Question

The accompanying table shows results of challenged referee calls in a major tennis tournament. Use a 0.05 significance level to test the claim that the gender of the tennis player is independent of whether a call is overturned. Click the icon to view the table. Hy: The gender of the tennis player is independent of whether a call is overturned. Determine the test statistic. x2 = 12.994 (Round to three decimal places as needed.) Determine the P-value of the test statistic. P-value = 0.0003 (Round to four decimal places as needed.)

Answer 1

I ran the tests again and every calculation and the values do match. I think the problem is cause the fields are demanding 4 decimal places and you're inputting 3 decimal place answers. So i'll add the fourth decimal in the images below so to make them more accurate. Check the final values and re enter.

The following cross tablulation have been provided. The row and column total have been calculated and they are shown below: Yes No Total Men 314 997 1311 Women 208 947 1155 Total 522 1944 2466 The expected values are computed in terms of row and column totals. In fact, the formula is Ejj Rixci, 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 Yes No Total = Men 522 x 1311 2466 = 277.511 1944x1311 2466 1033.489 1311 Women 522 x 1155 2466 = 244.489 1944x1155 2466 = 910.511 1155 Total 522 1944 2466 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E – 0)2 / E. The table with squared distances is shown below:


Squared Distances Yes No Men (314-277.511) 277.511 4.798 (997-1033.4892 1033.489 1.288 Women (208-244.4892 244.489 = 5.446 (947-910.511) 910.511 = 1.462 Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: The two variables are independent Ha: The two variables are dependent This corresponds to a Chi-Square test of independence. Rejection Region Based on the information provided, the significance level is a = 0.05 , the number of degrees of freedom is df = (2 – 1) (2 – 1) = 1, so then the rejection region for this test is R = = {x? : x2 > 3.841}.


Test Statistics The Chi-Squared statistic is computed as follows: n (Oij – Eij) =Σ Eij 4.798 + 5.446 + 1.288 + 1.462 = 12.9943 i=1 Decision about the null hypothesis = 12.994 > x2 = 3.841, it is then concluded that the null hypothesis is Since it is observed that x2 rejected. Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables are dependent, at the 0.05 significance level. The corresponding p-value for the test is p = Pr(xỉ > 12.994) = 0.0003 77