Question

Question 21 (4 points) A sales region has been divided into five territories, each of which was believed to have equal sales potential. The actual Sales Volume for several sampled days is logged in DATA. At α 0.05, do the territories have equal Sales Volume? None of the answers are correct HO: Territories have equal Sales Volume is not rejected with pvalue 0.334. The counts are consistent with the model of equal proportions. HO: Territories have equal Sales Volume is rejected with pvalue 0.041. The counts are not consistent with the model of equal proportions. Territories have unequal Sales Volume. H0: Territories have equal Sales Volume is not rejected with pvalue 0.074. The counts are consistent with the model of equal proportions HO: Territories have equal Sales Volume is rejected with pvalue 0.012. The counts are not consistent with the model of equal proportions. Territories have unequal Sales Volume. D View hint for Question 21 Page 21 of 27 Next Page Confidence Interv 0 40 0 1234567890|1234567890

Answer 1

Answer Date: 24/04/2019 To test the hypothesis is that the counts are not consistent with the model of equal proportions at 5 % level of significance The null and alternative hypothesis is, The nul hypothesis is that the counts are consistent with the model ofequal proportions The altemative hypothesis is that the counts are not consistent with the model of equal proportions The chi-square test statistic for goodness-of-fit test is First, find frequency expected then compute chi-square test statistics The average for the observed frequencies computation is the frequencies expected. The total for the expected frequencies must equal for the total observed frequencies The expected frequencies for category 1 are,

Total oberved frequency Total number of category 110+140 100106+110 Expected frequency - -113.20 If the frequencies are uniformly distributed, the expected frequency is the same for each category and for other categories which are 2 to 5 each has an equal expected frequency which is 113.20 The chi-square test statistic for goodness-of-fit test is, By using Excel software, find chi-square test statistics with the help of following steps: 1) Import the data. 2) Select Goodness-of-Fit Test from Chi-Square in MegaStat option from Add Ins menu 3) Seect Observed values and Expected values. 4) Click Ok.

Goodness of Fit Test observed xpected 113.200 113.200 113.200 113.200 113.200 566.000 O-E 3.200 26.800 13.200 7.200 3.200 0.000 (0-E)" / E 0.090 6.345 1.539 0.458 0.090 8.523 %ofchis 1.06 74.44 18.06 5.37 1.06 100.00 110 140 100 106 110 566 8.523 chi-square 4 df 0742 p-value From the Excel output, the chi-square test statistic for goodness-of-fit test is 8.523 The p-value is From the Excel output, the chi-square testp-value is 0.074 The p-value is 0.074

Decision The conclusion is that the significant p-value is greater than o.0s which is o.074, so the null hypothesis is not rejected at 5% level of significance. There is insufficient evidence to indicate that the counts are not consistent with the model of equal proportions. The result is not statistically significant The answer options