Question

This problem needs to be solved on Microsoft Excel!

METHODS AND APPLICATIONS 13.17 A marketing research firm wishes to study the relationship between wine consumption and whether a person likes to watch professional tennis on television. One hundred randomly selected people are asked whether they drink wine and whether they watch tennis. The following results are obtained: DS WineCons Do Not Watch Tennis Watch Tennis Totals Drink Wine 16 24 40 Do Not Drink Wine 4 56 60 Totals 20 80 100 a For each row and column total, calculate the corresponding row or column percentage. b For each cell, calculate the corresponding cell, row, and column percentages. c Test the hypothesis that whether people drink wine is independent of whether people watch tennis. Set a = .05. d Given the results of the chi-square test, does it make sense to advertise wine during a televised tennis match (assuming that the ratings for the tennis match are high enough)? Explain.

Answer 1

Watch Tennis Dont Watch tennis Total 16 24 40 4 56 60 Total 20 80 100 The expected values are computed in terms of row and column totals. In fact, the formula is E.; = R XC, where R; 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 Watch Tennis Dont Watch tennis Total 20240 = 8 ROX 40 = 32 100 40 100 20/80 = 12 ROX 6 = 48 60 100 100 Total 20 80 100 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E-OPE. The table with squared distances is shown below. Squared Distances Watch Tennis Dont Watch tennis (16-37 = 8 (24–32) = 2 6-12 = 5.333 (56-48 48 = 1.333 12 Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H,: The two variables are independent H. 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?:X'> 3.841). Test Statistics The Chi-Squared statistic is computed as follows: (0, E.;) Ej = 8+5.333+2+1.333 = 16.667 = Decision about the null hypothesis Since it is observed that x = 16.667 > x = 3.841, it is then concluded that the null hypothesis is 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 > 16.667) = 0.