Question

4 500 randomly selected automobile owners were current automobile. The results are given below. questioned on the main reason they had purchased their Total Styling Engineering Fuel Economy Male 340 68 132 140 160 Female 32 18 110 Total 500 250 100 150 State the null and alternative hypotheses for a contingency table test for independence of gender. a. b. State the decision rule, using a .01 level of significance. Calculate the x Give your conclusion for this test. test statistic for this data using the correct formula. c. d.

Answer 1

(a) The step-by-step instructions to obtain Chi-square test for Asscoication output using Minitab18: * Store the Gender in the column C1 and Automobile in the rest of the columns of the Minitab 18 worksheet. * Choose the Stat → Tables — Chi-square test for Association * Choose the summarized data in a two-way table. * Select all the columns (C2-C4) in the Columns containing the variable. * Labels for the table: Rows: Gender and Columns Automobile * Click ok. The Chi-square Test for Association using Minitab 18:

Chi-Square Test for Association: Gender, Automobile Rows: Gender Columns: Automobile Fuel Styling Engineering Economy All Male 68 140 340 170 Female 32 110 160 32 All 100 150 250 500 Cell Contents Count Expected count Chi-Square Test Chi-Square DF P-value Pearson 44.118 2 0.000 Likelihood Ratio 48.453 2 0.000

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Step1: To, test whether the current automobile is independent of gender or not at the 1% level of significance. The null and alternative hypotheses are as follows. Ho: The null hypothesis is that the current automobile is independent of gender. versus 4 : The alternative hypothesis is that the current automobile is dependent of gender. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (b) Decision rule: Step 2: The level of significance is a =0.01. If the test statistic is za > {012 = 9.21034. We reject the null hypothesis H. Otherwise, we do not reject the null hypothesis H. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (c) Step3: The formula for calculating the test statistic and its value is of the chi-square test of association output of the Minitab18 is (48;-(4B;).) z? = (4B;) = 44.118 22 = 44.118 Therefore, the chi-square test statistic is za s 44.12) ------ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (d) Step4: The degrees of freedom is (r - 1)(0-1) = (2-1)(3-1) =(1)(2) = 2 df = 2 The critical value is 2012 = 9.21034) Step5: The P-value approach: P(x > x) = 1- P(x2 < xo) =1-P(x* <44.118) = 1-0.99999999974 =0.00000000026 P(x2 > 22 ) 0.00000000026 <«=0.01. We reject the null hypothesis --------------- --- -- -- - - - - - Step 6: Since, the test statistic is za = 44.118 > 2.0L2 = 9.21034 We reject the null hypothesis H, Since, the P-value = 0.00000000026 <a=0.01. We reject the null hypothesis H, There is sufficient evidence to conclude that the current mobile is dependent of the gender at the 1% level of significance.