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Ch 12_Black-1 - Business Statistics Fifth Edition Ken Black...

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Unformatted text preview: Business Statistics Fifth Edition Ken Black Chapter 12: Analysis of Categorical Data Not to be reproduced without the written permission of RB Alt 12.1 CHI—SQUARE GOODNESS—OF—FIT TEST [PAGE 468] 0 Chi—Squared tests use Count Data 0 Goodness—of-Fit Test for k Proportions 0 Concept: Are the proportions for k mutually exclusive and completely exhaustive categories equal to specified values? 0 Case 1: The population proportions are equal [Page 470-471] Examgle: The Hershey Company wants to determine if customers have a preference for any of the following three candy bars: A. Mr. Goodbar B. Hershey’s Milk Chocolate C. Krackel From a random sample of 140 consumers, it was found that 43 preferred Mr. Goodbar, 53 preferred Hershey’s Milk Chocolate, and 44 preferred Krackel Test the null hypothesis that customers have no preference for any of the three candy bars. Null Hypothesis: H0: pA = , p3 = 9 PC = ___._ . Procedure: Compare the actual or observed counts (Oi) in each category with the expected counts (Ei), assuming H0 is true. 0 Black’s notation: f0 forO and fe forE k 0 —E. 2 - Goodness—of-Fit Test Statistic: 12 = ELL-175;) i=1 ‘ I 0 Degrees of Freedom for 12 test statistic: k -l o Reject H0 if 12 > 22a, k-19 obtained from Table AS or reject H0 if p—value < a. Not to be reproduced without the written permission of RB Alt 2 Example: a. Calculate the expected frequencies, assuming that consumers have no preference Thenullhypothesisis: HOZpA=l/3, pB=l/3, pC=l/3 _l ypothesized Expected Candy Bar) I'_ro.ortions Counts (Ei) _--I __-'I 46.66 = 140(.33) A I I - ounts (Bi) - ounts (Oi) arm-:- “—- M U‘I Rejection Region (R.R.): Conclusion: 0 In Minitab: Stat > Tables > Chi—Square Goodness—of—F it Test (One variable) Observed Counts (Enter in C1 or any column) Equal proportions Not to be reproduced without the written permission of F .B Alt 3 o Minitab output follows Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: 0 Test Contribution Category Observed Proportion Expected to Chi-Sq A 43 0.333333 46.6667 0.288095 B 53 0.333333 46.6667 0.859524 C 44 0.333333 46.6667 0.152381 N DF Chi-Sq P-Value 140 2 1.3 0.522 Conclusion using p—Value: 0 Caveat: Conservative requirement: Bi 2 5. Not to be reproduced without the written permission of F .B Alt 4 0 Case 2: The population proportions have specific values [page 469] Example: From historical data, such as sales records, the Hershey Company knows that 30% of its customers prefer Mr. Goodbar, 50% prefer Hershey’s Milk Chocolate, and 20% prefer Krackel. Suppose that marketing analysts recently sample customers and find that 49 prefer Mr. Goodbar, 97 prefer Hershey’s Milk Chocolate and 54 prefer Krackel. Have current preferences for these products changed from the known historical preferences? Null Hypothesis: H0: PA = , PB = , PC = Use a significance level of 5%. Value of the Test Statistic: 1/ = 7.01 R.R.: Reject H0 if 12 > Conclusion: 0 In Minitab: Stat > Tables > Chi-Square Goodness-of—Fit Test (One variable) Observed Counts (Enter in C1 or any column) Specific proportions (Enter in C2 or any column) Not to be reproduced without the written permission of RB Alt 5 o The Minitab output follows Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: 0 Test Contribution Category Observed Proportion Expected to Chi-Sq 1 49 0.3 60 2.01667 2 97 0.5 100 0.09000 3 54 0.2 40 4.90000 N DF Chi-Sq P—Value 200 2 7.00667 0.030 Conclusion using p—value: Action: Not to be reproduced without the written permission of F.B Alt 6 Example: Is the percentage of each color in "M&M's"® packages as stated? An online search revealed the following: "M&M's"® Milk Chocolate Candies: 30% brown, 20% each of yellow and red and 10% each of orange, green and blue. ...” Is this true? Null Hypothesis: H0 The frequencies for a sample of 200, together with the alleged proportions, follow. Color F reguency Specified Blue 13 0.1 Brown 28 0.3 Green 64 0.1 Red 32 0.2 Orange 52 0.1 Yellow 11 0.2 Test Contribution Category Observed Proportion Expected to Chi-Sq 1 13 0.1 20 2.4500 2 28 0.3 60 17.0667 3 64 0.1 20 96.8000 4 32 0.2 40 1.6000 5 52 0.1 20 51.2000 6 11 0.2 40 21.0250 N DF Chi~Sq P—Value 200 5 190.142 0.000 Dish-rm not / ChFSqmre. «i=5 Conclusion: Not to be reproduced without the written permission of RB Alt 7 0 Testing a Population Proportion by Using the Chi-Square Test [Page 474]. Example: A market research firm interviewed a large number of potential automobile buyers by phone. One of the questions asked was whether the individual would prefer 0% financing or a $2000 rebate on the price of the car. Of the 1586 customers interviewed, 847 stated a preference for the 0% financing option. State the null and alternative hypotheses (assuming that individuals are equally divided in preferring the 0% financing), the value of the test statistic, the rejection region and your conclusion. Use a significance level of 5%. p = population proportion of individuals preferring 0% financing. H0:p vs. Hazp f7 —— p0 .534 — p0 (1—p0)/n:‘/ /1586 T.S.:Z= 22.71 R.R.: At the .05 level, reject H0 if lZl > 1.96 (Table A5) ____.:_—Ri——»__ 1.96 2.71 Conclusion: H0: p = .50 at the .05 level of significance. The Minitab output follows: Test and CI for One Proportion Test ofp = 0.5 vs p not = 0.5 Sample X N Samplep 95% Cl Z-Value P—Value l 847 1586 0.534048 (0.509498, 0.558598) 2.71 0.007 Using the normal approximation. Not to be reproduced without the written permission of F .B Alt 8 0 Using Chi—Square Approach 0 E 0% Financing 847 793 = Rebate 739 793 Value of the Test Statistic: [2 = R.R.: Reject H0 if 12 > Conclusion: The Minitab output follows: Chi-Square Goodness-of—Fit Test for Observed Counts in Variable: 0 Test Contribution Category Observed Proportion Expected to Chi-Sq 1 847 0.5 793 3.67718 2 739 0.5 793 3.67718 N DF Chi-Sq P-Value 1586 1 7.35435 0.007 Not to be reproduced Without the written permission of PB Alt 9 12.2 Contingency Analysis: Chi-Square Test of Independence [Page 479] 0 Data are cross-classified according to two factors 0 Contingency table — presentation of sample data according to two factors. Example 1: A personnel director for a large, research—oriented firm categorizes colleges and universities as most desirable, good, adequate, and undesirable for purposes of hiring their graduates. Data are collected on a random sample of 156 recent graduates and each is rated by a supervisor as outstanding, average or poor. The results follow. Rating School Outstanding Average Poor Most Desirable 21 25 2 Good 20 3 6 1 O Adequate 4 14 7 Undesirable 3 8 L 6 Example 2: A finance company wishes to learn whether marital status has any bearing on whether or not a new car loan becomes delinquent within the first year. The marital and loan status of a random sample of 950 loans is summarized in the following table: Marital Status Unmarried Married Total Loan Delinquent 29 47 76 Status Not Delinquent 384 490 874 Total 413 537 950 Not to be reproduced without the written permission of F .B Alt 10 Example 3: The table below is from a survey of American attitudes where 1397 randomly sampled Americans have been cross—classified both by their attitude to the death penalty and their attitude to gun registration. Favors Death Penalty Yes No Total Favors fig Yes 784 236 1020 Regulation No 31 1 66 377 Total 1095 302 1397 o Are the two factors inde endent ? o “The chi-square test of independence can be used to analyze any level of data measurement, but it is particularly useful in analyzing nominal data.” [Page 479] 0 Generic Contingency Table Factor B Factor A 2 @ o If events A and B are independent, then P(A and B) = 0 To calculate expected frequencies, assume that Factors A and B are independen 0 Test Statistic: c r (0 _ E 2 2 if if) 0 df=(r—l)(c—~1) Not to be reproduced without the written permission of F .B Alt 1] Example 2: A finance company wishes to learn whether marital status has any bearing on whether or not a new car loan becomes delinquent within the first year. The marital and loan status of a random sample of 950 loans is summarized in the following table: Marital Status Unmarried Married Total Loan Delinquent L 29 47 l 76 Status Not Delinquent 384 490 874 Total 413 537 950 a. Precisely state the null and alternative hypogs. Ho: Loan status and marital status are events. Ha: Loan status and marital status are E events. b. Estimate the probability that a randomly selected individual has a non-delinquent loan. Please express your answer as a fraction. P (Non—Delinquent Loan) = E = 0.92 c. Estimate the probability that a randomly selected individual is married. Please express your answer as a fraction. P (Married) = E = 0.565 d. If the null hypothesis is true, estimate the probability that a randomly selected individual has a non—delinquent loan (NDL) and is married (M). Also, determine the expected fiequency for this cell (non-delinquent loan and married)? P (NDL and M) = E] = 0.52 9 E22 = 950 (0.52) = 494.04 6. What is the value of the term in the chi-square statistic corresponding to this cell (non- delinquent loan and married)? 2 _ (022 4222)2 = (490—49404)2 _ = 0.033 In 522 494.04 Not to be reproduced without the written permission of F .B Alt 12 f. Use the MTB output to state the value of the test statistic, the rejection region and your conclusion. Use a significance level of 5%. Chi-Square Test: U, M Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts U M Total 1 29 47 76 33.04 42.96 0.494 0.380 2 384 490 874 379.96 494.04 0.043 0.033 Total 413 537 950 Chi-Sq = 0.950, DF = l, P-Value = 0.330 Reject H0 if 12 = 0.950 > E] = 3.84 -) (Do, Do not) Reject H0 _, 51 Not to be reproduced without the written permission of RB Alt 13 Example 3: The table below is from a survey of American attitudes where 1397 randomly sampled Americans have been cross—classified both by their attitude to the death penalty and their attitude to gun registration. Favors Death PenalQ Yes No Total Favors _G_1_1_r_z Yes 784 23 6 1020 Regulation No 3 l 1 66 3 77 Total 1095 302 1397 a. For a randomly selected individual, what is the probability that the individual “Favors Gun Registration” ? P (Favors Gun Registration) = E] = .7301 b. For a randomly selected individual, what is the probability that the individual “Favors Death Penalty” ? P (Favors Death Penalty) = I'VE—l = .7838 c. If attitudes are independent, what is the probability that a randomly selected individual “Favors Gun Registration” and “Favors Death Penal ” P (Favors Gun Registration and Favors Death Penalty) =( ) ( ) = .5723 d. If attitudes are independent, what is the expected frequency for the (Yes, Yes) cell? 1511 = 1397 (.5723) = 799.46 e. Show how the corresponding term in the chi—square statistic is calculated for the (Yes, Yes) cell only. 2 _(011_Eii)2 : E] 111— E11 Not to be reproduced without the written permission of F .B Alt 14 f. Use the MTB output to state the value of the test statistic, the rejection region and your conclusion based on the test statistic approach. Use a significance level of 5%. Chi-Square Test: Yes, No Expected counts are printed below observed counts Chi—Square contributions are printed below expected counts Yes No Total 1 784 236 1020 799.50 220.50 0.300 1.089 2 311 66 377 295.50 81.50 0.813 2.947 Total 1095 302 1397 Chi-Sq = 5.150, DF = l, P-Value = 0.023 El @ T.S.: RR: Conclusion: (Do, Do not) Reject H0: Factors are independent g. Precisely stgour conclusion based on the p-value approach? Since p = .05, (Reject, Do Not Reject) H0: Factors are independent El Not to be reproduced without the written permission of RB Alt 15 Example 1: A personnel director for a large, research-oriented firm categorizes colleges and universities as most desirable, good, adequate, and undesirable for purposes of hiring their graduates. Data are collected on a random sample of 156 recent graduates and each is rated by a supervisor as outstanding, average or poor. The results follow. Rating School Outstanding Average Poor Most Desirable 21 25 2 Good 20 3 6 1 0 Adequate 4 l4 7 Undesirable 3 8 6 Chi-Square Test: 0, A, P Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts 0 A P Total 1 21 25 2 48 14.77 25.54 7.69 66 ‘ 2.629 0.011 4.212 2 20 36 10 20.31 35.12 10.58 0.005 0.022 0.031 3 4 14 7 25 7.69 13.30 4.01 1.772 0.037 2.237 4 3 8 6 17 5.23 9.04 2.72 0.951 0.121 3.938 Total 48 g 25 156 Chi-Sq 15.967 DF = 6, P-Value = 0.014 2 cells with expected counts less than 5. o In Minitab: Stat > Tables > Chi-Square Test (Two—Way Table in Worksheet) Observed Counts (Enter data in as many columns as necessary) Not to be reproduced without the written permission of F .B Alt 16 ...
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