Question

Perform a chi-square homogeneity test. An independent simple random sample of residents in three regions gave the data on race shown in the table. At the 1% signifcance level, do the data provide sufficient evidence to conclude that a difference exists in race distributions among the three regions? White Black Other Total 91 Midwest West East 14 114 117 16 10 143 115 16 139 Total 396 323 38 35 What are the null and alternative hypotheses? OA H: The racial distribution is the same in at least two of the three regions. H: The racial distribution is not the same in at least two of the three regions O B. Ho: The racial distribution is not the same in at least two of the three regions H: The racial distribution is the same in at least two of the three regions. OC. Ho: The racial distribution is the same in each of the three regions. H The racial distribution is not the same in each of the three regions. OD. Hg: The racial dstribution is not the same in each of the three regions. H, The racial distribution is the same in each of the three regions. Find the test statistic. -(Round to three decimal places as needed.) Find the Pvalun Round to four decimal places as needed.)

Answer 1

Solution- * null and alternative hypothesis D. Hoi The racial distribution is not the same in each of the three regions. HI: The racial distribution is same in each of the three regions test statistic I X2 = 5.227 P-value 0.2648

White Black Total East 91 14 114 Midwest 117 143 West 115 139 Total 323 396 The expected values are computed in terms of row and column totals. In fact, the formula is Ejj = Rxci, where Ri corresponds to the total sum of elements in row i, C; corresponds to the total sum of elements in column j, and T is the grand total. The table below shows the calculations to obtain the table with expected values: Expected Values White Black Other Total East 323 x 114 - 99 985 = 92.985 396 38x114 = 10.939 396 35x114 = 10.076 114 396 Midwest 323x143 = 116.639 38x143 = 13.722 396 35x143 = 12.639 396 143 396 West 323x139 = 113.376 396 38x139 = 13.338 396 35x139 = 12.285 139 396 Total 323 38 35 396 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E - 0)? /E. The table with squared distances is shown below: Squared Distances White Black Other (91-92.985) (14-10.939) = 0.042 East (9-10.076) = 0.856 = 0.115 92.985 10.939 10.076 Midwest (117-116.639) 116.639 0.001 (16-13.722) = 0.378 (10-12.639) = 0.551 13.722 12.639 West (115-113.376) 113.376 0.023 (8-13.338) = 2.137 (16-12-285)2 - = 1.123 12.285 13.338

Rejection Region Based on the information provided, the significance level is a = 0.01 , the number of degrees of freedom is df = (3 – 1) (3 – 1) = 4, so then the rejection region for this test is R = {x?: x2 > 13.277). Test Statistics The Chi-Squared statistic is computed as follows: X “ (Oij – E;;)? il = 0.042 +0.001 +0.023+0.856 + 0.378 + 2.137 + 0.115 +0.551 +1.123 = 5.227 = 2 Eij Decision about the null hypothesis Since it is observed that x2 = 5.227 < x2 = 13.277, it is then concluded that the null hypothesis is not rejected. Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that the two variables are dependent, at the 0.01 significance level. The corresponding p-value for the test is p= Pr(xź > 5.227) = 0.2648.

----------------------

Thank you...!!

Please like..