2 Measures of Variation
Outcome: 3
51) A store manager tracks the number of customer complaints each week. The following data reflect a random sample of ten weeks.
11 | 19 | 4 | 6 | 8 | 9 | 6 | 4 | 0 | 3 |
The range for these data is 8.
Answer: FALSE
Diff: 1
Keywords: range, variation
Section: 3-2 Measures of Variation
Outcome: 3
52) A store manager tracks the number of customer complaints each week. The following data reflect a random sample of ten weeks.
11 | 19 | 4 | 6 | 8 | 9 | 6 | 4 | 0 | 3 |
The variance for these data is approximately 27.78.
Answer: TRUE
Diff: 2
Keywords: sample, variance
Section: 3-2 Measures of Variation
Outcome: 3
53) A store manager tracks the number of customer complaints each week. The following data reflect a random sample of ten weeks.
11 | 19 | 4 | 6 | 8 | 9 | 6 | 4 | 0 | 3 |
The standard deviation for these data is approximately 27.78.
Answer: FALSE
Diff: 2
Keywords: standard, deviation, sample
Section: 3-2 Measures of Variation
Outcome: 3
54) The interquartile range contains the middle 50 percent of a data set.
Answer: TRUE
Diff: 2
Keywords: interquartile range
Section: 3-2 Measures of Variation
Outcome: 3
55) For a given set of data, if the data are treated as a population, the calculated standard deviation will be less than it would be had the data been treated as a sample.
Answer: TRUE
Diff: 2
Keywords: sample, standard, deviation, population
Section: 3-2 Measures of Variation
Outcome: 3
56) If a population standard deviation is computed to be 345, it will almost always be the case that a standard deviation computed from a random sample from that population will be larger than 345.
Answer: TRUE
Diff: 2
Keywords: population, sample, standard, deviation
Section: 3-2 Measures of Variation
Outcome: 3
57) The advantage of using the interquartile range as a measure of variation is that it utilizes all the data in its computation.
Answer: FALSE
Diff: 2
Keywords: variation, interquartile, range
Section: 3-2 Measures of Variation
Outcome: 3
58) Suppose the standard deviation for a given sample is known to be 20. If the data in the sample are doubled, the standard deviation will be 40.
Answer: FALSE
Diff: 3
Keywords: standard, deviation
Section: 3-2 Measures of Variation
Outcome: 3
59) Populations with larger means will also have larger standard deviations since the data will be more spread out for populations with larger means.
Answer: FALSE
Diff: 2
Keywords: mean, standard deviation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
60) In comparing two distributions with the same mean, the coefficient of variation is the only way to assess which distribution has the greatest relative variability.
Answer: FALSE
Diff: 1
Keywords: mean, coefficient, variation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
61) Consider a situation involving two populations where population 1 is known to have a higher coefficient of variation than population 2. In this situation, we know that population 1 has a higher standard deviation than population 2.
Answer: FALSE
Diff: 2
Keywords: coefficient, variation, standard deviation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
62) Acme Taxi has two taxi cabs. The manager tracks the daily revenue for each cab. Over the past 20 days, Cab A has averaged $76.00 per night with a standard deviation equal to $11.00. Cab B has averaged $200.00 per night with a standard deviation of $18.00. Based on this information, Cab B has the greatest relative variation.
Answer: FALSE
Diff: 3
Keywords: standard deviation, relative variation, coefficient of variation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
63) Acme Taxi has two taxi cabs. The manager tracks the daily revenue for each cab. Over the past 20 days, Cab A has averaged $76.00 per night with a standard deviation equal to $11.00. Cab B has averaged $200.00 per night with a standard deviation of $18.00. Based on this information, the coefficient of variation for Cab B is 9 percent.
Answer: TRUE
Diff: 2
Keywords: standard deviation, relative variation, coefficient of variation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
64) Based on the empirical rule we can assume that all bell-shaped distributions have approximately 95 percent of the values within ± 2 standard deviations of the mean.
Answer: TRUE
Diff: 1
Keywords: empirical rule
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
65) Suppose a distribution has a mean of 80 and standard deviation of 10. It is found that 84 percent of the values in the data set lie between 70 and 90. This implies that the distribution is not bell-shaped.
Answer: TRUE
Diff: 2
Keywords: mean, standard deviation, empirical rule
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
66) The credit card balances for customers at State Bank and Trust has a mean equal to $800 and a standard deviation equal to $60.00. Kevin Smith’s balance is $1,352. Based on this, his standardized value is 9.20.
Answer: TRUE
Diff: 2
Keywords: mean, standard deviation, z score, standardized value
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
67) Based on the empirical rule we can expect about 95 percent of the values in bell-shaped distributions to be within ± one standard deviation of the mean.
Answer: FALSE
Diff: 1
Keywords: mean, standard deviation, empirical rule
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
68) A major automobile maker has two models of sedans. The first model has been shown to get an average of 27 mpg on the highway with a standard deviation equal to 5 mpg. The second model gets 33 mpg on average with a standard deviation of 8 mpg. Based on this information the first car model is relatively more variable than the second car model.
Answer: FALSE
Diff: 2
Keywords: mean, standard deviation, coefficient of variation, relative variation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
69) The distribution of bankcard balances for customers is highly right-skewed with a mean of $1,100 and a standard deviation equal to $250. Based on this information, approximately 68 percent of the customers will have bank balances between $850 and $1,350.
Answer: FALSE
Diff: 2
Keywords: skewed, mean, standard deviation, Tchebysheff
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
70) The distribution of dollars paid for car insurance by car owners in a major east coast city is bell-shaped with a mean equal to $750 every six months and a standard deviation equal to $100. Based on this information we should use Tchebysheff’s theorem to determine the conservative percentage of car owners that will pay between $550 and $950 for car insurance.
Answer: FALSE
Diff: 2
Keywords: mean, standard deviation, Tchebysheff
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
71) A population measure, such as the population mean, is called a:
- A) statistic.
- B) parameter.
- C) prime number.
- D) sample value.
Answer: B
Diff: 1
Keywords: population, parameter
Section: 3-1 Measures of Center and Location
Outcome: 1
72) If a business manager selected a sample of customers and computed the mean income for this sample of customers, she has computed:
- A) a statistic.
- B) an ordinal value.
- C) a nominal value.
- D) a parameter.
Answer: A
Diff: 1
Keywords: sample, statistic
Section: 3-1 Measures of Center and Location
Outcome: 1
73) Which of the following statements is true?
- A) The mean of a population will always be larger than the population standard deviation.
- B) The mean of the population will generally be larger than the mean of the sample selected from that population.
- C) The population mean and a sample mean for a sample selected from that population will usually be different values.
- D) The population mean and sample mean will always be identical.
Answer: C
Diff: 2
Keywords: population, sample, mean
Section: 3-1 Measures of Center and Location
Outcome: 1
74) The most frequently used measure of central tendency is:
- A) median.
- B) mean.
- C) mode.
- D) middle value.
Answer: B
Diff: 1
Keywords: central tendency, mean
Section: 3-1 Measures of Center and Location
Outcome: 1
75) Consider the following sample data:
25 | 11 | 6 | 4 | 2 | 17 | 9 | 6 |
For these data the sample mean is:
- A) 8
- B) 10
- C) 3
- D) 12
Answer: B
Diff: 1
Keywords: sample, mean
Section: 3-1 Measures of Center and Location
Outcome: 1
76) Consider the following sample data:
25 | 11 | 6 | 4 | 2 | 17 | 9 | 6 |
For these data the median is:
- A) 7.5
- B) 3.5
- C) 10
- D) None of the above
Answer: A
Diff: 2
Keywords: sample, median
Section: 3-1 Measures of Center and Location
Outcome: 1
77) A small company has 7 employees. The numbers of years these employees have worked for this company are shown as follows:
4 | 14 | 3 | 16 | 9 | 8 | 16 |
Based upon this information, the mean number of years that employees have been with this company is:
- A) 16
B)
- C) 8.40
- D) 10
Answer: D
Diff: 1
Keywords: population, mean
Section: 3-1 Measures of Center and Location
Outcome: 1
78) A small company has 7 employees. The numbers of years these employees have worked for this company are shown as follows:
4 | 14 | 3 | 16 | 9 | 8 | 16 |
Based upon this information, the median number of years that employees have been with this company is:
- A) 9 years.
- B) 16 years.
- C) 10 years.
- D) 14 years.
Answer: A
Diff: 2
Keywords: population, median
Section: 3-1 Measures of Center and Location
Outcome: 1
79) A small company has 7 employees. The numbers of years these employees have worked for this company are shown as follows:
4 | 14 | 3 | 16 | 9 | 8 | 16 |
Based upon this information, the mode number of years that employees have been with this company is:
- A) 16
- B) 2
- C) 9
- D) 10
Answer: A
Diff: 1
Keywords: population, mode
Section: 3-1 Measures of Center and Location
Outcome: 1
80) A sample of people who have attended a college football game at your university has a mean = 3.2 members in their family. The mode number of family members is 2 and the median number is 2.0. Based on this information:
- A) the population mean exceeds 3.2.
- B) the distribution is bell-shaped.
- C) the distribution is right-skewed.
- D) the distribution is left-skewed.
Answer: C
Diff: 2
Keywords: skewed, mean, median
Section: 3-1 Measures of Center and Location
Outcome: 1
81) A major retail store has studied customer behavior and found that the distribution of time customers spend in a store per visit is symmetric with a mean equal to 17.3 minutes. Based on this information, which of the following is true?
- A) The distribution is right-skewed.
- B) The median is to the right of the mean.
- C) The median is approximately 17.3 minutes.
- D) The median is to the left of the mean.
Answer: C
Diff: 1
Keywords: symmetric, mean, median
Section: 3-1 Measures of Center and Location
Outcome: 1
82) A large retail company gives an employment screening test to all prospective employees. Franklin Gilman recently took the test and it was reported back to him that his score placed him at the 80th percentile. Therefore:
- A) 80 people who took the test scored below Franklin.
- B) Franklin scored as high or higher than 80 percent of the people who took the test.
- C) Franklin was in the bottom 20 percent of those that have taken the test.
- D) Franklin’s score has a z-score of 80.
Answer: B
Diff: 2
Keywords: percentile
Section: 3-1 Measures of Center and Location
Outcome: 1
83) A large retail company gives an employment screening test to all prospective employees. If a prospective employee receives a report saying that she scored at the 40th percentile:
- A) she scored above the median.
- B) she scored better than 40 percent of people who took the test.
- C) she scored in the top 40 percent of people who took the test.
- D) her z-score was a 40.
Answer: B
Diff: 2
Keywords: percentile, median
Section: 3-1 Measures of Center and Location
Outcome: 1
84) If a data set has 740 values that have been sorted from low to high, which value in the data set will be the 20th percentile?
- A) The average of the 148thand 149thvalues
- B) The 20thvalue
- C) The 148thvalue
- D) None of the above
Answer: A
Diff: 2
Keywords: percentile, location
Section: 3-1 Measures of Center and Location
Outcome: 1
85) If a data set has 1,133 sorted values, what value corresponds to the 3rd quartile?
- A) The 250thvalue
- B) The 850thvalue
- C) The 760thvalue
- D) The 849thvalue
Answer: B
Diff: 2
Keywords: percentile, quartile, value
Section: 3-1 Measures of Center and Location
Outcome: 1
86) At a sawmill in Oregon, a process improvement team measured the diameters for a sample of 1,500 logs. The following summary statistics were computed:
Given this information, the boundaries on the box in a box and whisker plot are:
- A) 8.9 in and 15.6 in.
- B) 13.5 in ± 1.5 (Q3-Q1).
- C) 14.2 in ± 1.5 (Q3-Q1).
- D) 8.9 in and 14.2 in.
Answer: A
Diff: 2
Keywords: box, whisker, plot, boundary
Section: 3-1 Measures of Center and Location
Outcome: 2
87) At a sawmill in Oregon, a process improvement team measured the diameters for a sample of 1,500 logs. The following summary statistics were computed:
Given this information, in a box and whisker plot, which of these four values will NOT appear?
- A) 8.9 in.
- B) 13.5 in.
- C) 15.6 in.
- D) 14.2 in.
Answer: D
Diff: 2
Keywords: box, whisker, plot
Section: 3-1 Measures of Center and Location
Outcome: 2
88) At a sawmill in Oregon, a process improvement team measured the diameters for a sample of 1,500 logs. The following summary statistics were computed:
Given this information, which of the following statements is correct?
- A) The distribution of log diameters is symmetric.
- B) A log that is over 20 inches in diameter can be considered an outlier.
- C) The distribution of log diameters is right-skewed.
- D) The distribution is left-skewed.
Answer: C
Diff: 2
Keywords: skewed, mean, median, quartile
Section: 3-1 Measures of Center and Location
Outcome: 1
89) At a sawmill in Oregon, a process improvement team measured the diameters for a sample of 1,500 logs. The following summary statistics were computed:
Given this information, for a box and whisker plot which of the following statements is appropriate?
- A) Seventy-five percent of the trees in the sample have values between 8.9 in. and 15.6 in.
- B) Virtually all of the data should fall between 0 in. and 25.65 in.
- C) No tree will have a diameter of more than 22.3 in.
- D) Fifty percent of the trees will have diameters between 13.5 in. and 15.6 in.
Answer: B
Diff: 3
Keywords: box, whisker, outlier
Section: 3-1 Measures of Center and Location
Outcome: 2
90) If a distribution for a quantitative variable is thought to be nearly symmetric with very little variation, and a box and whisker plot is created for this distribution, which of the following is true?
- A) The box will be quite wide but the whisker will be very short.
- B) The left and right-hand edges of the box will be approximately equal distance from the median.
- C) The whiskers should be about half as long as the box is wide.
- D) The upper whisker will be much longer than the lower whisker.
Answer: B
Diff: 2
Keywords: box, whisker, symmetric, median
Section: 3-1 Measures of Center and Location
Outcome: 2
91) The box and whisker plot CANNOT be used to identify:
- A) skewedness.
- B) centerness.
- C) outliers.
- D) symmetry.
Answer: B
Diff: 3
Keywords: box, whisker, outlier
Section: 3-1 Measures of Center and Location
Outcome: 2
92) For ordinal data, ________ is the preferred measure of central location.
- A) the mean
- B) the median
- C) the percentile
- D) the quartile
Answer: B
Diff: 3
Keywords: mean, median, percentile, quartile
Section: 3-1 Measures of Center and Location
Outcome: 1
93) Which of the following is the most frequently used measure of variation?
- A) The range
- B) The standard deviation
- C) The variance
- D) The mode
Answer: B
Diff: 1
Keywords: variation, standard deviation
Section: 3-2 Measures of Variation
Outcome: 3
94) Which of the following measures is not affected by extreme values in the data?
- A) The mean
- B) The median
- C) The range
- D) The standard deviation
Answer: B
Diff: 2
Keywords: extreme, insensitive, median
Section: 3-2 Measures of Variation
Outcome: 3
95) The following data reflect the number of customers who test drove new cars each day for a sample of 20 days at the Redfield Ford Dealership.
5 | 7 | 2 | 9 | 4 |
9 | 7 | 10 | 4 | 7 |
5 | 6 | 4 | 0 | 7 |
6 | 3 | 4 | 14 | 6 |
Given these data, what is the range?
- A) 14
- B) 1
- C) Approximately 3.08
- D) 5.95
Answer: A
Diff: 1
Keywords: range, sample
Section: 3-2 Measures of Variation
Outcome: 3
96) The following data reflect the number of customers who test drove new cars each day for a sample of 20 days at the Redfield Ford Dealership.
5 | 7 | 2 | 9 | 4 |
9 | 7 | 10 | 4 | 7 |
5 | 6 | 4 | 0 | 7 |
6 | 3 | 4 | 14 | 6 |
Given these data, what is the variance?
- A) 0.69
- B) Approximately 3.08
- C) Approximately 9.52
- D) Approximately 181
Answer: C
Diff: 2
Keywords: variance, sample
Section: 3-2 Measures of Variation
Outcome: 3
97) The following data reflect the number of customers who test drove new cars each day for a sample of 20 days at the Redfield Ford Dealership.
5 | 7 | 2 | 9 | 4 |
9 | 7 | 10 | 4 | 7 |
5 | 6 | 4 | 0 | 7 |
6 | 3 | 4 | 14 | 6 |
Given these data, what is the interquartile range?
- A) 3
- B) 7
- C) 4
- D) 14
Answer: A
Diff: 3
Keywords: interquartile range, sample
Section: 3-2 Measures of Variation
Outcome: 3
98) The advantage of using the interquartile range versus the range as a measure of variation is:
- A) it is easier to compute.
- B) it utilizes all the data in its computation.
- C) it gives a value that is closer to the true variation.
- D) it is less affected by extremes in the data.
Answer: D
Diff: 1
Keywords: range, variation, interquartile range
Section: 3-2 Measures of Variation
Outcome: 3
99) The following data reflect the number of customers who return merchandise for a refund on Monday. Note these data reflect the population of all 10 Mondays for which data are available.
40 | 12 | 17 | 25 | 9 |
46 | 13 | 22 | 16 | 7 |
Based on these data, what is the standard deviation?
- A) 13.03
- B) 12.36
- C) 39
- D) 152.8
Answer: B
Diff: 2
Keywords: population, standard deviation
Section: 3-2 Measures of Variation
Outcome: 3
100) The following data reflect the number of customers who return merchandise for a refund on Monday. Note these data reflect the population of all 10 Mondays for which data are available.
40 | 12 | 17 | 25 | 9 |
46 | 13 | 22 | 16 | 7 |
Assume that this same exact pattern of data were replicated for the next ten days. How would this affect the standard deviation for the new population with 20 items?
- A) The standard deviation would be doubled.
- B) The standard deviation would be cut in half.
- C) The standard deviation would not be changed.
- D) There is no way of knowing the exact impact without knowing how the mean is changed.
Answer: C
Diff: 3
Keywords: standard deviation, population
Section: 3-2 Measures of Variation
Outcome: 3
101) In order to compute the mean and standard deviation, the level of data measurement should be:
- A) ratio or interval.
- B) qualitative.
- C) nominal.
- D) ordinal.
Answer: A
Diff: 1
Keywords: mean, standard deviation, ratio, interval
Section: 3-2 Measures of Variation
Outcome: 3
102) Consider the following data, which represent the number of miles that employees commute from home to work each day. There are two samples: one for males and one for females.
Males:
13 | 5 | 2 | 23 | 14 | 5 |
Females:
15 | 6 | 3 | 2 | 4 | 6 |
Which of the following statements is true?
- A) The female distribution is more variable since the range for the females is greater than for the males.
- B) Females in the sample commute farther on average than do males.
- C) The males in the sample commute farther on average than the females.
- D) Males and females on average commute the same distance.
Answer: C
Diff: 2
Keywords: range, mean
Section: 3-2 Measures of Variation
Outcome: 3
103) Consider the following data, which represent the number of miles that employees commute from home to work each day. There are two samples: one for males and one for females.
Males:
13 | 5 | 2 | 23 | 14 | 5 |
Females:
15 | 6 | 3 | 2 | 4 | 6 |
The coefficient of variation of commute miles for the males is:
- A) approximately 76 percent.
- B) about 7.8.
- C) approximately 61.5.
- D) about 67 percent.
Answer: A
Diff: 2
Keywords: coefficient of variation, mean, standard deviation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
104) Consider the following data, which represent the number of miles that employees commute from home to work each day. There are two samples: one for males and one for females.
Males:
13 | 5 | 2 | 23 | 14 | 5 |
Females:
15 | 6 | 3 | 2 | 4 | 6 |
Which of the following statements is true?
- A) Females have the larger mean.
- B) The coefficient of variation is larger for females than for males.
- C) The coefficient of variation is larger for males than for females.
- D) Females have the larger range.
Answer: B
Diff: 2
Keywords: sample, coefficient of variation, relative variability
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
105) If the age distribution of customers at a major retail chain is thought to be bell-shaped with a mean equal to 43 years and a standard deviation equal to 7 years, the percentage of customers between the ages of 29 and 57 years is:
- A) approximately 81.5.
- B) approximately 68.
- C) at least 75.
- D) approximately 95.
Answer: D
Diff: 1
Keywords: empirical rule, distribution
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
106) Under what circumstances is it necessary to use the coefficient of variation to compare relative variability between two or more distributions?
- A) When the means of the distributions are equal
- B) When the means of the distributions are not equal
- C) When the standard deviations of the distributions are not equal
- D) When the standard deviations of the distributions are equal
Answer: B
Diff: 2
Keywords: coefficient of variation, mean, standard deviation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
107) In the annual report, a major food chain stated that the distribution of daily sales at its Detroit stores is known to be bell-shaped, and that 95 percent of all daily sales fell between $19,200 and $36,400. Based on this information, what were the mean sales?
- A) Around $20,000
- B) Close to $30,000
- C) Approximately $27,800
- D) Can’t be determined without more information.
Answer: C
Diff: 2
Keywords: empirical rule, mean, standard deviation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
108) The number of days that homes stay on the market before they sell in Houston is bell-shaped with a mean equal to 56 days. Further, 95 percent of all homes are on the market between 40 and 72 days. Based on this information, what is the standard deviation for the number of days that houses stay on the market in Houston?
- A) 8
B)
- C) 16
- D) 4
Answer: A
Diff: 2
Keywords: empirical rule, mean, standard deviation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
109) Incomes in a particular market area are known to be right-skewed with a mean equal to $33,100. In a report issued recently, a manager stated that at least 89 percent of all incomes are in the range of $26,700 to $39,500, and this was based on Tchebysheff’s theorem. Given these facts, what is the standard deviation for the incomes in this market area?
- A) Approximately $6,400
- B) Approximately $3,200
- C) Approximately $2,133
- D) Approximately $4266
Answer: C
Diff: 3
Keywords: standard deviation, mean, Tchebysheff
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
110) A distribution has a coefficient of variation of 65 percent and mean of 74. What is the value of the standard deviation?
- A) 0.65
- B) 4810
- C) 113.8
- D) 48.1
Answer: D
Diff: 2
Keywords: mean, standard deviation, coefficient of variation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
111) The asking price for homes on the real estate market in Baltimore has a mean value of $286,455 and a standard deviation of $11,200. Four homes are listed by one real estate company with the following prices:
Home 1: | $456,900 |
Home 2: | $306,000 |
Home 3: | $266,910 |
Home 4: | $201,456 |
Based upon this information, which house has a standardized value that is relatively closest to zero?
- A) Home 1
- B) Home 2
- C) Home 3
- D) Home 2 and home 3
Answer: D
Diff: 2
Keywords: mean standard deviation, standardized value, z score
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
112) The asking price for homes on the real estate market in Baltimore has a mean value of $286,455 and a standard deviation of $11,200. The mean and standard deviation in asking price for homes in Denver are $188,468 and $8,230, respectively. Recently, one home sold in each city where the asking price for each home was $193,000. Based on these data, which of the following conclusions can be made?
- A) The two homes have approximately the same standardized values.
- B) The distribution of asking prices in the two cities is bell-shaped.
- C) The house in Baltimore is relatively farther from the mean than the house in Denver.
- D) The asking prices of homes in Denver is less variable than those in Baltimore.
Answer: C
Diff: 2
Keywords: mean, standard deviation, standardized value, z score
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
113) The asking price for homes on the real estate market in Baltimore has a mean value of $286,455 and a standard deviation of $11,200. The mean and standard deviation in asking price for homes in Denver are $188,468 and $8,230, respectively. Recently, one home sold in each city where the asking price for each home was $193,000. Assuming that both distributions are bell-shaped, which of the following statements is true?
- A) The Baltimore home has the higher standard z-value.
- B) The coefficient of variation for Denver is less than for Baltimore.
- C) The Denver home has a higher standard z-value.
- D) Both cities have the same coefficient of variation.
Answer: C
Diff: 2
Keywords: mean, standard deviation, standardized value, z score
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
114) A report on spending by adults on recreation stated the following: At least 75 percent of the people in the survey spend between $750 and $1,250 per year. The report also said that at least 88 percent spend between $625 and $1,375 per year. Given this information, which of the following is most apt to be true?
- A) The standard deviation is approximately $125.
- B) The distribution of spending on recreation can be assumed to be bell-shaped.
- C) The standard deviation is approximately $187.5.
- D) The standard deviation is approximately $250.
Answer: A
Diff: 3
Keywords: standard deviation, Tchebysheff
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
115) The distribution of the actual weight of potato chips in a 16 ounce sack is thought to be bell-shaped with a mean equal to 16 ounces and a standard deviation equal to 0.45 ounces. Based on this, between what two limits could we expect 95 percent of all sacks to weigh?
- A) 14 to 18 ounces
- B) 15.10 to 16.90 ounces
- C) 15.55 to 16.45 ounces
- D) 14.65 to 17.35 ounces
Answer: B
Diff: 2
Keywords: standard deviation, empirical rule
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
116) A recent study in the restaurant business determined that the mean tips for male waiters per hour of work are $6.78 with a standard deviation of $2.11. The mean tips per hour for female waiters are $7.86 with a standard deviation of $2.20. Based on this information, which of the following statements do we know to be true?
- A) The distribution of tips for both males and females is right-skewed.
- B) The variation in tips received by females is more variable than males.
- C) The median tips for females exceeds that of males.
- D) On a relative basis, males have more variation in tips per hour than do females.
Answer: D
Diff: 2
Keywords: mean, standard deviation, coefficient of variation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
117) Data was collected on the number of television sets in a household, and it was found that the mean was 3.5 and the standard deviation was 0.75.
Based on these sample data, what is the standardized value corresponding to 5 televisions?
- A) -2.00
- B) 1.5
- C) 2.00
- D) 1.125
Answer: C
Diff: 2
Keywords: mean, standard deviation, sample, z score, standardized value
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
118) Portfolio A of a collection of stocks is considered more risky than portfolio B if:
- A) portfolio A has a higher mean than portfolio B.
- B) portfolio A has a higher variance than portfolio B.
- C) portfolio A has a higher standard deviation.
- D) portfolio A has a higher coefficient of variation than portfolio B.
Answer: D
Diff: 3
Keywords: mean, standard deviation, coefficient of variation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
119) When comparing data measured by substantially different scales, we must use:
- A) standardized data values.
- B) standardized data scales.
- C) standardized data variations.
- D) standardized data scores.
Answer: A
Diff: 2
Keywords: z-score, standardized value
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
120) The Empirical Rule states that for a bell-shaped distribution, approximately 95 percent of data should lie within:
- A) one standard deviation from either side of the mean.
- B) two standard deviations from either side of the mean.
- C) three standard deviations from either side of the mean.
- D) four standard deviations from either side of the mean.
Answer: B
Diff: 2
Keywords: mean, standard deviation, bell-shaped, empirical rule
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
121) A professor wishes to develop a numerical method for giving grades. He intends to base the grade on homework, two midterms, a project, and a final examination. He wishes the final exam to have the largest influence on the grade. He wants the project to have 10%, each midterm to have 20%, and the homework to have 10% of the influence of the semester grade.
Determine the weights the professor should use to produce a weighted average for grading purposes.
A)
Instrument | Final | Project | Midterm 1 | Midterm 2 | Homework |
Weight | 40 | 10 | 20 | 20 | 10 |
B)
Instrument | Final | Project | Midterm 1 | Midterm 2 | Homework |
Weight | 50 | 10 | 20 | 20 | 10 |
C)
Instrument | Final | Project | Midterm 1 | Midterm 2 | Homework |
Weight | 40 | 15 | 15 | 15 | 15 |
D)
Instrument | Final | Project | Midterm 1 | Midterm 2 | Homework |
Weight | 30 | 10 | 20 | 20 | 10 |
Answer: A
Diff: 3
Keywords: measure of variation
Section: 3-2 Measures of Variation
Outcome: 1
122) A professor wishes to develop a numerical method for giving grades. He intends to base the grade on homework, two midterms, a project, and a final examination. He wishes the final exam to have the largest influence on the grade. He wants the project to have 10%, each midterm to have 20%, and the homework to have 10% of the influence of the semester grade.
For a student with the following grades during the quarter, calculate a weighted average for the course:
Instrument | Final | Project | Midterm 1 | Modterm 2 | Homework |
Percentage Grade | 64 | 98 | 67 | 63 | 89 |
- A) 68.50
- B) 73.30
- C) 68.30
- D) 70.30
Answer: D
Diff: 1
Keywords: measure of center and location
Section: 3-1 Measures of Center and Location
Outcome: 1
123) Todd Lindsey & Associates, a commercial real estate company located in Boston, owns six office buildings in the Boston area that it leases to businesses. The lease price per square foot differs by building due to location and building amenities. Currently, all six buildings are fully leased at the prices shown here.
Price per Square Foot | Number of Square Feet | |
Building 1 | $ 75 | 125,000 |
Building 2 | $ 85 | 37,500 |
Building 3 | $ 90 | 77,500 |
Building 4 | $ 45 | 35,000 |
Building 5 | $ 55 | 60,000 |
Building 6 | $110 | 130,000 |
Compute the weighted average (mean) price per square foot for these buildings.
- A) 86.25
- B) 83.25
- C) 80.15
- D) 86.15
Answer: B
Diff: 1
Keywords: measure of center and location
Section: 3-1 Measures of Center and Location
Outcome: 1
124) Each year, Business Week publishes information and rankings of master of business administration (MBA) programs. The data file MBA Analysis contains data on several variables for eight reputable MBA programs as presented in the October 2, 2000, issue of Business Week. The variables include pre- and post-MBA salary, percentage salary increase, undergraduate GPA, average Graduate Management Admission Test (GMAT) score, annual tuition, and expected annual student cost. Compute the mean and median for each of the variables in the database.
A)
Mean | Median | |
Pre-MBA Salary | 39077 | 43337.63 |
Post-MBA Salary | 82203 | 98902 |
Percentage Increase in Salary | 116 | 123 |
Undergraduate GPA | 3.46 | 3.40 |
GMAT Score | 635.00 | 631.13 |
Annual Tuition | 13163.50 | 15967.50 |
Expected Annual Student Cost | 23169.5 | 27980.75 |
B)
Mean | Median | |
Pre-MBA Salary | 46667.63 | 39077 |
Post-MBA Salary | 98902 | 84403 |
Percentage Increase in Salary | 113 | 116 |
Undergraduate GPA | 3.40 | 3.56 |
GMAT Score | 661.16 | 635.00 |
Annual Tuition | 15967.50 | 19163.70 |
Expected Annual Student Cost | 28980.75 | 23179.5 |
C)
Mean | Median | |
Pre-MBA Salary | 43337.63 | 39077 |
Post-MBA Salary | 98902 | 82203 |
Percentage Increase in Salary | 123 | 116 |
Undergraduate GPA | 3.40 | 3.46 |
GMAT Score | 631.13 | 635.00 |
Annual Tuition | 15967.50 | 13163.50 |
Expected Annual Student Cost | 27980.75 | 23169.5 |
D)
Mean | Median | |
Pre-MBA Salary | 39077 | 46667.63 |
Post-MBA Salary | 84403 | 98902 |
Percentage Increase in Salary | 116 | 113 |
Undergraduate GPA | 3.56 | 3.40 |
GMAT Score | 635.00 | 661.16 |
Annual Tuition | 19163.70 | 15967.50 |
Expected Annual Student Cost | 23179.5 | 28980.75 |
Answer: C
Diff: 2
Keywords: measure of center and location
Section: 3-1 Measures of Center and Location
Outcome: 1
125) Dynamic random-access memory (DRAM) memory chips are made from silicon wafers in manufacturing facilities through a very complex process called wafer fabs. The wafers are routed through the fab machines in an order that is referred to as a recipe. The wafers may go through the same machine several times as the chip is created. The data file DRAM Chips contains a sample of processing times, measured in fractions of hours, at a particular machine center for one chip recipe.
Compute the mean processing time.
- A) 0.24 minutes
- B) 0.22 minutes
- C) 0.31 minutes
- D) 0.33 minutes
Answer: D
Diff: 1
Keywords: measure of center and location
Section: 3-1 Measures of Center and Location
Outcome: 1
126) Dynamic random-access memory (DRAM) memory chips are made from silicon wafers in manufacturing facilities through a very complex process called wafer fabs. The wafers are routed through the fab machines in an order that is referred to as a recipe. The wafers may go through the same machine several times as the chip is created. The data file DRAM Chips contains a sample of processing times, measured in fractions of hours, at a particular machine center for one chip recipe.
Compute the median processing time.
- A) 0.31 minutes
- B) 0.24 minutes
- C) 0.21 minutes
- D) 0.44 minutes
Answer: A
Diff: 1
Keywords: measure of center and location
Section: 3-1 Measures of Center and Location
Outcome: 1
127) Dynamic random-access memory (DRAM) memory chips are made from silicon wafers in manufacturing facilities through a very complex process called wafer fabs. The wafers are routed through the fab machines in an order that is referred to as a recipe. The wafers may go through the same machine several times as the chip is created. The data file DRAM Chips contains a sample of processing times, measured in fractions of hours, at a particular machine center for one chip recipe.
Determine what the mode processing time is.
- A) 0.22
- B) 0.24
- C) 0.33
- D) 0.34
Answer: B
Diff: 1
Keywords: measure of center and location
Section: 3-1 Measures of Center and Location
Outcome: 1
128) Dynamic random-access memory (DRAM) memory chips are made from silicon wafers in manufacturing facilities through a very complex process called wafer fabs. The wafers are routed through the fab machines in an order that is referred to as a recipe. The wafers may go through the same machine several times as the chip is created. The data file DRAM Chips contains a sample of processing times, measured in fractions of hours, at a particular machine center for one chip recipe.
Calculate the 80th percentile for processing time.
- A) 0.40 minutes
- B) 0.35 minutes
- C) 0.45 minutes
- D) 0.20 minutes
Answer: A
Diff: 2
Keywords: measure of variation
Section: 3-2 Measures of Variation
Outcome: 1
129) Suppose that the distribution of grocery purchases is thought to be symmetric. If the mean purchase is $23.14, what would the median purchase be?
Answer: For a symmetric distribution, the mean and median are equal. Therefore, the median purchase should also be $23.14.
Diff: 1
Keywords: mean, median, symmetric, distribution
Section: 3-1 Measures of Center and Location
Outcome: 1
130) The AMI Company has two assembly lines in its Kansas City plant. Line A produces an average of 335 units per day with a standard deviation equal to 11 units. Line B produces an average of 145 units per day with a standard deviation equal to 8 units. Based on this information, which line is relatively more consistent?
Answer: At first glance it may appear that Line B is more consistent since it has a smaller standard deviation. However, if we wish to compare relative variability when the means of two distributions are different, then we need to compute the coefficient of variation for each. The one with the smallest coefficient of variation is the more consistent. The coefficient of variation is given by:
CV = 100
Then for Line A we get: CV = 100 = 100 = 3.28%
For Line B we get: CV = 100 = 100 = 5.52%
Thus, Line A is the more consistent of the two lines with respect to daily production output.
Diff: 2
Keywords: mean standard deviation, relative variation, coefficient of variation
Section: 3-1 Measures of Center and Location
Outcome: 1
131) The following sample data reflect electricity bills for ten households in San Diego in March.
$118.20 | $67.88 | $133.40 | $88.42 | $110.34 |
$76.90 | $144.56 | $127.89 | $89.34 | $129.10 |
Determine three measures of central tendency for these sample data. Then, based on these measures, determine whether the sample data are symmetric or skewed.
Answer: The three measures of central tendency are the mean, median, and mode. These are computed as follows:
Mean: = = = $108.60
Median: Arrange the data in order from low to high. Since we have an even number of values, the median is the mean of the 5th and 6th values.
$67.88 | $76.90 | $88.42 | $89.34 | $110.34 | $118.20 | $127.89 | $129.10 | $133.40 | $144.56 |
The median is found as: Md ≈ = $114.27
Mode: The mode is the value in the data that occurs most frequently. Since no value in this sample occurs more frequently than one time, there is no mode.
Data are symmetric if the mean and the median are equal. Since the sample mean = $108.60 and the median equals $114.27 the data are not symmetric. Since the mean is less than median, we conclude that the sample data are left-skewed.
Diff: 2
Keywords: central tendency, mean, median, mode
Section: 3-1 Measures of Center and Location
Outcome: 1
132) The following sample data reflect electricity bills for ten households in San Diego in March.
$118.20 | $67.88 | $133.40 | $88.42 | $110.34 |
$76.90 | $144.56 | $127.89 | $89.34 | $129.10 |
Compute the range, variance, and standard deviation for these sample data. Discuss which of these three measures you would prefer to use as a measure of variation.
Answer: The range is found as follows:
Range = High – Low = $144.56 – $67.88 = $76.68
The sample variance is found using:
S2 =
We begin by determining the sample mean:
= = = 108.60
We then sum the deviations of the individual values from the sample mean giving:
We now divide this sum by n-1 giving:
S2 = = = 692.99
Thus, the sample variance is 692.99.
The sample deviation is found by taking the square root of the sample variance:
S = = = $26.32
Thus, the sample standard deviation is $26.32.
Although the range is far easier to compute, it contains information only from the extremes in the data. The variance is in squared units and therefore does not have any meaning in the context of money spent on electricity. The standard deviation is preferred since it uses all the data in its calculation and is expressed in the original units.
Diff: 2
Keywords: mean, standard deviation, range, variance
Section: 3-2 Measures of Variation
Outcome: 3
133) Why is it that when we find the sample standard deviation, we divide by n-1 but when we find the population standard deviation we divide by n?
Answer: The technical answer to this question is beyond the scope of the text. However, we can think of it this way. If our objective in computing the sample standard deviation is to estimate the population standard deviation, we would want an estimate that would be correct on the average. That means that if we took repeated random samples from the population and for each sample we computed the standard deviation, we would want the average of the sample standard deviations to equal the population standard deviation. If our formula for the sample standard deviation uses n-1, this occurs.
Diff: 3
Keywords: sample, population, standard deviation
Section: 3-2 Measures of Variation
Outcome: 3
134) Explain how the empirical rule can be used to help describe data in a population or a sample.
Answer: The empirical rule applies when the data distribution is bell-shaped. When this is the case, we know that approximately 68 percent of the data will fall with ± 1 standard deviation of the mean, approximately 95 percent will fall within ± 2 standard deviations of the mean, and virtually all of the data will fall within ± 3 standard deviations of the mean. Thus, by knowing that the distribution is bell-shaped and by knowing the mean and standard deviation, we know more about how the individual data are arranged.
Diff: 2
Keywords: empirical rule, bell-shaped, standard deviation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
135) Explain how Tchebysheff’s theorem can be used to help describe data in a population or a sample.
Answer: If the data in a sample or a population are known to be bell-shaped, the empirical rule can be used to help us describe the data. However, when the sample or the population is not bell-shaped, Tchebysheff’s theorem is very useful. It is a conservative theorem because it applies to any distribution. If we know the mean and standard deviation, Tchebysheff tells us that at least 75 percent of the data values will fall with ± 2 standard deviations of the mean and at least 88 percent will lie within 3 standard deviations of the mean. Keep in mind that in most instances, the percentage of observations will exceed these minimum amounts, but at least it gives us some idea of how the data are distributed without actually looking at all the data.
Diff: 2
Keywords: Tchebysheff, data, population, sample
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
136) Explain what is meant by percentiles and quartiles.
Answer: Percentiles and quartiles are measures that help us understand how the data are distributed. Percentiles divide the data into 100 parts so that the same number of observations fall in each percentile. In order to construct the percentiles, the data must be arranged in order from low to high. The percentile value is determined by determining the location of the data value that corresponds to the percent of the way through the ordered data that we wish to go. For example, if we want to find the 80th percentile, we locate the value using:
i = n where P = 80 and n equal the number of values in the data set. If we have 1400 data values, then the 80th percentile value is:
i = n = (1400 ) = 1,120. But since this is an integer, we would compute the 80th percentile to be the average of the 1,120th and 1,121st values in the data.
Quartiles are similar to percentages except that the ordered data are divided into four segments with an equal number of values in each. The 1st quartile corresponds to the 25th percentile, the 2nd quartile corresponds to the 50th percentile, and the 3rd quartile corresponds to the 75th percentile.
Diff: 2
Keywords: percentile, quartile
Section: 3-2 Measures of Variation
Outcome: 3
137) Consumer products are required by law to contain at least as much as the amount printed on the package. For example a bag of potato chips that is labeled as 10 ounces should contain at least 10 ounces. Assume that the standard deviation of the packaging equipment yields a bag weight standard deviation of 0.2 ounces. Explain what average bag weight must be used to achieve at least 97.5 percent of the bags having 10 or more ounces in the bag. Assume the bag weight distribution is bell-shaped.
Answer: If the average bag weight were 10 ounces this would mean only 50 percent of the bags would weigh enough and the other 50 percent would be underweight, so the average must be set higher to achieve 97.5 percent having at least 10 ounces. We want to allow only 2.5 percent of the bags be underweight. Based on the empirical rule, about 95 percent of the bags will be within 2 standard deviations of the mean. This means that 5 percent of the bags will be farther from the mean, and since the distribution is symmetrical that puts 2.5 percent in each tail. Therefore the mean needs to be 2 standard deviations above 10 to have only 2.5 percent in the lower tail below 10 ounces.
Z = -2.0 = so, μ = 10 + 2.0(0.2) = 10.4 ounces
Diff: 3
Keywords: standardized value, empirical distribution
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 5
138) What is meant by the concept, standardizing the data? Explain why a decision maker may wish to compute a standardized value.
Answer: The concept, standardized data, refers to the number of standard deviations a value is from the mean of the sample or population from which it was selected. The population standardized value is referred to as the z-value and is computed as follows:
z =
The reason that we might want to use the standardized value rather than the original value is if we are interested in comparing individual values from two or more distributions that have different means and standard deviations. By comparing z-values, we are able to determine which original value falls relatively more closely to its mean or which value is relatively more extreme compared to the other data in the sample or population.
Diff: 2
Keywords: standardized value, z-score, mean, standard deviation
Section: 3-3 Using the Mean and Standard Deviation Together
Outcome: 4
Business Statistics, 9e (Groebner/Shannon/Fry)
Chapter 13 Goodness-of-Fit Tests and Contingency Analysis
1) A goodness-of-fit test can be used to determine whether a set of sample data comes from a specific hypothesized population distribution.
Answer: TRUE
Diff: 1
Keywords: fit, goodness-of-fit, distribution, population
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
2) If the test statistic for a chi-square goodness-of-fit test is larger than the critical value, the null hypothesis should be rejected.
Answer: TRUE
Diff: 1
Keywords: fit, goodness-of-fit, test statistic, critical value, null
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
3) The logic behind the chi-square goodness-of-fit test is based on determining how far the actual observed frequencies are from the expected frequencies.
Answer: TRUE
Diff: 2
Keywords: fit, goodness-of-fit, chi-square
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
4) The goodness-of-fit test is always a one-tail test with the rejection region in the upper tail.
Answer: TRUE
Diff: 2
Keywords: fit, goodness-of-fit, chi-square, rejection region
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
5) When the expected cell frequencies are smaller than 5, the cells should be combined in a meaningful way such that the expected cell frequencies do exceed 5.
Answer: TRUE
Diff: 2
Keywords: fit, goodness-of-fit, cell, chi-square, frequency
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
6) The reason that a decision maker might want to combine groups before performing a goodness-of-fit test is to avoid accepting the null hypothesis due to an inflated value of the test statistic.
Answer: FALSE
Diff: 2
Keywords: fit, goodness-of-fit, null, group, test statistic
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
7) In a goodness-of-fit test, when the null hypothesis is true, the expected value for the chi-square test statistic is zero.
Answer: TRUE
Diff: 2
Keywords: fit, goodness-of-fit, test statistic
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
8) The Conrad Real Estate Company recently conducted a statistical test to determine whether the number of days that homes are on the market prior to selling is normally distributed with a mean equal to 50 days and a standard deviation equal to 10 days. The sample of 200 homes was divided into 8 groups to form a grouped data frequency distribution. The degrees of freedom for the test will be 7.
Answer: TRUE
Diff: 2
Keywords: fit, goodness-of-fit, normal, degrees of freedom
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
9) The Conrad Real Estate Company recently conducted a statistical test to determine whether the number of days that homes are on the market prior to selling is normally distributed with a mean equal to 50 days and a standard deviation equal to 10 days. The sample of 200 homes was divided into 8 groups to form a grouped data frequency distribution. If a chi-square goodness-of-fit test is to be conducted using an alpha = .05, the critical value is 14.0671.
Answer: TRUE
Diff: 2
Keywords: fit, goodness-of-fit, chi-square, critical value, degrees of freedom
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
10) A business with 5 copy machines keeps track of how many copy machines need service on a given day. It believes this is binomially distributed with a probability of p = 0.2 of each machine needing service on any given day. It has collected the following based on a random sample of 100 days.
X | Frequency |
0 | 28 |
1 | 38 |
2 | 22 |
3 | 7 |
4 or 5 | 5 |
Total | 100 |
Given this information, assuming that all expected values are sufficiently large to use the classes as shown above, the critical value for testing the hypothesis will be based on 5 degrees of freedom.
Answer: FALSE
Diff: 2
Keywords: fit, goodness-of-fit, degrees of freedom
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
11) A business with 5 copy machines keeps track of how many copy machines need service on a given day. It believes this is binomially distributed with a probability of p = 0.2 of each machine needing service on any given day. It has collected the following based on a random sample of 100 days.
X | Frequency |
0 | 28 |
1 | 38 |
2 | 22 |
3 | 7 |
4 or 5 | 5 |
Total | 100 |
Given this information the expected number of days on which exactly 1 machine breaks down is 40.96.
Answer: TRUE
Diff: 3
Keywords: fit, goodness-of-fit, chi-square, expected value
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
12) A business with 5 copy machines keeps track of how many copy machines need service on a given day. It believes this is binomially distributed with a probability of p = 0.2 of each machine needing service on any given day. It has collected the following based on a random sample of 100 days.
X | Frequency |
0 | 28 |
1 | 38 |
2 | 22 |
3 | 7 |
4 or 5 | 5 |
Total | 100 |
Given this information, assuming that all expected values are sufficiently large to use the classes as shown above, the critical value based on a 0.05 level of significance is 9.4877.
Answer: TRUE
Diff: 2
Keywords: fit, goodness-of-fit, critical value
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
13) It is believed that the number of drivers who are ticketed for speeding on a particular stretch of highway is a Poisson distribution with a mean of 3.5 per hour. A random sample of 100 hours is selected with the following results:
X | Frequency |
0 | 5 |
1 | 10 |
2 | 20 |
3 | 18 |
4 | 20 |
5 | 15 |
6 | 4 |
7 | 6 |
8 | 1 |
9 | 2 |
100 |
Given this information, and without regard to whether there is a need to combine cells due to expected cell frequencies, the critical value for testing whether the distribution is Poisson with a mean of 3.5 per hour at an alpha level of .05 is x2 = 15.5073.
Answer: FALSE
Diff: 2
Keywords: fit, goodness-of-fit, chi-square, critical value
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
14) It is believed that the number of drivers who are ticketed for speeding on a particular stretch of highway is a Poisson distribution with a mean of 3.5 per hour. A random sample of 100 hours is selected with the following results:
X | Frequency |
0 | 5 |
1 | 10 |
2 | 20 |
3 | 18 |
4 | 20 |
5 | 15 |
6 | 4 |
7 | 6 |
8 | 1 |
9 | 2 |
100 |
Given this information, it can be seen that the cells will need to be combined since the actual number of occurrences at some levels of x is less than 5.
Answer: FALSE
Diff: 2
Keywords: fit, goodness-of-fit, chi-square, cell
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
15) If the sample size is large, the standard normal distribution can be used in place of the chi-square in a goodness-of-fit test for testing whether the population is normally distributed.
Answer: FALSE
Diff: 2
Keywords: fit, goodness-of-fit, normal, chi-square, standard normal
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
16) By combining cells we guard against having an inflated test statistic that could have led us to incorrectly accept the null hypothesis.
Answer: FALSE
Diff: 3
Keywords: fit, goodness-of-fit, null, test statistic
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
17) If any of the observed frequencies are smaller than 5, then categories should be combined until all observed frequencies are at least 5.
Answer: FALSE
Diff: 2
Keywords: fit, goodness-of-fit, frequency
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
18) A lube and oil change business believes that the number of cars that arrive for service is the same each day of the week. If the business is open six days a week (Monday – Saturday) and a random sample of n = 200 customers is selected, the expected number that will arrive on Monday is about 33.33.
Answer: TRUE
Diff: 1
Keywords: fit, goodness-of-fit, expected value, random sample
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
19) A lube and oil change business believes that the number of cars that arrive for service is the same each day of the week. If the business is open six days a week (Monday – Saturday) and a random sample of n = 200 customers is selected, the sum of the expected frequencies over the six days cannot be determined without seeing the actual sample data.
Answer: FALSE
Diff: 2
Keywords: fit, goodness-of-fit, expected frequency
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
20) A lube and oil change business believes that the number of cars that arrive for service is the same each day of the week. If the business is open six days a week (Monday – Saturday) and a random sample of n = 200 customers is selected, the critical value for testing the hypothesis using a goodness-of-fit test is x2 = 9.2363 if the alpha level for the test is set at .10.
Answer: TRUE
Diff: 2
Keywords: fit, goodness-of-fit, chi-square, critical value
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
21) A goodness-of-fit test can decide whether a set of data comes from a specific hypothesized distribution.
Answer: TRUE
Diff: 2
Keywords: fit, goodness-of-fit, distribution
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
22) If the calculated chi-square statistic is large, this is evidence to suggest the fit of the actual data to the hypothesized distribution is not good, and H0 should be rejected.
Answer: TRUE
Diff: 2
Keywords: fit, goodness-of-fit, chi-square, null
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
23) The goodness-of-fit test is essentially determining if the test statistic is significantly larger than zero.
Answer: FALSE
Diff: 2
Keywords: fit, goodness-of-fit, test statistic
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
24) By combining cells we guard against having an inflated test statistic that could have led us to incorrectly reject the H0.
Answer: TRUE
Diff: 2
Keywords: fit, goodness-of-fit, test statistic, null
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
25) If one or more parameters are left unspecified in a goodness-of-fit test, they must be estimated from the sample data and one degree of freedom is lost for each parameter that must be estimated.
Answer: TRUE
Diff: 2
Keywords: fit, goodness-of-fit, parameter, degrees of freedom
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
26) The sampling distribution for a goodness-of-fit test is the Poisson distribution.
Answer: FALSE
Diff: 2
Keywords: goodness-of-fit, sampling distribution
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
27) Contingency analysis helps to make decisions when multiple proportions are involved.
Answer: TRUE
Diff: 1
Keywords: contingency analysis, proportions
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
28) Contingency analysis is used only for numerical data.
Answer: FALSE
Diff: 1
Keywords: contingency analysis, interval, ratio, measurement
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
29) Managers use contingency analysis to determine whether two categorical variables are independent of each other.
Answer: TRUE
Diff: 1
Keywords: contingency analysis, categorical, independent
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
30) A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed:
Males | Females | |
Have Laptop | 120 | 70 |
No Laptop | 50 | 60 |
Given this information, the sample size in the survey was 300 people.
Answer: TRUE
Diff: 1
Keywords: contingency analysis, sample size
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
31) A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed:
Males | Females | |
Have Laptop | 120 | 70 |
No Laptop | 50 | 60 |
Given this information, if having a laptop is independent of gender, the expected number of males with laptops in this survey is 150.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, expected value
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
32) A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed:
Males | Females | |
Have Laptop | 120 | 70 |
No Laptop | 50 | 60 |
Given this information, if an alpha level of .05 is used, the critical value for testing whether the two variables are independent is x2 = 3.8415.
Answer: TRUE
Diff: 2
Keywords: contingency analysis, critical value, degrees of freedom
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
33) A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed:
Males | Females | |
Have Laptop | 120 | 70 |
No Laptop | 50 | 60 |
Given this information, if an alpha level of .05 is used, the sum of the expected cell frequencies will be equal to the sum of the observed cell frequencies.
Answer: TRUE
Diff: 2
Keywords: contingency analysis, expected, observed, cell frequencies
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
34) A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed:
Males | Females | |
Have Laptop | 120 | 70 |
No Laptop | 50 | 60 |
Given this information, if an alpha level of .05 is used, the test statistic for determining whether having a laptop is independent of gender is approximately 14.23.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, test statistic
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
35) When the variables of interest are both categorical and the decision maker is interested in determining whether a relationship exists between the two, a statistical technique known as contingency analysis is useful.
Answer: TRUE
Diff: 1
Keywords: contingency analysis, categorical
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
36) In conducting a test of independence for a contingency table that has 4 rows and 3 columns, the number of degrees of freedom is 11.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, degrees of freedom, contingency, table
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
37) A study was recently conducted in which people were asked to indicate which new medium was their preferred choice for national news. The following data were observed:
radio | television | newspaper | |
under 21 | 30 | 50 | 5 |
21-40 | 20 | 25 | 30 |
41 and over | 30 | 30 | 50 |
Given this data, if we wish to test whether the preferred news source is independent of age, the expected frequency in the cell, radio—under 21 cell is 30.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, expected frequency
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
38) A cell phone company wants to determine if the use of text messaging is independent of age. The follow data has been collected from a random sample of their customers.
Regularly use text messaging | Do not regularly use text messaging | |
Under 21 | 82 | 38 |
21-39 | 57 | 34 |
40 and over | 6 | 83 |
Using the data above, in order to test for the independence of age and the use of text messaging, the expected value for the “under 21 and regularly use text messaging” cell is 82.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, hypothesis, expected number
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
39) A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed:
radio | television | newspaper | |
under 21 | 30 | 50 | 5 |
21-40 | 20 | 25 | 30 |
41 and over | 30 | 30 | 50 |
Given this data, if we wish to test whether the preferred news source is independent of age with an alpha equal to .05, the critical value will be a chi-square value with 9 degrees of freedom.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, chi-square, degrees of freedom
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
40) A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed:
radio | television | newspaper | |
under 21 | 30 | 50 | 5 |
21-40 | 20 | 25 | 30 |
41 and over | 30 | 30 | 50 |
Given this data, if we wish to test whether the preferred news source is independent of age, the cell with the largest expected cell frequency is also the cell with the largest observed frequency.
Answer: FALSE
Diff: 3
Keywords: contingency analysis, expected, observed, frequency
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
41) A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed:
radio | television | newspaper | |
under 21 | 30 | 50 | 5 |
21-40 | 20 | 25 | 30 |
41 and over | 30 | 30 | 50 |
Given this data, if we wish to test whether the preferred news source is independent of age, for an alpha = .05 level, the critical value from the chi-square table is based on 8 degrees of freedom.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, chi-square, degrees of freedom, critical value
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
42) A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed:
radio | television | newspaper | |
under 21 | 30 | 50 | 5 |
21-40 | 20 | 25 | 30 |
41 and over | 30 | 30 | 50 |
Given this data, if we wish to test whether the preferred news source is independent of age, for an alpha = .05 level, the critical value from the chi-square table is 9.4877.
Answer: TRUE
Diff: 2
Keywords: contingency analysis, critical value, chi-square
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
43) A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed:
radio | television | newspaper | |
under 21 | 30 | 50 | 5 |
21-40 | 20 | 25 | 30 |
41 and over | 30 | 30 | 50 |
Given this data, if we wish to test whether the preferred news source is independent of age, for an alpha = .05 level, the test statistic is computed to be approximately 40.70.
Answer: TRUE
Diff: 3
Keywords: contingency analysis, test statistic, chi-square
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
44) A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers.
Regularly use text messaging | Do not regularly use text messaging | |
Under 21 | 82 | 38 |
21-39 | 57 | 34 |
40 and over | 6 | 8 |
Using this data, if we wish to test whether the preferred news source is independent of age using a 0.05 level of significance, the critical value is 5.9915.
Answer: TRUE
Diff: 2
Keywords: contingency analysis, chi-square, null, critical value
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
45) In order to apply the chi-square contingency methodology for quantitative variables, we must first break the quantitative variable down into discrete categories.
Answer: TRUE
Diff: 2
Keywords: contingency analysis, variables, chi-square
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
46) A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded.
US | Japanese | German | |
East Coast | 200 | 200 | 50 |
Central | 250 | 100 | 20 |
West Coast | 80 | 300 | 40 |
Given this situation, the sample size used in this study was nine.
Answer: FALSE
Diff: 1
Keywords: contingency analysis, sample size
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
47) A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded.
US | Japanese | German | |
East Coast | 200 | 200 | 50 |
Central | 250 | 100 | 20 |
West Coast | 80 | 300 | 40 |
Given this situation, the null hypothesis to be tested is that the car origin is dependent on the geographical location of the buyer.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, null, hypothesis
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
48) A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded.
US | Japanese | German | |
East Coast | 200 | 200 | 50 |
Central | 250 | 100 | 20 |
West Coast | 80 | 300 | 40 |
Given this situation, to test whether the car origin is independent of the geographical location of the buyer, the sum of the expected cell frequencies will equal 1,240.
Answer: TRUE
Diff: 2
Keywords: contingency analysis, cell, cell frequencies, sample size
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
49) A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded.
US | Japanese | German | |
East Coast | 200 | 200 | 50 |
Central | 250 | 100 | 20 |
West Coast | 80 | 300 | 40 |
Given this situation, to test whether the car origin is independent of the geographical location of the buyer, the critical value for alpha = .10 is 14.6837.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, critical value, chi-square
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
50) A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded.
US | Japanese | German | |
East Coast | 200 | 200 | 50 |
Central | 250 | 100 | 20 |
West Coast | 80 | 300 | 40 |
Given this situation, to test whether the car origin is independent of the geographical location of the buyer, the expected number of people in the sample who bought a German made car and who lived on the East Coast is just under 40 people.
Answer: TRUE
Diff: 2
Keywords: contingency analysis, expected value
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
51) A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers.
Regularly use text messaging | Do not regularly use text messaging | |
Under 21 | 82 | 38 |
21-39 | 57 | 34 |
40 and over | 6 | 83 |
To conduct a test of independence, the difference expected value for the “40 and over and regularly use text messaging” cell is just over 43 people.
Answer: TRUE
Diff: 2
Keywords: contingency analysis, chi-square, expected frequency
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
52) Contingency analysis can be used when the level of data measurement is nominal or ordinal.
Answer: TRUE
Diff: 1
Keywords: contingency analysis, nominal, ordinal, data
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
53) To employ contingency analysis, we set up a 2-dimensional table called a contingency table.
Answer: TRUE
Diff: 1
Keywords: contingency analysis, table
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
54) A contingency table and a cross-tabulation table are two separate things and should not be used for the same purpose.
Answer: FALSE
Diff: 1
Keywords: contingency analysis, cross-tabulation
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
55) In a contingency analysis, we expect the actual frequencies in each cell to approximately match the corresponding expected cell frequencies when H0 is true.
Answer: TRUE
Diff: 2
Keywords: contingency analysis, observed, expected, frequency, null
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
56) In a chi-square contingency test, the number of degrees of freedom is equal to the number of cells minus 1.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, degrees of freedom, cell
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
57) In a chi-square contingency analysis application, the expected cell frequencies will be equal in all cells if the null hypothesis is true.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, expected frequency, null, hypothesis
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
58) Unlike the case of goodness-of-fit testing, with contingency analysis there is no restriction on the minimum size for an expected cell frequency.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, goodness-of-fit, expected frequency
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
59) In a contingency analysis the expected values are based on the assumption that the two variables are independent of each other.
Answer: TRUE
Diff: 2
Keywords: contingency analysis, chi-square, expected values
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
60) If a contingency analysis test is performed with a 4 × 6 design, and if alpha = .05, the critical value from the chi-square distribution is 24.9958
Answer: TRUE
Diff: 2
Keywords: contingency analysis, critical value, chi-square
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
61) If a contingency analysis test performed with a 4 × 6 design results in a test statistic value of 18.72, and if alpha = .05, the null hypothesis that the row and column variable are independent should be rejected.
Answer: FALSE
Diff: 2
Keywords: contingency analysis, chi-square, test statistic, critical value, null
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
62) If the null hypothesis is not rejected, you do not need to worry when the expected cell frequencies drop below 5.0
Answer: TRUE
Diff: 3
Keywords: contingency analysis, chi-square, null
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
63) The degrees of freedom for the chi-square goodness-of-fit test are equal to ________, where k is the number of categories.
- A) k + 1
- B) k – 1
- C) k + 2
- D) k – 2
Answer: B
Diff: 1
Keywords: chi-square, goodness of fit, degrees of freedom
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
64) Which of the following statements is true in the context of a chi-square goodness-of-fit test?
- A) The degrees of freedom for determining the critical value will be the number of categories minus 1.
- B) The critical value will come from the standard normal table if the sample size exceeds 30.
- C) The null hypothesis will be rejected for a small value of the test statistic.
- D) A very large test statistic will result in the null not being rejected.
Answer: A
Diff: 2
Keywords: chi-square, goodness of fit, degrees of freedom
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
65) A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collected the following data based on a random sample of 100 days.
Frequency | |
Mon | 25 |
Tue | 22 |
Wed | 19 |
Thu | 18 |
Fri | 16 |
Total | 100 |
Based on this information how many degrees for freedom are involved in this goodness of fit test?
- A) 99
- B) 100
- C) 4
- D) 5
Answer: C
Diff: 1
Keywords: chi-square, goodness of fit, degrees of freedom
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
66) A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collected the following data based on a random sample of 100 days.
Frequency | |
Mon | 25 |
Tue | 22 |
Wed | 19 |
Thu | 18 |
Fri | 16 |
Total | 100 |
Assuming that a goodness-of-fit test is to be conducted using a 0.10 level of significance, the critical value is:
- A) 9.4877
- B) 11.0705
- C) 7.7794
- D) 9.2363
Answer: C
Diff: 2
Keywords: chi-square, goodness-of-fit, critical value
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
67) A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collected the following data based on a random sample of 100 days.
Frequency | |
Mon | 25 |
Tue | 22 |
Wed | 19 |
Thu | 18 |
Fri | 16 |
Total | 100 |
To conduct a goodness-of-fit test, what is the expected value for Friday?
- A) 20
- B) 25
- C) 16
- D) 100
Answer: A
Diff: 1
Keywords: chi-square, goodness-of-fit, expected frequency
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
68) A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collected the following data based on a random sample of 100 days.
Frequency | |
Mon | 25 |
Tue | 22 |
Wed | 19 |
Thu | 18 |
Fri | 16 |
Total | 100 |
What is the value of the test statistic needed to conduct a goodness-of-fit test?
- A) 8.75
- B) 7.7794
- C) 2.46
- D) 2.50
Answer: D
Diff: 2
Keywords: chi-square, goodness-of-fit, test statistic
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
69) A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collected the following data based on a random sample of 100 days.
Frequency | |
Mon | 25 |
Tue | 22 |
Wed | 19 |
Thu | 18 |
Fri | 16 |
Total | 100 |
Based on these data, conduct a goodness-of-fit test using a 0.10 level of significance. Which conclusion is correct?
- A) Arrivals are not uniformly distributed over the weekday because (test statistic) > (critical value).
- B) Arrivals are uniformly distributed over the weekday because (test statistic) > (critical value).
- C) Arrivals are not uniformly distributed over the weekday because (test statistic) < (critical value).
- D) Arrives are uniformly distributed over the weekday because (test statistic) < (critical value).
Answer: D
Diff: 3
Keywords: chi-square, goodness-of-fit, critical value, test statistic
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
70) In a chi-square goodness-of-fit test, by combining cells we guard against having an inflated test statistic that could have caused us to:
- A) incorrectly reject the H0.
- B) incorrectly accept the H0.
- C) incorrectly reject the H1.
- D) incorrectly accept the H1.
Answer: A
Diff: 2
Keywords: chi-square, goodness-of-fit, null, cells
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
71) In a goodness-of-fit test about a population distribution, if one or more parameters are left unspecified in H0, they must be estimated from the sample data. This will reduce the degrees of freedom by ________ for each estimated parameter.
- A) 1
- B) 2
- C) 3
- D) None of the above
Answer: A
Diff: 2
Keywords: chi-square, goodness-of-fit, degrees of freedom, parameter
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
72) If a sample with n = 60 subjects distributed over 3 categories was selected, a chi-square test for goodness-of-fit will be used. How many degrees of freedom will be used in determining the chi-square test statistic?
- A) 1
- B) 2
- C) 16
- D) 64
Answer: B
Diff: 2
Keywords: chi-square, goodness-of-fit, degrees of freedom
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
73) Consider a goodness-of-fit test with a computed value of chi-square = 1.273 and a critical value = 13.388, the appropriate conclusion would be to:
- A) reject H0.
- B) fail to reject H0.
- C) take a larger sample.
- D) take a smaller sample.
Answer: B
Diff: 2
Keywords: chi-square, goodness-of-fit, null, critical value, test statistic
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
74) A researcher is using a chi-square test to determine whether there are any preferences among 4 brands of orange juice. With alpha = 0.05 and n = 30, the critical region for the hypothesis test would have a boundary of:
- A) 7.81
- B) 8.71
- C) 8.17
- D) 42.25
Answer: A
Diff: 2
Keywords: chi-square, goodness-of-fit, critical value
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
75) A chi-square test for goodness-of-fit is used to test whether or not there are any preferences among 3 brands of peas. If the study uses a sample of n = 60 subjects, then the expected frequency for each category would be:
- A) 20
- B) 30
- C) 60
- D) 33
Answer: A
Diff: 2
Keywords: chi-square, goodness-of-fit, expected frequency
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
76) We are interested in determining whether the opinions of the individuals on gun control (as to Yes, No, and No Opinion) are uniformly distributed.
A sample of 150 was taken and the following data were obtained.
Do you support gun control | Number of Responses |
Yes | 40 |
No | 60 |
No Opinion | 50 |
The conclusion of the test with alpha = 0.05 is that the views of people on gun control are:
- A) uniformly distributed.
- B) not uniformly distributed.
- C) inconclusive.
- D) None of the above
Answer: A
Diff: 2
Keywords: chi-square, goodness-of-fit, expected frequency
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
77) To use contingency analysis for numerical data, which of the following is true?
- A) Contingency analysis cannot be used for numerical data.
- B) Numerical data must be broken up into specific categories.
- C) Contingency analysis can be used for numerical data only if both variables are numerical.
- D) Contingency analysis can be used for numerical data only if it is interval data.
Answer: B
Diff: 2
Keywords: contingency analysis, nominal, ordinal, data
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
78) What does the term observed cell frequencies refer to?
- A) The frequencies found in the population being examined
- B) The frequencies found in the sample being examined
- C) The frequencies computed from H0
- D) The frequencies computed from H1
Answer: B
Diff: 2
Keywords: contingency analysis, observed, cell frequency
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
79) What does the term expected cell frequencies refer to?
- A) The frequencies found in the population being examined
- B) The frequencies found in the sample being examined
- C) The frequencies computed from H0
- D) the frequencies computed from H1
Answer: C
Diff: 2
Keywords: contingency analysis, expected, cell frequency
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
80) We expect the actual frequencies in each cell to approximately match the corresponding expected cell frequencies when:
- A) H0is false.
- B) H0is true.
- C) H0is falsely accepted.
- D) the variables are related to each other.
Answer: B
Diff: 1
Keywords: contingency analysis, null, hypothesis, observed, actual, expected, frequency
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
81) In a contingency analysis, the greater the difference between the actual and the expected frequencies, the more likely:
- A) H0should be rejected.
- B) H0should be accepted.
- C) we cannot determine H
- D) the smaller the test statistic will be.
Answer: A
Diff: 1
Keywords: contingency analysis, actual, expected, observed, null
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
82) In a chi-square contingency analysis, when expected cell frequencies drop below 5, the calculated chi-square value tends to be inflated and may inflate the true probability of ________ beyond the stated significance level.
- A) committing a Type I error
- B) committing a Type II error
- C) Both A and B
- D) All of the above
Answer: A
Diff: 2
Keywords: contingency analysis, expected, frequency, type I, alpha
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
83) In performing chi-square contingency analysis, to overcome a small expected cell frequency problem, we:
- A) combine the categories of the row and/or column variables.
- B) increase the sample size.
- C) Both A and B
- D) None of the above
Answer: C
Diff: 2
Keywords: contingency analysis, combine, sample size, expected, frequency
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
84) How can the degrees of freedom be found in a contingency table with cross-classified data?
- A) When df are equal to rows minus columns
- B) When df are equal to rows multiplied by columns
- C) When df are equal to rows minus 1 multiplied by columns minus 1
- D) Total number of cell minus 1
Answer: C
Diff: 2
Keywords: contingency analysis, degrees of freedom, table, cross-classified
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
85) A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers.
Regularly use text messaging | Do not regularly use text messaging | |
Under 21 | 82 | 38 |
21-39 | 57 | 34 |
40 and over | 6 | 83 |
Based on the data above what is the expected value for the “under 21 and regularly use text messaging” cell?
- A) 82
- B) 50
- C) 120
- D) 58
Answer: D
Diff: 2
Keywords: contingency analysis, expected frequency
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
86) A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers.
Regularly use text messaging | Do not regularly use text messaging | |
Under 21 | 82 | 38 |
21-39 | 57 | 34 |
40 and over | 6 | 83 |
To conduct a contingency analysis, the number of degrees of freedom is:
- A) 6
- B) 5
- C) 3
- D) 2
Answer: D
Diff: 2
Keywords: contingency analysis, critical value, chi-square
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
87) A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers.
Regularly use text messaging | Do not regularly use text messaging | |
Under 21 | 82 | 38 |
21-39 | 57 | 34 |
40 and over | 6 | 83 |
To conduct a contingency analysis using a 0.01 level of significance, the value of the critical value is:
- A) 15.0863
- B) 5.9915
- C) 9.2104
- D) 11.0705
Answer: C
Diff: 2
Keywords: contingency analysis, chi-square, critical value
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
88) A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers.
Regularly use text messaging | Do not regularly use text messaging | |
Under 21 | 82 | 38 |
21-39 | 57 | 34 |
40 and over | 6 | 83 |
To conduct a contingency analysis, the value of the test statistic is:
- A) 9.2104
- B) 88.3
- C) 275.02
- D) 14.6
Answer: B
Diff: 2
Keywords: contingency analysis, test statistic, chi-square
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
89) For a chi-square test involving a contingency table, suppose H0 is rejected. We conclude that the two variables are:
- A) curvilinear.
- B) linear.
- C) related.
- D) not related.
Answer: C
Diff: 2
Keywords: contingency analysis, chi-square, null, alternative, hypothesis
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
90) When testing for independence in a contingency table with 3 rows and 4 columns, there are ________ degrees of freedom.
- A) 5
- B) 6
- C) 7
- D) 12
Answer: B
Diff: 1
Keywords: contingency analysis, degrees of freedom
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
91) In testing a hypothesis that two categorical variables are independent using the x2 test, the expected cell frequencies are based on assuming:
- A) the null hypothesis.
- B) the alternative hypothesis.
- C) the normal distribution.
- D) the variable are related.
Answer: A
Diff: 1
Keywords: contingency analysis, chi-square, expected, frequency, null
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
92) A study published in the American Journal of Public Health was conducted to determine whether the use of seat belts in motor vehicles depends on ethnic status in San Diego County. A sample of 792 children treated for injuries sustained from motor vehicle accidents was obtained, and each child was classified according to (1) ethnic status (Hispanic or non-Hispanic) and (2) seat belt usage (worn or not worn) during the accident. The number of children in each category is given in the table below.
Hispanic | Non-Hispanic | |
Seat belts worn | 31 | 148 |
Seat belts not worn | 283 | 330 |
Referring to these data, which test would be used to properly analyze the data in this experiment?
- A) x2test for independence in a two-way contingency table
- B) x2test for equal proportions in a one-way table
- C) ANOVA F-test for interaction in a 2 × 2 factorial design
- D) x2goodness-of-fit test
Answer: A
Diff: 2
Keywords: contingency analysis, chi-square, independence
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
93) A study published in the American Journal of Public Health was conducted to determine whether the use of seat belts in motor vehicles depends on ethnic status in San Diego County. A sample of 792 children treated for injuries sustained from motor vehicle accidents was obtained, and each child was classified according to (1) ethnic status (Hispanic or non-Hispanic) and (2) seat belt usage (worn or not worn) during the accident. The number of children in each category is given in the table below.
Hispanic | Non-Hispanic | |
Seat belts worn | 31 | 148 |
Seat belts not worn | 283 | 330 |
Referring to these data, the calculated test statistic is:
- A) approximately -0.9991
- B) nearly -0.1368
- C) about 48.1849
- D) approximately 72.8063
Answer: C
Diff: 3
Keywords: contingency analysis, test statistic, chi-square
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
94) A study published in the American Journal of Public Health was conducted to determine whether the use of seat belts in motor vehicles depends on ethnic status in San Diego County. A sample of 792 children treated for injuries sustained from motor vehicle accidents was obtained, and each child was classified according to (1) ethnic status (Hispanic or non-Hispanic) and (2) seat belt usage (worn or not worn) during the accident. The number of children in each category is given in the table below.
Hispanic | Non-Hispanic | |
Seat belts worn | 31 | 148 |
Seat belts not worn | 283 | 330 |
Referring to these data, which of the following conclusions should be reached if the appropriate hypothesis is conducted using an alpha = .05 level?
- A) The mean value for Hispanics is the same as for Non-Hispanics.
- B) There is no relationship between whether someone is Hispanic and whether they wear a seat belt.
- C) The use of seat belts and whether a person is Hispanic or not is statistically related.
- D) None of the above
Answer: C
Diff: 3
Keywords: contingency analysis, chi-square, hypothesis
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
95) Many companies use well-known celebrities as spokespeople in their TV advertisements. A study was conducted to determine whether brand awareness of female TV viewers and the gender of the spokesperson are independent. Each in a sample of 300 female TV viewers was asked to identify a product advertised by a celebrity spokesperson. The gender of the spokesperson and whether or not the viewer could identify the product was recorded. The numbers in each category are given below.
Male Celebrity | Female Celebrity | |
Identified product | 41 | 61 |
Could not identify | 109 | 89 |
Referring to these sample data, which test would be used to properly analyze the data in this experiment?
- A) x2test for independence in a two-way contingency table
- B) x2test for equal proportions in a one-way table
- C) ANOVA F-test for main treatment effect
- D) x2goodness-of-fit test
Answer: A
Diff: 2
Keywords: contingency analysis, chi-square, table
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
96) Many companies use well-known celebrities as spokespeople in their TV advertisements. A study was conducted to determine whether brand awareness of female TV viewers and the gender of the spokesperson are independent. Each in a sample of 300 female TV viewers was asked to identify a product advertised by a celebrity spokesperson. The gender of the spokesperson and whether or not the viewer could identify the product was recorded. The numbers in each category are given below.
Male Celebrity | Female Celebrity | |
Identified product | 41 | 61 |
Could not identify | 109 | 89 |
Referring to these sample data, if the appropriate hypothesis test is to be conducted using a .05 level of significance, which of the following is correct critical value?
- A) 9.4877
- B) 3.8415
- C) 1.96
- D) 7.8147
Answer: B
Diff: 2
Keywords: contingency analysis, chi-square, critical value
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
97) Many companies use well-known celebrities as spokespeople in their TV advertisements. A study was conducted to determine whether brand awareness of female TV viewers and the gender of the spokesperson are independent. Each in a sample of 300 female TV viewers was asked to identify a product advertised by a celebrity spokesperson. The gender of the spokesperson and whether or not the viewer could identify the product was recorded. The numbers in each category are given below.
Male Celebrity | Female Celebrity | |
Identified product | 41 | 61 |
Could not identify | 109 | 89 |
Referring to these sample data, which of the following values is the correct value of the test statistic?
- A) Approximately 9.48
- B) Nearly 23.0
- C) About 3.84
- D) Approximately 5.94
Answer: D
Diff: 3
Keywords: contingency analysis, chi-square, test statistic
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
98) Many companies use well-known celebrities as spokespeople in their TV advertisements. A study was conducted to determine whether brand awareness of female TV viewers and the gender of the spokesperson are independent. Each in a sample of 300 female TV viewers was asked to identify a product advertised by a celebrity spokesperson. The gender of the spokesperson and whether or not the viewer could identify the product was recorded. The numbers in each category are given below.
Male Celebrity | Female Celebrity | |
Identified product | 41 | 61 |
Could not identify | 109 | 89 |
Referring to these sample data, if the appropriate null hypothesis is tested using a significance level equal to .05, which of the following conclusions should be reached?
- A) There is a relationship between gender of the celebrity and product identification.
- B) There is no relationship between gender of the celebrity and product identification.
- C) The mean number of products identified for males is different than the mean number for females.
- D) Females have higher brand awareness than males.
Answer: A
Diff: 3
Keywords: contingency analysis, chi-square, hypothesis, null
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
99) The degrees of freedom for a contingency table with 11 rows and 10 columns is:
- A) 11
- B) 10
- C) 110
- D) 90
Answer: D
Diff: 1
Keywords: contingency analysis, chi-square
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
100) We want to test whether type of car owned (domestic or foreign) is independent of gender. A contingence table is obtained from a sample of 990 people as
At alpha = 0.05 level, we conclude that:
- A) x2= 3.34 and type of car owned is independent of gender.
- B) x2= 3.34 and type of car owned is dependent of gender.
- C) x2= 3.84 and type of car owned is independent of gender.
- D) x2= 3.84 and type of car owned is dependent of gender.
Answer: A
Diff: 3
Keywords: contingency analysis, chi-square, null, critical value
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
101) The billing department of a national cable service company is conducting a study of how customers pay their monthly cable bills. The cable company accepts payment in one of four ways: in person at a local office, by mail, by credit card, or by electronic funds transfer from a bank account. The cable company randomly sampled 400 customers to determine if there is a relationship between the customer’s age and the payment method used. The following sample results were obtained:
Based on the sample data, can the cable company conclude that there is a relationship between the age of the customer and the payment method used? Conduct the appropriate test at the alpha= 0.01 level of significance.
- A) Because x2= 42.2412 > 21.666, do not reject the null hypothesis. Based on the sample data conclude that age and type of payment are independent.
- B) Because x2= 42.2412 > 21.666, reject the null hypothesis. Based on the sample data conclude that age and type of payment are not independent
- C) Because x2= 50.3115 > 21.666, do not reject the null hypothesis. Based on the sample data conclude that age and type of payment are
- D) Because x2= 50.3115 > 21.666, reject the null hypothesis. Based on the sample data conclude that age and type of payment are not independent.
Answer: D
Diff: 2
Keywords: contingency analysis
Section: 13-2 Introduction to Contingency Analysis
Outcome: 1
102) Explain the basic logic behind the chi-square goodness-of-fit test.
Answer: The logic of the chi-square goodness-of-fit test is as follows: If the sample data selected randomly from a hypothesized population distribution have a distribution that “closely” matches the hypothesized distribution, then the hypothesis should be “accepted.” If the sample data do not closely fit the hypothesized distribution, then we would reject the hypothesis. The decision is based on the chi-square test statistic, which compares the actual frequency of sample occurrences as prescribed levels of the variable against the expected frequencies at those same levels assuming the hypothesized distribution is correct. The test statistics is computed as follows:
x2 =
where:
oi = Observed cell frequency for category i
ei = Expected cell frequency for category i
k = Number of categories.
If the test statistic is larger than the chi-square critical value for a specified level of significance, the null hypothesis is rejected.
Diff: 2
Keywords: chi-square, goodness-of-fit
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
103) The Bradfield Container Company makes “cardboard” boxes for commercial use (i.e., pizza boxes). One of the big issues for the company is the set-up time required to change over from one order to the next. At one particular machine, the set-up time is thought to be uniformly distributed between 10 and 21 minutes. To test whether this is true or not, a random sample of 180 set-ups on this machine was selected with set-up time rounded to the nearest two-minute intervals. The following results occurred:
Set-up Time Frequency
10-11 minutes 13
12-13 minutes 23
14-15 minutes 40
16-17 minutes 44
18-19 minutes 40
20-21 minutes 20
- What are the appropriate null and alternative hypothesis to be tested?
- Based on the null and alternative hypotheses stated in part a, determine the expected frequencies for each set-up time category.
- Assuming that we wish to conduct the hypothesis test at the .05 level, what is the critical value that should be used?
- Compute the test statistic and carry out the hypothesis test.
Answer:
- The null and alternative hypotheses are:
H0 : set-up times on this machine are uniformly distributed between 10 and 21 minutes
HA : set-up times are not uniformly distributed between 10 and 21 minutes.
- Because the hypothesized distribution is a uniform distribution, the expected number of values at each set-up time level is found by dividing number of levels by the sample size:
Expected = 180/6 = 30
Set-up Time Frequency Expected
10-11 minutes 13 30
12-13 minutes 23 30
14-15 minutes 40 30
16-17 minutes 44 30
18-19 minutes 40 30
20-21 minutes 20 30
- The critical value will be a chi-square value with degrees of freedom equal to the number of levels minus one. The degrees of freedom is 6 – 1 = 5. From the chi-square table with alpha = .05 and 5 degrees of freedom, the critical value is x2= 11.0705.
Thus, if the test statistic > 11.0705, reject the null hypothesis.
- The test statistic is found using:
x2 =
The computations are shown as follows:
Set-up Time Frequency Expected (O-E)2 (O-E)2/E
10-11 minutes 13 30 (13-30)2=289 289/30=9.63
12-13 minutes 23 30 (23-30)2 = 49 49/30=1.63
14-15 minutes 40 30 (40-30)2 =100 100/30 = 3.33
16-17 minutes 44 30 (44-30)2=196 196/30 = 6.53
18-19 minutes 40 30 (40-30)2=100 100/30 = 3.33
20-21 minutes 20 30 (20-30)2=100 100/30 = 3.33
Total =27.78
Since x2 = 27.78 > 11.07, we reject the null hypothesis and conclude that the distribution of set-up times is not uniform.
Diff: 3
Keywords: null, alternative, hypothesis, test statistic, chi-square, critical value
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
104) Explain why, in performing a goodness-of-fit test, it is sometimes necessary to combine categories.
Answer: Because of the way in which the chi-square test statistic is computed by squaring the difference between the observed and expected frequencies, when the expected frequencies are small (less than 5), the calculated test statistic can become artificially large and therefore may lead to an increased chance of committing a Type I statistical error. That is, a true null hypothesis may be rejected at a higher rate than indicated by the selected significance level. By combining categories, the small expected frequencies are grouped to become larger than five and thus the issue of inflated Type I error probability dissolves.
Note: An alternative to combining categories is to increase the sample size. Large sample sizes result in greater expected cell frequencies in all categories.
Diff: 2
Keywords: goodness-of-fit, combine, categories, cell
Section: 13-1 Introduction to Goodness-of-fit Tests
Outcome: 1
105) Recently a survey was conducted involving customers of a fitness center in Dallas, Texas. Participants were asked to indicate how often they use the club by checking one of the following categories: 0-1 time per week; 2-3 times per week; 4-5 times per week; more than 5 times. The following data show how males and females responded to this question.
0-1 | 2-3 | 4-5 | over 5 | |
Males | 41 | 61 | 50 | 20 |
Females | 109 | 89 | 60 | 45 |
One of the purposes of the survey was to determine whether there is a relationship between the gender of the customer and the number of visits made each week.
- State the appropriate null and alternative hypothesis.
- What test procedure is appropriate to use to conduct this test?
- Conduct the hypothesis test using an alpha = .05 level.
Answer:
- H0: number of visits is independent of gender
HA : number of visits is related to gender
- Because the data are observed frequencies in various discrete categories, the appropriate test to use is chi-square contingency analysis. This involves determining the expected frequencies assuming the null hypothesis is true and then comparing these expected frequencies, cell by cell, to the observed frequencies. If these closely match, then the null hypothesis should not be rejected. However, if there is a big difference between the expected and observed cell frequencies, we should reject the null hypothesis.
- The test statistic for performing a chi-square contingency analysis is computed as follows:
x2 = with d.f. = (r – 1)(c – 1).
The first step needed is to compute the expected cell frequencies. This is done under the assumption that the null hypothesis is true and that the proportion of customers in each use level is the same regardless of gender. The expected frequencies can be computed using:
Expected Frequency =
For example, for the cell corresponding to males who use the center 0-1 times per week, we get:
Expected Frequency = = 54.3158.
The following shows the expected cell frequencies for each cell:
0-1 | 2-3 | 4-5 | over 5 | |
Males | 54.31579 | 54.34179 | 39.83158 | 23.53684 |
Females | 96.68421 | 95.68421 | 70.16842 | 41.46316 |
Next for each cell we compute: . For example in the cell for males and use between 0 and 1, we get:
3.26
Below we show the computation for each cell:
0-1 | 2-3 | 4-5 | over 5 | |
Males | 3.264433 | 0.822572 | 2.59585 | 0.531475 |
Females | 1.853077 | 0.466939 | 1.473552 | 0.301696 |
The chi-square test statistic is computed by summing these values giving
x2 = 11.309
The critical value for the contingency analysis test with (2 – 1) × (4 – 1) = 3 degrees of freedom and alpha equal .05 is found in the chi-square table to be 7.8147.
The decision rule is:
If x2 > 7.8147, reject the null hypothesis
Otherwise, do not reject.
Since x2 = 11.309 > 7.8147, reject the null hypothesis and conclude that use rate is related to gender of the customer.
Diff: 3
Keywords: null, alternative, hypothesis, contingency table, crosstabulation
Section: 13-2 Introduction to Contingency Analysis
Outcome: 2
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