Question

Passed 19 Falled 15 Results from a civil servant exam are shown in the table to the right. Is there sufficient evidence to support the claim that the results from the tost are White candidates Minority candidates 9 O A Hoi A white candidate is more likely to pass the tool than a minority candidato Hy: A white candidate is not more likely to pass the lost than a minority candidate B. Hy: White and minority candidates do not have the same chance of passing the test. He: White and minority candidates have the same chance of passing the lost. OC Hoi A white candidate is not more likely to pass the lost than a minority candidate Hy: A white candidate is more likely to pass the test than a minority candidate OD. Hy: White and minority candidates have the same chance of passing the test. He: White and minority candidates do not have the same chance of passing the best Determine the best statistio. 7- (Round to three decimal places as needed) Determine the value of the test statistic. Palue - Round to four decimal places as needed.) Is there sufficient evidence to support the claim that the results from the test are discriminatory O A. There is sufficient evidence to support the claim that the results are discriminatory OB. There is not suficient evidence to support the claim that the results are discriminatory. OC. There is soient evidence lo rejecte daim that a white candidate is more likely to pass the festman a minority candidate OD. There is not suficient evidence to reject the claim that a white candidate is more likely to pass the best than a minority candidate

Answer 1

0. Hy: White and minority candidates do not have the same chance of passing the test. Hy: White and minority candidates have the same chance of passing the tost.

Expected Values Column 1 Column 2 Total Row 1 28 x 34 67 - 14.209 39x34 67 19.791 34 Row 2 28433 = 13.791 39x33 67 = 19.209 33 Total 28 39 67

Squared Distances Column 1 Column 2 Row 1 (19–14.209) 14.209 1.615 (15-19.791) 19.791 1.16 Row 2 (9-13.791) 13.791 -1.664 (24–19.209) 19.209 1.195

The Chi-Squared statistic is computed as follows: n x- (Oij – E;;)? = 1.615 +1.664 + 1.16 + 1.195 = 5.635 Eij i=1

Based on the information provided, the significance level is a = 0.01 , the number of degrees of freedom is df = (2 – 1) (2 - 1) = 1, so then the rejection region for this test is R= {x?:x> 6.635}

The corresponding p-value for the test is p = Pr(xỉ >5.635) = 0.0176.

Since it is observed that y’ = 5.635 < x = 6.635, it is then concluded that the null hypothesis is not rejected.

there is not sufficient evidence to support the claim that results are discriminatory