Question

Write clear & neat

Several UTA alums have been pushing for UTA to restart football. You have data from a random sample of faculty/admin and alum/students and wonder if there is a relationship between which group someone belongs to and their support for restarting football. 2. Oppose 30 50 Neutral 15 15 Favor 17 30 UTA faculty and admin UTA alum and students a) Provide null and alternative hypotheses in formal terms and layperson's terms for the chi-square two sample test. Remember to apply Yates if necessary. (5) b) Do the math and reject/accept at a-.05 (15) c) Explain the results in layperson's terms (20) d) Explain in layterms the row percentage for the cell "UTA alum and students oppose (5)

Answer 1

Answer a)

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

H0​: The two variables are independent

Ha​: The two variables are dependent

Layperson's Term

H0​: There is no relationship between which group someone belongs to and their support for restarting football

Ha​: There is relationship between which group someone belongs to and their support for restarting football

Answer b)

The following cross tablulation have been provided. The row and column total have been calculated and they are shown below: Total 62 95 157 Column1 17 30 47 Column 2 15 15 30 Column 3 30 50 80 Row 1 Row 2 Total The expected values are computed in terms of row and column totals. In fact, the formula is = , where Ri corresponds to the total sum of elements in row i, Ci 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 Total 62 95 157 Expected Values Row 1 Row 2 Total Column1 Column 2 Column 3 2 18.561 2=11.847 2 = 31.592 18.153 48.408 80 30 47 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E O)-/E. The table with squared distances is shown below. Squared Distances Row 1 Row 2 Column1 Column 2 Column 3 0.131.830.08 50 48408 408L = 0.052 130 28 나WL LitwL = 0.548 = 0.086

Based on the information provided, the significance level is a 0.05, the number of degrees of freedom is df (2-1) x (3-1) 2, so then the rejection region for this test is R x?:x> 5.991) Test Statistics The Chi-Squared statistic is computed as follows: 2-y-ї Ч- 0.131+ 0.0864 0.8394 0.5484 0.084 0.052-1.736 10 Decision about the null hypothesis Since it is observed that χ-1.736く not rejected. = 5.991, it is then concluded that the null hypothesis is 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 0.05 significance level.

Answer c)

Based on hypothesis test, we can say that there is no relationship between which group someone belongs to and their support for restarting football. In other words, we do not have enough evidence to claim that there is difference in opinion between two groups regarding restarting football.

Answer d)

Row percentage for cell "UTA alum and students, oppose" = 50*100/(30+15+50)

Row percentage for cell "UTA alum and students, oppose" = 52.63%

Out of the total UTA alum and students surveyed, 52.63% opposed proposal of restarting the football.