Chapter 15 Chi-Square Tests

True / False Questions1.In a chi-square test of a 5 5 contingency table at = .05, the critical value is 37.65.FALSE2.05 = 26.30 for d.f. = (5 - 1)(5 - 1) = 16.

2.If two variables are independent, we would anticipate a chi-squaretest statistic close to zero.TRUEThe difference between observed and expected should be near zero.

3.The null hypothesis for a chi-square test on a contingency table is that the variables are dependent.FALSEThe null hypothesis is independence (not dependence).

4.The shape of the chi-square distribution depends only on its degrees of freedom.TRUEThe chi-square distribution has only one parameter (called degrees of freedom).

5.In a chi-square test for independence, expected frequencies must be integers (or rounded to the nearest integer).FALSEExpected frequencies are integers only in unusual situations (if frequencies are "nice").

6.In a chi-square test for independence, observed frequencies must be at least 5 in every cell.FALSESmall expected (not observed) frequencies are to be avoided.

7.In a chi-square test for independence, observed and expected frequencies must sum across to the same row totals and down to the same column totals.TRUEExpected frequencies reallocate the row (or column) total, so they must sum to the total.

8.In samples drawn from a population in which the row and column categories are independent, the value of the chi-square test statistic will be zero.FALSESampling variation exists even if the null hypothesis is true for the population.

9.In a hypothesis test using chi-square, if the null hypothesis is true, the sample value of the sample chi-square test statistic will be exactly zero.FALSESampling variation exists even if the null hypothesis is true for the population.

10.The chi-square test for independence is a nonparametric test (no parameters are estimated).TRUEThe chi-square test does not estimate a parameter.

11.Cochran's Rule requires observed frequencies of 5 or more in each cell of a contingency table.FALSESmall expected (not observed) frequencies are to be avoided.

12.A large negative chi-square test statistic would indicate that the null hypothesis should be rejected.FALSEIt is a sum of squares divided by a positive expected frequency so it cannot be negative.

13.The degrees of freedom in a 3 4 chi-square contingency table would equal 11.FALSEd.f. = (3 - 1)(4 - 1) = 6.

14.The null hypothesis for a chi-square contingency test of independence for two variables always assumes that the variables are independent.TRUEThe null hypothesis must be phrased like this so there is only one way it can be true.

15.The chi-square test is unreliable when there are any cells with small observed frequency counts.FALSESmall expected (not observed) frequencies are to be avoided.

16.The chi-square test can only be used to assess independence between two variables.FALSEChi-square tests can be used to test for goodness of fit, for example.

17.The chi-square test is based on the analysis of frequencies.TRUEIts attraction is that the test can be performed on categorical data (counts).

18.A chi-square distribution is always skewed right.TRUEEspecially for small degrees of freedom, the chi-square distribution is right-skewed.

19.A chi-square test for independence is called a distribution-free test since the test is based on categorical data rather than on populations that follow any particular distribution.TRUEThe lack of assumed population shape is an attraction of this test.

20.Observed frequencies in a chi-square goodness-of-fit test for normality may be less than 5 or even 0 in some cells, as long as the expected frequencies are large enough.TRUESmall expected (not observed) frequencies are to be avoided.

21.In a chi-square goodness-of-fit test, a small p-value would indicate a good fit to the hypothesized distribution.FALSEWhen the p-value is small, we are inclined to reject the hypothesized distribution.

22.For a chi-square goodness-of-fit test for a uniform distribution with 5 categories, we would use the critical value for 4 degrees of freedom.TRUEd.f. = k - 1 - m = 5 - 1 - 0 = 4 where m = 0 parameters are estimated and k = 5 categories.

23.For a chi-square goodness-of-fit test for a uniform distribution with 7 categories, we would use the critical value for 6 degrees of freedom.TRUEd.f. = k - 1 - m = 7 - 1 - 0 = 6 where m = 0 parameters are estimated and k = 7 categories.

24.For a chi-square goodness-of-fit test for a normal distribution using 8 categories with estimated mean and standard deviation, we would use the critical value for 7 degrees of freedom.FALSEd.f. = k - 1 - m = 8 - 1 - 2 = 5 where m = 2 parameters are estimated and k = 8 categories.

25.For a chi-square goodness-of-fit test for a normal distribution using 7 categories with estimated mean and standard deviation, we would use the critical value for 4 degrees of freedom.TRUEd.f. = k - 1 - m = 7 - 1 - 2 = 4 where m = 2 parameters are estimated and k = 7 categories.

26.A probability plot usually allows outliers to be detected.TRUEOutliers will be seen as unusual (tail) points far from the main body of data.

27.In a goodness-of-fit test, a linear probability plot suggests that the null hypothesis should be rejected.FALSEData that follow a straight line would support the null hypothesis (and conversely).

28.In a chi-square goodness-of-fit test, we lose one degree of freedom for each parameter estimated.TRUEd.f. = k - 1 - m where m = number of parameters that are estimated and k = number of categories.

29.In a chi-square goodness-of-fit test, we gain one degree of freedom if n increases by 1.FALSEd.f. = k - 1 - m for m parameters and k categories (n does not enter this formula).

30.In a chi-square goodness-of-fit test, a sample of n observations has n - 1 degrees of freedom.FALSEd.f. = k - 1 - m for m parameters and k categories (n does not enter this formula).

31.The Poisson goodness-of-fit test is inappropriate for continuous data.TRUEPoisson data are integers.

32.The Kolmogorov-Smirnov and Anderson-Darling tests are based on the ECDF (Empirical Cumulative Distribution Function).TRUEThe ECDF provides the basis for several such tests.

33.Goodness-of-fit tests using the ECDF (Empirical Cumulative Distribution Function) compare the actual cumulative frequencies with expected cumulative frequencies for each observation under the assumption that the data came from the hypothesized distribution.TRUEYes, with each data value considered separately (no grouping into categories).

34.An attraction of the Kolmogorov-Smirnov test is that it is fairly easy to do without a computer.FALSEThe K-S test is done with a computer.

35.An attraction of the Anderson-Darling test is that it is fairly easy to do without a computer.FALSEThe A-D test is done with a computer (it requires an inverse distribution function).

36.ECDF tests have an advantage over the chi-square goodness-of-fit test on frequencies because an ECDF test treats observations individually.TRUEEach data value is considered separately (no grouping into categories) so more power.

37.In an ECDF test for goodness-of-fit, the n observations are grouped into categories rather than being treated individually.FALSEIn ECDF tests, each data value is considered separately (no grouping into categories).

38.When raw data are available, ECDF tests usually surpass the chi-square test in their ability to detect departures from the distribution specified in the null hypothesis.TRUEEach data value is considered separately (no grouping into categories) so more power.

39.The Anderson-Darling test is used to test the assumption of normality.TRUEMost software packages have the A-D normality test because normality tests are popular.

40.Probability plots are used to test the assumption of normality.TRUEMost software packages have the normal P-P because normality tests are popular.

41.In a test for a uniform distribution with k categories, the expected frequency is n/k in each cell.TRUEFor uniformity we expect n/k in each category.

Multiple Choice Questions42.If samples are drawn from a population that is normal, a goodness-of-fit test for normality could yield:

A.Type I error but not Type II error.

B.Type II error but not Type I error.

C.Either Type I error or Type II error.

D.Both Type I and Type II errors.

If the hypothesis (H0: population is normal) is true, we cannot commit Type II error (failing to reject a false hypothesis). But in reality, we would not know that H0 is true.

43.The number of cars waiting at a certain residential neighborhood stop light is observed at 6:00 a.m. on 160 different days. The observed sample frequencies are shown here:

Under the null hypothesis of a uniform distribution, the expected number of days we would see 0 cars is:

A.10.

B.20.

C.30.

D.40.

n/k = 160/4 = 40.

44.A chi-square goodness of fit test for a normal distribution used 40 observations, and the mean and standard deviation were estimated from the sample. The test used six categories. We would use how many degrees of freedom in looking up the critical value for the test?

A.39

B.37

C.5

D.3

d.f. = k - 1 - m = 6 - 1 - 2 = 3 where m = 2 parameters are estimated and k = 6 categories (n is not in the formula).

45.A chi-square goodness of fit test for a normal distribution used 60 observations, and the mean and standard deviation were estimated from the sample. The test used seven categories. We would use how many degrees of freedom in looking up the critical value for the test?

A.6

B.4

C.59

D.57

d.f. = k - 1 - m = 7 - 1 - 2 = 4 where m = 2 parameters are estimated and k = 7 categories (n is not in the formula).

46.Which of these statements concerning a chi-square goodness-of-fit test is correct?

A.Data could be ratio or interval measurements.

B.Population must be normally distributed.

C.All the expected frequencies must be equal.

Distances between data values must be meaningful.

47.Which of the following is not a potential solution to the problem that arises when not all expected frequencies are 5 or more in a chi-square test for independence?

A.Combine some of the columns

B.Combine some of the rows

C.Increase the sample size

D.Add more rows or columns

Subdividing rows or columns would make the expected frequencies smaller.

48.Which of these statements concerning a chi-square goodness-of-fit test is correct?

A.It is inapplicable to test for a normal distribution with open-ended top and bottom classes.

B.It is generally a better test than the chi-square test of independence.

C.There is no way to get the degrees of freedom since the right tail goes to infinity.

D.It can be used to test whether a sample follows a specified distribution.

The GOF test asks whether the sample contradicts a proposed population distribution.

49.A proofreader checked 160 ads for grammatical errors. The sample frequency distribution is shown:

Under the null hypothesis of a uniform distribution, the expected number of times we would get 0 errors is:

A.10.

B.20.

C.30.

D.40.

n/k = 160/4 = 40.

50.A proofreader checked 160 ads for grammatical errors. The sample frequency distribution is shown:

Using a goodness-of-fit test to determine whether this distribution is uniform would result in a chi-square test statistic of approximately:

A.55.

B.79.

C.85.

D.161.

(10 - 40)2/40 + (65 - 40)2/40 + (71 - 40)2/40 + (14 - 40)2/40 = 79.05.

51.A proofreader checked 160 ads for grammatical errors. The distribution obtained is shown:

At = .01, what decision would we reach in a goodness-of-fit test to see whether this sample came from a uniform distribution?

A.Reject the null and conclude the distribution is not uniform.

B.Conclude that there is insufficient evidence to reject the null.

C.No conclusion can be made due to small expected frequencies.

D.No conclusion can be made due to inadequate sample size.

(10 - 40)2/40 + (65 - 40)2/40 + (71 - 40)2/40 + (14 - 40)2/40 = 79.05 > 2.01 = 11.34 for d.f. = 3.

52.A chi-square test of independence is a one-tailed test. The reason is that:

A.we are testing whether the frequencies exceed their expected values.

B.we square the deviations so the test statistic lies at or above zero.

C.hypothesis tests are one-tailed tests when dealing with sample data.

D.the chi-square distribution is positively skewed.

The chi-square test statistic contains (Obs - Exp)2, so differences in either direction are positive.

53.We sometimes combine two row or column categories in a chi-square test when:

A.observed frequencies are more than 5.

B.observed frequencies are less than 5.

C.expected frequencies are more than 5.

D.expected frequencies are less than 5.

Consolidating two rows (or columns) would increase expected frequencies (but fewer d.f.).

54.To determine how well an observed set of frequencies fits an expected set of frequencies from a Poisson distribution we must estimate:

A.no parameters.

B.one parameter ().

C.two parameters (, ).

D.three parameters (, , n).

We lose one extra degree of freedom when we estimate the Poisson mean .

55.The critical value in a chi-square test for independence depends on:

A.the normality of the data.

B.the variance of the data.

C.the number of categories.

D.the expected frequencies.

2 depends on d.f. = k - 1 - m for m estimated parameters and k categories.

56.In a chi-square test of independence, the number of degrees of freedom equals the:

A.number of observations minus one.

B.number of categories minus one.

C.number of rows minus one times the number of columns minus one.

D.number of sample observations minus the missing observations.

2 depends on d.f. = (r - 1)(c - 1).

57.In order to apply the chi-square test of independence, we prefer to have:

A.at least 5 observed frequencies in each cell.

B.at least 5 expected observations in each cell.

C.at least 5 percent of the observations in each cell.

D.not more than 5 observations in each cell.

Larger expected frequencies are desirable (at least 5 according to Cochran's Rule).

58.Kortholts that fail to meet certain precise specifications must be reworked on the next day until they are within the desired specifications. A sample of one day's output of kortholts from the Melodic Kortholt Company showed the following frequencies:

Find the chi-square test statistic for a hypothesis of independence.

A.7.22

B.4.17

C.5.13

D.6.08

The test statistic is 2calc = (Obs - Exp)2/Exp where Exp = [(row sum) (col sum)]/n.

59.Kortholts that fail to meet certain precise specifications must be reworked on the next day until they are within the desired specifications. A sample of one day's output of kortholts from the Melodic Kortholt Company showed the following frequencies:

Find the p-value for the chi-square test statistic for a hypothesis of independence.

A.Less than .01

B.Between .01 and .025

C.Between .025 and .05

D.Greater than .05

2calc = 4.167 with d.f. = 1 is between 2.05 = 3.841 and 2.025 = 5.024. Using Excel, the p-value is =CHISQ.DIST.RT(4.167,1) = .0412.

60.An operations analyst counted the number of arrivals per minute at a bank ATM in each of 30 randomly chosen minutes. The results were: 0, 3, 3, 2, 1, 0, 1, 0, 0, 1, 1, 1, 2, 1, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 0, 1, 0, 1. Which goodness-of-fit test would you recommend?

A.Uniform.

B.Poisson.

C.Normal.

D.Binomial.

Arrivals per unit of time with a small mean would resemble a Poisson distribution.

61.An operations analyst counted the number of arrivals per minute at an ATM in each of 30 randomly chosen minutes. The results were: 0, 3, 3, 2, 1, 0, 1, 0, 0, 1, 1, 1, 2, 1, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 0, 1, 0, 1. For the Poisson goodness-of-fit test, what is the expected frequency of the data value X = 1?

A.Impossible to determine.

B.11.04

C.1.00

D.2.47

The sample mean is 1.00 so n P(X = 1 | = 1.00) = (30)(.3679) = 11.037.

62.The table below is a tabulation of opinions of employees of Axolotl Corporation, who were sampled at random from pay records and asked to complete an anonymous job satisfaction survey.

For a chi-square test of independence, degrees of freedom would be:

A.2

B.3

C.4

D.6

Feedback: d.f. = (2 - 1)(3 - 1) = 2.

63.The table below is a tabulation of opinions of employees of Axolotl Corporation, who were sampled at random from pay records and asked to complete an anonymous job satisfaction survey.

For a chi-square test of independence, the critical value for = .01 is:

A.9.210.

B.4.605.

C.11.34.

D.16.81.

2.01 = 9.210 for d.f. = (2 - 1)(3 - 1) = 2.

64.The table below is a tabulation of opinions of employees of Axolotl Corporation, who were sampled at random from pay records and asked to complete an anonymous job satisfaction survey.

Assuming independence, the expected frequency of satisfied hourly employees is:

A.80.

B.90.

C.75.

D.60.

e21 = (R2)(C1)/n = (180)(120)/240 = 90.

65.To carry out a chi-square goodness-of-fit test for normality you need at least:

A.5 categories altogether.

B.5 observations in each category.

C.5 expected observations in each category.

D.50 samples or more.

Because d.f. = k - 1 - m = k - 3 since m = 2, we need at least k = 4 groups each with e 5.

66.Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.

Under the assumption of independence, the expected frequency in the upper left cell is:

A.15.09.

B.24.00.

C.19.72.

D.20.22.

e11 = (R1)(C1)/n = (46)(42)/128 = 15.09.

67.Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.

For a chi-square test of independence, degrees of freedom would be:

A.2.

B.9.

C.4.

D.127.

d.f. = (3 - 1)(3 - 1) = 4.

68.Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.

For a chi-square test of independence, the critical value for = .05 is:

A.5.991.

B.7.815.

C.9.488.

D.16.92.

2.05 = 9.488 for d.f. = (3 - 1)(3 - 1) = 4.

69.Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.

Assuming independence, the expected frequency of very uncertain students with 60 credits or more is:

A.12.47.

B.2.00

C.14.56.

D.11.09.

e31 = (R3)(C1)/n = (38)(42)/128 = 12.47.

70.Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.

Which statement is most nearly correct?

A.The contingency table violates Cochran's Rule.

B.Visual inspection of column frequencies suggests independence.

C.At = .05 we would easily reject the null hypothesis of independence.

D.At = .05 we cannot reject he null hypothesis of independence.

2calc = 29.528 > 2.05 = 9.488 for d.f. = (3 - 1)(3 - 1) = 4.

71.As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls.

For a chi-square test of independence, degrees of freedom would be:

A.20.

B.12.

C.399.

D.6.

d.f. = (5 - 1)(4 - 1) = 12.

72.As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls.

For a chi-square test of independence, the critical value for = .10 is:

A.10.64.

B.14.68.

C.28.41.

D.18.55.

2.10 = 18.55 for d.f. = (5 - 1)(4 - 1) = 12.

73.As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls.

Assuming independence, the expected frequency of SUVs in Jamestown is:

A.12.

B.21.

C.75.

D.60.

e44 = (R4)(C4)/n = (84)(100)/400 = 21.

74.Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey.

The expected frequency for the shaded cell in the table would be:

A.163.

B.158.

C.165.

D.160.

e22 = (R2)(C2)/n = (400)(320)/800 = 160.

75.Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey.

Degrees of freedom for this test (shaded cell below the table) would be:

A.6.

B.7.

C.799.

D.12.

d.f. = (3 - 1)(4 - 1) = 6.

76.Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey.

The appropriate conclusion would be:

A.do not reject H0.

B.reject H0 at = .10.

C.reject H0 at = .05.

D.reject H0 at = .01.

2calc = 6.206 does not even exceed 2.10 = 10.64 for d.f. = (3 - 1)(4 - 1) = 6.

77.You test a hypothesis of independence of two variables. The number of observations is 500 and you have classified the data into a 4 by 4 contingency table. The test statistic has __________ degrees of freedom.

A.16

B.9

C.499

D.498

Sample size does not enter into the calculation: d.f. = (4 - 1)(4 - 1) = 9.

78.You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods:

Using = .05, what is the critical value of the test statistic that you would use?

A.3.841

B.12.59

C.5.991

D.7.815

2.05 = 5.991 for d.f. = (2 - 1)(3 - 1) = 2.

79.You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods:

The expected frequency for the shaded cell is:

A.22.5.

B.30.

C.40.

D.40.5.

e23 = (R2)(C3)/n = (45)(75)/150 = 22.5.

80.You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods:

What is the value of the test statistic?

A.306.25

B.0.00

C.54.44

D.13.61

2calc = (40 - 22.5)2/22.5 + (30 - 30)2/30 + (5 - 22.5)2/22.5 + (5 - 22.5)2/22.5 + (30 - 30)2/30 + (40 - 22.5)2/22.5 = 54.444.

81.You want to test the hypothesis that the prime rate and inflation are independent. The following table of frequencies is prepared from a random sample, collected in various countries and various time periods:

Based on an analysis of the data in this table, which conclusion can be made at = .01?

A.The prime rate and inflation rate are independent.

B.The prime rate and inflation rate are not independent.

C.Small observed frequencies in some cells suggest that no reliable conclusion can be made.

D.Small expected frequencies in some cells suggest that no reliable conclusion can be made.

2calc = 54.444 greatly exceeds 2.01 = 9.210 for d.f. = (2 - 1)(3 - 1) = 2.

82.You want to sell your house, and you decide to obtain an appraisal on it. Looking at past data you discover that actual prices obtained for houses and the appraisals given for them prior to their sale were as follows:

Based on these data we can say that:

A.no conclusion is possible without knowing .

B.appraisal and actual price are not independent at = .05.

C.appraisal and actual price are independent at any .

D.the degrees of freedom are insufficient for a decision.

Column frequencies are all in the same ratio 3:2 so perfect independence exists (e = f).

83.Preferences for the type of diet drink from a random sample of 121 shoppers are in the table below. A researcher is interested in determining if there is a relationship between the type of diet drink preferred and the age of the shoppers.

In performing a chi-square test of independence on these data, how many degrees of freedom will the test statistic have?

A.1

B.2

C.4

D.6

d.f. = (3 - 1)(2 - 1) = 2.

84.Preferences for the type of diet drink from a random sample of 121 shoppers are in the table below. A researcher is interested in determining if there is a relationship between the type of diet drink preferred and the age of the shoppers.

Using = .025, what is the critical value of the test statistic that you would use in a decision rule to test an appropriate hypothesis?

A.5.02

B.5.99

C.7.38

D.14.45

2.025 = 7.378 for d.f. = (3 - 1)(2 - 1) = 2.

85.Preferences for the type of diet drink from a random sample of 121 shoppers are in the table below. A researcher is interested in determining if there is a relationship between the type of diet drink preferred and the age of the shoppers.

What can you conclude for the data analysis at = .05?

A.The means are equal for all three groups.

B.There is insufficient evidence to conclude that the type of drink and age are dependent.

C.Conclude that the type of drink and age are dependent.

D.No conclusion is possible without knowing the p-value.

2calc = 21.21 greatly exceeds 2.05 = 5.991 for d.f. = (3 - 1)(2 - 1) = 2.

86.A taste test of randomly selected students was conducted to see if there was a difference in preferences among four popular drinks. The following table shows the frequency of responses:

The expected number of students preferring Dr. Pepper is:

A.25.

B.40.

C.50.

D.60.

n = 51 + 66 + 43 + 40 = 200 so, assuming a uniform distribution, e = n/k = 200/4 = 50.

87.A taste test of randomly selected students was conducted to see if there was a difference in preferences among four popular drinks. The following table shows the frequency of responses:

Using = .025, the critical value of the test you would use in determining whether the preferences are the same among the drinks is:

A.5.991.

B.7.378.

C.9.348.

D.11.07.

2.025 = 9.348 for d.f. = k -1 = 4 - 1 = 3.

88.A taste test of randomly selected students was conducted to see if there was a difference in preferences among four popular drinks. The following table shows the frequency of responses:

The value of the chi-square test statistic you would use in testing whether the preferences are the same among the drinks is:

A.7.54.

B.8.12.

C.10.76.

D.D.12.56.

(51 - 50)2/50 + (66 - 50)2/50 + (43 - 50)2/50 + (40 - 50)2/50 = 8.12.

89.A taste test of randomly selected students was conducted to see if there was a difference in preferences among four popular drinks. The following table shows the frequency of responses:

Using = .025, what can you conclude from your analysis?

A.Reject the null and conclude some drinks are preferred more than others.

B.There is not enough evidence to say that a preference exists.

C.Pepsi is the preferred drink.

D.Form no conclusion because Cochran's Rule is violated.

2calc = (51 - 50)2/50 + (66 - 50)2/50 + (43 - 50)2/50 + (40 - 50)2/50 = 8.12 does not exceed 2.025 = 9.348 for d.f. = k -1 = 4 - 1 = 3.

90.The Oxnard Retailers Anti-Theft Alliance (ORATA) published a study that claimed the causes of disappearance of inventory in retail stores were 30 percent shoplifting, 50 percent employee theft, and 20 percent faulty paperwork. The manager of the Melodic Kortholt Outlet performed an audit of the disappearance of 80 items and found the frequencies shown below. She would like to know if her store's experience follows the same pattern as other retailers.

Under the null hypothesis that her store follows the published pattern, the expected number of items that disappeared due to shoplifting is:

A.16.

B.40.

C.24.

D.27.

n = 32 + 38 + 11 = 80 so for shoplifting e = .30 80 = 24.

91.The Oxnard Retailers Anti-Theft Alliance (ORATA) published a study that claimed the causes of disappearance of inventory in retail stores were 30 percent shoplifting, 50 percent employee theft, and 20 percent faulty paperwork. The manager of the Melodic Kortholt Outlet performed an audit of the disappearance of 80 items and found the frequencies shown below. She would like to know if her store's experience follows the same pattern as other retailers.

Using = .05, the critical value you would use in determining whether the Melodic Kortholt's pattern differs from the published study is:

A.7.815.

B.5.991.

C.1.960.

D.1.645.

2.05 = 5.991 for d.f. = k -1 = 3 - 1 = 2.

92.The Oxnard Retailers Anti-Theft Alliance (ORATA) published a study that claimed the causes of disappearance of inventory in retail stores were 30 percent shoplifting, 50 percent employee theft, and 20 percent faulty paperwork. The manager of the Melodic Kortholt Outlet performed an audit of the disappearance of 80 items and found the frequencies shown below. She would like to know if her store's experience follows the same pattern as other retailers.

The value of the chi-square test statistic you would use in testing whether there is a difference from the published pattern is:

A.7.54.

B.5.02.

C.9.76.

D.9.22.

(32 - 24)2/24 + (38 - 40)2/40 + (10 - 16)2/16 = 5.0167 with 1 = .30, 2 = .50, 3 = .20 and n = 80.

93.The Oxnard Retailers Anti-Theft Alliance (ORATA) published a study that claimed the causes of disappearance of inventory in retail stores were 30 percent shoplifting, 50 percent employee theft, and 20 percent faulty paperwork. The manager of the Melodic Kortholt Outlet performed an audit of the disappearance of 80 items and found the frequencies shown below. She would like to know if her store's experience follows the same pattern as other retailers.

Using = .05, what can you conclude from your analysis?

A.The store's pattern is clearly significantly different from the published data.

B.The store's pattern is almost, but not quite, significantly different from the published data.

C.The store's pattern is very close to the published data.

D.We can form no conclusion because Cochran's Rule is violated.

2calc = (32 - 24)2/24 + (38 - 40)2/40 + (10 - 16)2/16 = 5.0167 with 1 = .30, 2 = .50, 3 = .20 and n = 80 and 2.05 = 5.991 for d.f. = k -1 = 3 - 1 = 2, so we cannot quite reject H0: 1 = .30, 2 = .50, 3 = .20.

94.A contingency table shows:

A.frequency counts.

B.means of the data.

C.event probabilities.

D.chi-square values.

Contingency tables contain count data.

95.We would create a contingency table by:

A.summing the probabilities of two variables.

B.cross-tabulating frequencies of two variables.

C.applying the chi-square distribution to a sample.

D.using Cochran's Rule to estimate frequencies.

A contingency table is a two-way frequency distribution.

96.Which data set is consistent with the hypothesis of a normal population?

A.Data Set A.

B.Data Set C.

C.Neither data set.

D.Both data sets.

Data Set A has a linear probability plot. Data Set C is nonlinear and has a small Anderson-Darling p-value, which suggests rejection of the hypothesis of normality.

97.Which statement is most nearly correct regarding ECDF tests?

A.An attraction of the Anderson-Darling test is that it is fairly easy to do without a computer.

B.In an ECDF test for goodness-of-fit test, the n observations are grouped into categories rather than being treated individually.

C.When raw data are available, ECDF tests usually surpass the chi-square test in their ability to detect departures from the distribution specified in the null hypothesis.

ECDF tests (e.g., Anderson-Darling, Kolmogorov-Smirnov, probability plot) generally gain power by considering each data point separately. However, these tests are not easy without a computer.

Short Answer Questions98.Based on some ideas expressed in his psychology class, John decided to test a hypothesis about the possible relationship between parent dominance and political views. He used a survey of 189 statistics students to prepare the cross-tabulation and chi-square analysis shown below. Discuss John's results using concepts you learned in this chapter. Note any potential problems or concerns in the analysis, and suggest possible improvements.

Most cells have observed frequencies that are quite close to what would be expected under the hypothesis of independence. The chi-square test statistic (5.51) is nowhere near the critical value (15.51) for = .05. The p-value (.7014) says that such a sample could happen by chance about 70 times in 100 samples if the two variables were actually independent, so the data do not permit rejection of the hypothesis of independence. The lower left cell has a slightly small expected frequency (4.86), but the other expected frequencies are all at least 5 (Cochran's Rule) so a larger sample seems unlikely to change the conclusion.

Feedback: Most cells have observed frequencies that are quite close to what would be expected under the hypothesis of independence. The chi-square test statistic (5.51) is nowhere near the critical value (15.51) for = .05. The p-value (.7014) says that such a sample could happen by chance about 70 times in 100 samples if the two variables were actually independent, so the data do not permit rejection of the hypothesis of independence. The lower left cell has a slightly small expected frequency (4.86), but the other expected frequencies are all at least 5 (Cochran's Rule) so a larger sample seems unlikely to change the conclusion.

99.Based on some ideas expressed in her psychology class, Frieda decided to test a hypothesis about the possible relationship between political views and the number of traffic tickets received. She used a survey of 189 statistics students to prepare the cross-tabulation and chi-square analysis shown below. Discuss Frieda's results using concepts you learned in this chapter. Note any potential problems or concerns in the analysis, and suggest possible improvements.

Except in the first column, most of the cells have observed frequencies that are close to what would be expected under the hypothesis of independence. The chi-square test statistic (5.70) is well below the critical value (9.488) for = .05. The p-value (.2224) says that such a sample could happen by chance about 22 times in 100 samples if the two variables were actually independent, so the data do not permit rejection of the hypothesis of independence at the usual levels of significance. The lower left cell has a small expected frequency (3.26), as does the lower right cell (4.00). The other expected frequencies are all at least 5 (Cochran's Rule). The sample is fairly large, but if Frieda wants to increase the expected frequencies, she might take a larger sample.

Feedback: Except in the first column, most of the cells have observed frequencies that are close to what would be expected under the hypothesis of independence. The chi-square test statistic (5.70) is well below the critical value (9.488) for = .05. The p-value (.2224) says that such a sample could happen by chance about 22 times in 100 samples if the two variables were actually independent, so the data do not permit rejection of the hypothesis of independence at the usual levels of significance. The lower left cell has a small expected frequency (3.26), as does the lower right cell (4.00). The other expected frequencies are all at least 5 (Cochran's Rule). The sample is fairly large, but if Frieda wants to increase the expected frequencies, she might take a larger sample.