15: CHI SQUARED TESTS

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1 15: CHI SQUARED ESS MULIPLE CHOICE QUESIONS In the following multiple choice questions, please circle the correct answer. 1. Which statistical technique is appropriate when we describe a single population of qualitative data with exactly two categories? a. z test of a population proportion b. he chi squared test of a multinomial experiment c. he chi squared test of a contingency table d. Both a and b e. Both b and c d. If we want to conduct a two tail test of a population proportion, we can employ: a. z test of a population proportion b. the chi squared test of a binomial experiment since z = χ c. the chi squared test of a contingency table d. Both a and b e. Both b and c d 3. Which statistical technique is appropriate when we wish to analyze the relationship between two qualitative variables with two or more categories? a. he chi squared test of a multinomial experiment b. he chi squared test of a contingency table c. he t test of the difference between two means d. Both a and b e. Both b and c b 4. If we want to conduct a one tail test of a population proportion, we can employ: a. z test of a population proportion b. the chi squared test of a binomial experiment since z = χ c. the chi squared test of a contingency table d. Both a and b e. Both b and c a 5. In a goodness of fit test, suppose that the value of the test statistic is and df = 6. At the 5% significance level, the null hypothesis is

2 a. rejected and p value for the test is smaller than 0.05 b. not rejected, and p value for the test is greater than 0.05 c. rejected, and p value for the test is greater than 0.05 d. not rejected, and p value for the test is smaller than 0.05 a 6. In a goodness of fit test, suppose that a sample showed that the observed frequency f i and expected frequency ei were equal for each cell i. hen, the null hypothesis is a. rejected at α = 0.05 but is not rejected at α = 0.05 b. not rejected at α = 0.05 but is rejected at α = 0.05 c. rejected at any level α d. not rejected at any α level d 7. he president of a state university collected data from students concerning building a new library, and classified the responses into different categories (strongly agree, agree, undecided, disagree, strongly disagree) and according to whether the student was male or female. o determine whether the data provide sufficient evidence to indicate that the responses depend upon gender, the most appropriate test is: a. chi squared goodness of fit test b. chi squared test of a contingency table (test of independence) c. chi squared test of normality d. chi squared test for comparing five proportions b 8. he number of degrees of freedom for a contingency table with 6 rows and 6 columns is a. 36 b. 5 c. 1 d. 6 b 9. A chi squared test of a contingency table with 4 rows and 5 columns shows that the value of the test statistic is.18. he most accurate statement that can be made about the p value for this test is that a. p value is greater than 0.05 b. p value is smaller than 0.05 c. p value is greater than 0.05 but smaller than 0.05 d. p value is greater than 0.10 c 10. he number of degrees of freedom for a contingency table with 4 rows and 8 columns is a. 3 b. 8 c. 4 d. 1 d

3 11. he chi squared distribution is used: a. in a goodness of fit test b. in a test of a contingency table c. in making inferences about a single population variance d. All of the above answers are correct d 1. o determine whether a single coin is fair, the coin was tossed 100 times, and head was observed 60 times. he value of the test statistic is a. 40 b. 4 c. 60 d. 6 b 13. o determine whether data were drawn from any distribution, we use a. a chi squared goodness of fit test b. a chi squared test of a contingency table c. a chi square test for normality d. None of the above answers is correct a 14. A chi squared goodness of fit test is always conducted as: a. a lower tail test b. an upper tail test c. a two tail test d. Either a or b b 15. Contingency tables are used in: a. testing independence of two samples b. testing dependence in matched pairs c. testing independence of two qualitative variables in a population d. describing a single population c 16. he number of degrees of freedom in testing for normality is the a. number of intervals used to test the hypothesis minus 1 b. number of parameters estimated minus 1 c. number of intervals used to test the hypothesis minus number of parameters estimated minus 1 d. number of intervals used to test the hypothesis minus number of parameters estimated minus c 17. If the expected frequency ei for any cell i is less than 5, we a. must choose another sample of five or more observations

4 b. should use the normal distribution instead of the chi squared distribution c. should combine the cells such that each observed frequency f i is 5 or more d. increase the number of degrees of freedom for the test by 5 c 18. If each element in a population is classified into one and only one of several categories, the population is a: a. normal population b. multinomial population c. chi squared population d. binomial population b 19. o determine the critical values in the chi squared distribution table, the process requires the following: a. degrees of freedom b. probability of ype I error c. probability of ype II error d. Both a and b d 0. Of the values for a chi squared test statistic listed below, which one is likely to lead to rejecting the null hypothesis in a goodness of fit test? a. 0 b. 1 c. d. 40 d 1. he number of degrees of freedom in a test of a contingency table with 4 rows and 3 columns equals: a. 4 b. 7 c. 6 d. 3 c. Which statistical technique is appropriate when we describe a single population of qualitative data with two or more categories? a. z test of the difference between two proportions b. he chi squared test of a multinomial experiment c. he chi squared test of a contingency table d. Both a and b e. Both b and c b 3. he sampling distribution of the test statistic for a goodness of fit test with k categories is: a. Student t distribution with k 1 degrees of freedom

5 b. normal distribution c. chi squared distribution with k 1 degrees of freedom d. approximately chi squared distribution with k 1 degrees of freedom d 4. he chi squared test of a contingency table is based upon: a. two qualitative variables b. two quantitative variables c. three or more qualitative variables d. three or more quantitative variables a 5. Which of the following statements is not correct? a. he chi squared distribution is symmetrical b. he chi squared distribution is skewed to the right c. All values of the chi squared distribution are positive d. he critical region for a goodness of fit test with k categories is χ > χα,k 1 a 6. Which statistical technique is appropriate when we compare two populations of qualitative data with exactly two categories? a. z test of a population proportion b. z test of the difference between two proportions c. he chi squared test of a contingency table d. Both a and b e. Both b and c e Which of the following statements is not correct? a. he chi squared test of independence is a one sample test b. Both variables in the chi squared test of independence are qualitative variables c. he chi squared goodness of fit test involves two categorical variables d. he chi squared distribution is skewed to the right c A left tail area in the chi squared distribution equals or df = 8, the table value equals: a b c..736 d a 9. Which of the following statements is true for the chi squared tests? a. esting for equal proportions is identical to testing for goodness of fit b. he number of degrees of freedom in a test of a contingency table with r rows and c columns is (r 1)(c 1). c. he number of degrees of freedom in a goodness of fit test with k categories is k 1 d. All of the above statements are true.

6 d 30. he degrees of freedom in a chi squared test for normality, where the number of standardized intervals is 5 and there are population parameters to be estimated from the data, is equal to: a. 5 b. 4 c. 3 d. d 31. A chi squared test for independence with 6 degrees of freedom results in a test statistic χ = Using the χ tables, the most accurate statement that can be made about the p value for this test is that: a. p value > 0.10 b. p value > 0.05 c < p value < 0.10 d < p value < 0.05 d 3. In a goodness of fit test, the null hypothesis states that the data came from a normally distributed population. he researcher estimated the population mean and population standard deviation from a sample of 500 observations. In addition, the researcher used 6 standardized intervals to test for normality. Using a 5% level of significance, the critical value for this test is: a b c d c 33. In a chi squared test of a contingency table, the value of the test statistic was χ =1.678, and the critical value at α = 0.05 was hus, a. we fail to reject the null hypothesis at α = 0.05 b. we reject the null hypothesis at α = 0.05 c. we don t have enough evidence to accept or reject the null hypothesis at α = 0.05 d. we should decrease the level of significance in order to reject the null hypothesis a 34. Which statistical technique is appropriate when we compare two or more populations of qualitative data with two or more categories? a. z test of the difference between two proportions b. he chi squared test of a multinomial experiment c. he chi squared test of a contingency table d. Both a and b e. Both b and c c 35. Which of the following tests does not use the chi squared distribution?

7 a. est of a contingency table b. Goodness of fit test c. Difference between two population means test d. All of the above tests use the chi squared distribution c 36. Which statistical technique is appropriate when we compare two populations of qualitative data with two or more categories? a. z test of the difference between two proportions b. he chi squared test of a multinomial experiment c. he chi squared test of a contingency table d. Both a and b e. Both b and c c 37. Which of the following is not a characteristic of a multinomial experiment? a. he experiment consists of a fixed number, n, of trials. b. he outcome of each trial can be classified into one of two categories called successes and failure. c. he probability pi that the outcome will fall into cell i remain constant for each trial. d. Each trial of the experiment is independent of the other trials b 38. In a chi squared goodness of fit test, if the expected frequencies ei and the observed frequencies f i were quite different, we would conclude that: a. the null hypothesis is false, and we would reject it b. the null hypothesis is true, and we would not reject it c. the alternative hypothesis is false, and we would reject it d. the chi squared distribution is invalid, and we would use the t distribution instead a 39. In chi squared tests, the conventional and conservative rule known as the rule of five is to require that the: a. observed frequency for each cell be at least five b. degrees of freedom for the test be at least five c. expected frequency for each cell be at least five d. difference between the observed and expected frequency for each cell be at least five c 40. Consider a multinomial experiment with 00 trials, and the outcome of each trial can be classified into one of 5 categories. he number of degrees of freedom associated with the chi squared goodness of fit test equals: a. 195 b. 05 c. 199 d. 4 d

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9 RUE/ALSE QUESIONS 41. he null hypothesis states that the sample data came from a normally distributed population. he researcher calculates the sample mean and the sample standard deviation from the data. he data arrangement consisted of five categories. Using a 0.05 significance level, the appropriate critical value for this chi square test for normality is A test for independence is applied to a contingency table with 3 rows and 4 columns for two qualitative variables. he degrees of freedom for this chi square test must equal A chi square test for independence with 6 degrees of freedom results in a test statistic of Using the chi square table, the most accurate statement that can be made about the p value for this test is that p value is greater than 0.05 but smaller than Whenever the expected frequency of a cell is less than 5, one remedy for this condition is to increase the significance level. 45. In testing a population mean or constructing a confidence interval for the population mean, an essential assumption is that expected frequencies are at least five. 46. A right tailed area in the chi square distribution equals or 6 degrees of freedom the table value equals Whenever the expected frequency of a cell is less than 5, one remedy for this condition is to increase the size of the sample. 48. A left tailed area in the chi square distribution equals or 5 degrees of freedom the table value equals or a chi square distributed random variable with 10 degrees of freedom and a level of significance of 0.05, the chi square value from the table is he computed value of the test statistics is his will lead us to reject the null hypothesis. 50. A test for independence is applied to a contingency table with 4 rows and 4 columns for two qualitative variables. he degrees of freedom for this test will be 9.

10 51. A chi square test for independence with 10 degrees of freedom results in a test statistic of Using the chi square table, the most accurate statement that can be made about the p value for this test is that 0.05 < p value < Whenever the expected frequency of a cell is less than 5, one remedy for this condition is to decrease the size of the sample. 53. he middle 0.95 portion of the chi square distribution with 9 degrees of freedom has table values of and , respectively. 54. In applying the chi square goodness of fit test, the rule of thumb for all expected frequencies is that each expected frequency equal or exceeds In a chi squared test of independence, the value of the test statistic was χ = 15.65, and the critical value at α = 0.05 was hus, we must reject the null hypothesis at α = he chi squared test of independence is based upon three or more quantitative variables. 57. In a goodness of fit test, the null hypothesis states that the data came from a normally distributed population. he researcher estimated the population mean and population standard deviation from a sample of 300 observations. In addition, the researcher used 6 standardized intervals to test for normality. Using a.5% level of significance, the critical value for this test is A chi square goodness of fit test is always conducted as a two tail test. 59. A left tailed area in the chi square distribution equals or 10 degrees of freedom the table value equals or a chi square distributed random variable with 1 degrees of freedom and a level of significance of 0.05, the chi square value from the table is he computed value of the test statistics is his will lead us to reject the null hypothesis. 61. In a goodness of fit test, the null hypothesis states that the data came from a normally distributed population. he researcher estimated the population mean and population standard deviation

11 from a sample of 00 observations. In addition, the researcher used 5 standardized intervals to test for normality. Using a 10% level of significance, the critical value for this test is In chi square tests, the conventional and conservative rule known as the Rule of ive is to require that difference between the observed and expected frequency for each cell be at least five. 63. Whenever the expected frequency of a cell is less than 5, one remedy for this condition is to decrease the significance level. 64. he area to the right of a chi square value is or 8 degrees of freedom, the table value is A multinomial experiment, where the outcome of each trial can be classified into one of two categories, is identical to the binomial experiment. 66. he chi squared goodness of fit test is usually used as a test of multinomial parameters, but it can also be used to determine whether data were drawn from any distribution. 67. he chi squared test of a contingency table is used to determine if there is enough evidence to infer that two nominal variables are related, and to infer that differences exist among two or more populations of nominal variables. 68. he number of degrees of freedom for a contingency table with r rows and c columns is ν = rc, provided that both r and c are greater than or equal to. 69. When the problem objective is to describe a population of nominal data with exactly two categories, we can employ either the z test of population proportion p, or the chi squared goodness of fit test. 70. If we want to perform a one tail test of a population proportion p, we can employ either the z test of p, or the chi squared goodness of fit test. 71. If we want to perform a two tail test of a population proportion p, we must employ the z test of p.

12 7. If we want to test for differences between two populations of nominal data with exactly two categories, we can employ either the z test of p1 p, or the chi squared test of a contingency table (squaring the value of the z statistic yields the value of χ statistics). 73. If we want to perform a two tail test for differences between two populations of nominal data with exactly two categories, we can employ either the z test of p1 p, or the chi squared test of a contingency table (squaring the value of the z statistic yields the value of χ statistics). 74. If we want to perform a one tail test for differences between two populations of nominal data with exactly two categories, we must employ the z test of p1 p. 75. he number of degrees of freedom associated with the chi squared test for normality is the number of intervals used minus the number of parameters estimated from the data.

13 ES QUESIONS 76. he following data are believed to have come from a normal probability distribution he mean of this sample equals 6.80, and the standard deviation equals Use the goodness of fit test at the 5% significance level to test this claim. H 0 : he population has a normal probability distribution H 1 : he population does not have a normal probability distribution Since the sample size is less than 80, we employ the minimum number of intervals 4. Selected Intervals X X X X > Standardized Intervals Z 1 1 Z 0 Probability pi Observed requency fi Expected requency ei ( f i ei ) ei Z 1 Z>1 Rejection region: χ > χ 0.05,1 = est statistic: χ = p value = Conclusion: Reject the null hypothesis; concluding that the population does not have a normal probability distribution 77. Conduct a test to determine whether the two classifications A and B are independent, using the data in the accompanying table and α = 0.05 A1 A B B 5 0 H 0 : he two variables are independent H 1 : he two variables are dependent Rejection region: χ > χ 0.05, = est statistic: χ =.11, and p value = Conclusion: Don t reject the null hypothesis. Yes B3 0 5

14 78. he personnel manager of a consumer product company asked a random sample of employees how they felt about the work they were doing. he following table gives a breakdown of their responses by gender. Do the data provide sufficient evidence to conclude that the level of job satisfaction is related to gender? Use α = Response Gender Male emale Very Interesting airly Interesting Not Interesting 9 11 H 0 : he two variables are independent H 1 : he two variables are dependent Rejection region: χ > χ 0.10, = est statistic: χ = p value = Conclusion: Reject the null hypothesis. Yes 79. he personnel manager of a consumer products company asked a random sample of employees how they felt about the work they were doing. he following table gives a breakdown of their responses by age. Is there sufficient evidence to conclude that the level of job satisfaction is related to age? Use α = Response Age Under 30 Between 30 and 50 Over 50 Very Interesting airly Interesting Not Interesting H 0 : he two variables are independent H 1 : he two variables are dependent Rejection region: χ > χ 0.10,4 = est statistic: χ = 9.69 p value = Conclusion: Reject the null hypothesis. Yes 80. A firm has been accused of engaging in prejudicial hiring practices. According to the most recent census, the percentages of whites, blacks, and Hispanics in a certain community are 7%, 10%,

15 and 18%, respectively. A random sample of 00 employees of the firm revealed that 165 were white, 14 were black, and 1 were Hispanic. Do the data provide sufficient evidence to conclude at the 5% level of significance that the firm has been engaged in prejudicial hiring practices? H 0 : p1 = 0.7, p = 0.10, p3 = 0.18 Rejection region: χ > χ 0.05, = est statistic: χ = p value = Conclusion: Reject the null hypothesis. Yes 81. ive brands of orange juice are displayed side by side in several supermarkets in a large city. It was noted that in one day, 180 customers purchased orange juice. Of these, 30 picked Brand A, 40 picked Brand B, 5 picked Brand C, 35 picked Brand D, and 50 picked brand E. In this city, can you conclude at the 5% significance level that there is a preferred brand of orange juice? H 0 : p1 = p = p3 = p 4 = p 5 Rejection region: χ > χ 0.05,4 = est statistic: χ = p value = Conclusion: Reject the null hypothesis. Yes 8. A sport preference poll showed the following data for men and women: avorite Sport Gender Male emale Baseball 4 1 Basketball 17 0 ootball 30 Golf 18 1 ennis 8 Use the 5% level of significance and test to determine whether sport preferences depend on gender. H 0 : he two variables (gender and favorite sport) are independent H 1 : he two variables are dependent

16 Rejection region: χ > χ 0.05,4 = est statistic: χ = 3.30 p value = Conclusion: Don t reject the null hypothesis. No 83. Last year, Brand A microwaves had 45% of the market, Brand B had 35%, and Brand C had 0%. his year the makers of brand C launched a heavy advertising campaign. A random sample of appliance stores shows that of 10,000 microwaves sold, 4350 were Brand A, 3450 were Brand B, and 00 were Brand C. Has the market changed? est at α = H 0 : p1 = 0.45, p = 0.35, p3 = 0.0 Rejection region: χ > χ 0.01, = 9.10 est statistic: χ = p value = 0.0 Conclusion: Reject the null hypothesis. Yes 84. A study of educational levels of 500 voters and their political party affiliations in a particular state in the USA showed the following results: Party Affiliation Educational level Didn t Complete High School High School Diploma Has College Degree Democrat Republican Independent Use the 1% level of significance and test to see if party affiliation is independent of the educational level of the voters. H 0 : he two variables (educational level and political party affiliation) are independent H 1 : he two variables are dependent Rejection region: χ > χ 0.01,4 = est statistic: χ = p value = 0.0 Conclusion: Reject the null hypothesis. No 85. Consider a multinomial experiment involving 100 trials and 3 categories (cells). he observed frequencies resulting from the experiment are shown in the accompanying table. Category requency

17 Use the 5% significance level to test the hypotheses H 0 : p1 = 0.45, p = 0.30, p3 = 0.5 Rejection region: χ > χ 0.05, = est statistic: χ =.08 p value = Conclusion: Don t reject the null hypothesis. 86. In 000, the student body of a state university in Michigan consists of 30% freshmen, 5% sophomores, 7% juniors, and 18% seniors. A sample of 400 students taken from the 001 student body showed that there are 138 freshmen, 88 sophomores, 94 juniors, and 80 seniors. est with 5% significance level to determine whether the student body proportions have changed. H 0 : p1 = 0.30, p = 0.5, p3 = 0.7, p4 = 0.18 Rejection region: χ > χ 0.05,3 = est statistic: χ = p value = Conclusion: Don t reject the null hypothesis. No 87. Consider a multinomial experiment involving 160 trials 4 categories (cells). he observed frequencies resulting from the experiment are shown in the accompanying table. Category requency Use the 10% significance level to test the hypotheses H 0 : p1 = p = p3 = p 4 Rejection region: χ > χ 0.10,3 = 6.51 est statistic: χ = p value = Conclusion: Reject the null hypothesis. No 88. A statistics professor posted the following grade distribution guidelines for his elementary statistics class: 8% A, 35% B, 40% C, 1% D, and 5%. A sample of 100 elementary statistics grades at the end of last semester showed 1 As, 30 Bs, 35 Cs, 15 Ds, and 8 s. est at the 5%

18 significance level to determine whether the actual grades deviate significantly from the posted grade distribution guidelines. H 0 : p1 = 0.08, p = 0.35, p3 = 0.40, p4 = 0.1, p4 = 0.05 Rejection region: χ > χ 0.05,4 = est statistic: χ = p value = Conclusion: Don t reject the null hypothesis. No 89. Conduct a test to determine whether the two classifications A and B are independent, using the data in the accompanying table and α =.01 A1 A B1 4 3 B 8 57 H 0 : he two variables are independent H 1 : he two variables are dependent Rejection region: χ > χ 0.01,1 = est statistic: χ = p value = Conclusion: Reject the null hypothesis. No 90. Explain what is meant by the rule of five. he rule of five requires that the expected frequency for each cell be at least 5. Where necessary, cells should be combined in order to satisfy this condition. he choice of cells to be combined should be made in such a way that meaningful categories (cells) result from the combination. 91. In 1996, computers of Brand A controlled 5% of the market, Brand B 0%, Brand C 10%, and brand D 45%. In 000, sample data was collected from many randomly selected stores throughout the country. Of the 100 computers sold, 80 were Brand A, 70 were Brand B, 90 were Brand C, and 560 were Brand D. Has the market changed since 1996? est at the 1% significance level. H 0 : p1 = 0.5, p = 0.0, p3 = 0.10, p4 = 0.45 Rejection region: χ > χ 0.01,3 = est statistic: χ = 13.34

19 p value = Conclusion: Reject the null hypothesis. Yes 9. A major insurance firm interviewed a random sample of 100 college students to find out the type of life insurance preferred, if any. he results follow: Insurance Preference Gender emale Male erm Whole Life No Insurance Is there evidence that life insurance preference of male students is different than that of female students? est using the 5% level of significance. H 0 : he two variables (gender and insurance preference) are independent H 1 : he two variables are dependent Rejection region: χ > χ 0.05, = est statistic: χ = p value = Conclusion: Reject the null hypothesis. Yes 93. he number of cars sold by three salespersons over a 3 month period are shown below: Brand of Car Salesperson David Edward rank Brand A Brand B 4 5 Brand C Use the 5% level of significance and test for the independence of salesperson and type of product sold H 0 : he two variables (salesperson and brand of car) are independent H 1 : he two variables are dependent Rejection region: χ > χ 0.05,4 = est statistic: χ =.66 p value = Conclusion: Don t reject the null hypothesis. Yes 94. A telephone company prepared four versions of a set of instructions for placing collect calls. he company asked a sample of 1600 people which one of the four forms was easiest to understand. In the sample, 45 people preferred orm A, 385 preferred orm B, 375 preferred orm C, and

20 415 preferred orm D. At the 5% level of significance, can one conclude that in the population there is a preferred form? H 0 : p1 = p = p3 = p 4 Rejection region: χ > χ 0.05,3 = est statistic: χ = 4.5 p value = Conclusion: Don t reject the null hypothesis. No 95. Suppose that a random sample of 60 observations was drawn from a population. After calculating the mean and standard deviation, each observation was standardized and the number of observations in each of the intervals below was counted. Can we infer at the 10% significance level that the data were drawn from a normal population? Intervals Z 1 1 < Z 0 0<Z 1 Z>1 requency H 0 : p1 = , p = , p3 = , p4 = (he population is normal) (he population is not normal) Rejection region: χ > χ 0.10,1 =.7055 est statistic: χ = p value = Conclusion: Reject the null hypothesis. No 96. Suppose that a random sample of 150 observations was drawn from a population. After calculating the mean and standard deviation, each observation was standardized and the number of observations in each of the intervals below was counted. Can we infer at the 5% significance level that the data were drawn from a normal population? Intervals Z < Z.5.5 Z.5.5 < Z 1.5 Z > 1.5 requency

21 H 0 : p1 = , p = 0.417, p3 = , p4 = 0.417, p5 = (he population is normal) (he population is not normal) Rejection region: χ > χ 0.05, = est statistic: χ = p value = Conclusion: Reject the null hypothesis. No 97. he president of a large university has been studying the relationship between male/female supervisory structures in his institution and the level of employees job satisfaction. he results of a recent survey are shown in the table below. Conduct a test at the 5% significance level to determine whether the level of job satisfaction depends on the boss/employee gender relationship. Boss/Employee Level of Satisfaction Satisfied Neutral Dissatisfied Male/ emale emale/male Male/Male emale/emale H 0 : he two variables are independent H 1 : he two variables are dependent Rejection region: χ > χ 0.05,6 = 1.59 est statistic: χ = p value = 0.0 Conclusion: Reject the null hypothesis. Yes 98. Consumer panel preferences for three proposed fast food restaurants are as follows: Restaurant A 48 Restaurant B 6 Restaurant C 40 Use 0.05 level of significance and test to see if there is a preference among the three restaurants. H 0 : p1 = p = p3 Rejection region: χ > χ 0.05, = est statistic: χ = 4.96

22 p value = Conclusion: Don t reject the null hypothesis. No 99. A cafeteria proposes to serve 4 main entrees. or planning purposes, the manager expects that the proportions of each that will be selected by his customers will be: Selection Chicken Roast Beef Steak ish Proportion Of the first 100 customers, 44 selected chicken, 4 selected roast beef, 13 select steak, and 10 selected fish. Should the manager revise his estimates? Use α = H 0 : p1 = 0.50, p = 0.0, p3 = 0.10, p4 = 0.0 Rejection region: χ > χ 0.01,3 = est statistic: χ = 7.64 p value = Conclusion: Don t reject the null hypothesis. No 100. A large carpet store wishes to determine if the brand of carpet purchased is related to the purchaser s family income. As a sampling frame, they mailed a survey to people who have a store credit card. ive hundred customers returned the survey and the results follow: Brand of Carpet amily Income High Income Middle Income Low Income Brand A Brand B Brand C At the 5% level of significance, can you conclude that the brand of carpet purchased is related to the purchaser s family income? H 0 : he two variables (family income and brand of carpet) are independent

23 H 1 : he two variables are dependent Rejection region: χ > χ 0.05,4 = est statistic: χ = 7.37 p value = 0.0 Conclusion: Reject the null hypothesis. Yes 101. o determine whether a single coin is fair, the coin was tossed 00 times. he observed frequencies with which each of the two sides of the coin turned up are recorded as 11 heads and 88 tails. Is there sufficient evidence at the 5% significance level to allow you to conclude that the coin is not fair? H 0 : p1 = 0.50, p = 0.50 (the coin is fair) H 1 : At least one pi is not equal to its specified value (the coin is not fair) Rejection region: χ > χ 0.05,1 = est statistic: χ =.88 p value = Conclusion: Don t reject the null hypothesis. Yes QUESIONS 10 HROUGH 105 ARE BASED ON HE OLLOWING INORMAION: Consider a multinomial experiment involving n = 00 trials and k = 5 cells. he observed frequencies resulting from the experiment are shown in the following table: Cell requency he null hypothesis to be tested is as follows. H 0 : p1 = 0.10, p = 0.5, p3 = 0.30, p4 = 0.0, p5 = 0.15 p5 = est the hypothesis at the 5% level of significance Rejection region: χ > χ 0.05,4 = est statistic: χ = 4.587

24 p value = Conclusion: Don t reject the null hypothesis Re do Question 10 with the following frequencies: Cell requency Rejection region: χ > χ 0.05,4 = est statistic: χ =.93 p value = Conclusion: Don t reject the null hypothesis Re do Question 10 with the following frequencies: Cell requency Rejection region: χ > χ 0.05,4 = est statistic: χ = p value = Conclusion: Don t reject the null hypothesis Review the results of Questions 10 to 104. What is the effect of decreasing the sample size? As the sample size decreased by 50%, the value of the test statistic also decreased by 50% and the p value became much larger. QUESIONS 106 HROUGH 109 ARE BASED ON HE OLLOWING INORMAION: Consider the data in the accompanying table with classifications A and B: A1 A 106. B B Conduct a test to determine whether the two classifications A and B are independent, using α = 0.05 H 0 : he two variables are independent

25 H 1 : he two variables are dependent Rejection region: χ > χ 0.05,1 = est statistic: χ = 3.93 p value = Conclusion: Reject the null hypothesis. No 107. Re do Question 106 using the following table: A1 A B1 0 8 B 30 4 H 0 : he two variables are independent H 1 : he two variables are dependent Rejection region: χ > χ 0.05,1 = est statistic: χ = 1.96 p value = Conclusion: Don t reject the null hypothesis. Yes 108. Re do Question 106 using the following table: A1 A B B 15 1 H 0 : he two variables are independent H 1 : he two variables are dependent Rejection region: χ > χ 0.05,1 = est statistic: χ = p value = 0.3 Conclusion: Don t reject the null hypothesis. Yes 109. Review the results of Questions 106 to 108. What is the effect of decreasing the sample size? As the sample size decreased by 50%, the value of the test statistic also decreased by 50 % and the p value became much larger. QUESIONS 110 HROUGH 114 ARE BASED ON HE OLLOWING INORMAION:

26 A biology professor claimed that the proportions of grades in his classes are the same. A sample of 100 students showed the following frequencies: Grade requency 110. A 18 B 0 C 8 D 3 11 State the null and alternative hypotheses to be tested. H 0 : p1 = 0.0, p = 0.0, p3 = 0.0, p4 = 0.0, p5 = Determine the rejection region at the 5% significance level. Rejection region: χ > χ 0.05,4 = Compute the value of the test statistics. est statistic: χ = Use statistical software to compute the p value for this test. p value = Do the data provide enough evidence to support the professor s claim? Don t reject the null hypothesis. Yes, the data provide enough evidence to support the professor s claim. QUESIONS 115 AND 116 ARE BASED ON HE OLLOWING INORMAION: A salesperson makes five calls per day. A sample of 00 days gives the frequencies of sales volumes listed below Number of Sales Observed requency (days)

27 Assume the population is binomial distribution with a probability of purchase p equal to Compute the expected frequencies for x = 0, 1,, 3, 4, and 5 by using the binomial probability function or the binomial tables. Combine categories if necessary to satisfy the rule of five. x fi pi ei Should the assumption of a binomial distribution be rejected at the 5% significance level? H 0 : he population has a binomial probability distribution H 1 : he population does not have a binomial probability distribution Rejection region: χ > χ 0.05,5 = est statistic: χ = p value = Conclusion: Reject the null hypothesis. Yes QUESIONS 117 HROUGH 10 ARE BASED ON HE OLLOWING INORMAION: A tire manufacturer operates a plant in New York and another plant in New Jersey. Employees at each plant have been evenly divided among three issues (wages, working conditions, and pension benefits) in terms of which one they feel should be the primary issue in the upcoming contract negotiations. he president of the union has recently circulated pamphlets among the employees, attempting to convince them that pension benefits should be the primary issue. A subsequent survey revealed the following breakdown of the employees according to the plant at which they worked and the issue that they felt should be supported as the primary one. Issues Plant Location New York New Jersey 117. Very Interesting airly Interesting 6 56 Not Interesting Can you infer at the 5% significance level that the proportional support by the employees at both plants for the issues has changed since the pamphlet was circulated?

28 H 0 : p1 = p = p3 Rejection region: χ > χ 0.05, = est statistic: χ = p value = Conclusion: Don t reject the null hypothesis. No 118. Can you infer at the 5% significance level that the proportional support by the New York employees for the three issues has changed since the pamphlet was circulated? H 0 : p1 = p = p3 Rejection region: χ > χ 0.05, = est statistic: χ =.9 p value = 0.3 Conclusion: Don t reject the null hypothesis. No 119. Can you infer at the 5% significance level that the proportional support by the New Jersey employees for the three issues has changed since the pamphlet was circulated? H 0 : p1 = p = p3 Rejection region: χ > χ 0.05, = est statistic: χ =.68 p value = Conclusion: Don t reject the null hypothesis. No 10. Do the data indicate at the 5% significant level that there are differences between the two plants regarding which issue should be the primary one? H 0 : he two variables (plant location and issues) are independent H 1 : he two variables are dependent Rejection region: χ > χ 0.05, = est statistic: χ = 1.18 p value = Conclusion: Don t reject the null hypothesis. No

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