Question

[5 marks] In a market research study conducted by a local winery on white wine preference, the following results were found. Of a sample of 500 men, 120 preferred white wine and of a sample of 500 women, 210 preferred white wine. Setup the null and alternative hypothesis that will test the claim that the percentage of women preferring white wine is 25% higher than men. Calculate the test statistic and test the claim at a = .05. What do you conclude?

Answer 1

Let $p_1,p_2$ denote the true proportions of women and men who prefereed white wine.

Hypotheses : To test null hypothesis H. : P1 - P2 = 0.25 against alternative hypothesis H1 :P1 - P2 > 0.25 Let P1, P2 denote the respective sample proportions and n1, n2 denotes the sample sizes respectively. Here, 210 pi = 0 = 0.4200, n1 = 500, P2 = 120 60 = 0.2400, n2 = 500 500 The test statistic can be written as: (P1 -2) – 0.25 Pi*(1-) + P?#(1–P2) V=nTn2 which under H, follows a standard normal distribution. Decision rule / Rejection region : We reject Ho at 0.05 level of significance if P-value < 0.05 or if Zobs > 20.05 = 1.644854 density 0.0 - Now, 4 Value of test sta The value of the test statistic = -2.398160 The critical value = 1.644854 1.64 -2.40 P-value = P(Z > Zobs) = P(Z < -2.398160) = (1 – 0.008239) = 0.991761 ~ 0.9918 Conclusion : Since p-value > 0.05 and zobs critical = 1.644854, so We fail to reject Ho at 0.05 level of significance Hence, we can conclude that the the first population proportion is not significantly greater than second population proportion.

Conclusion in problem context :

There is not sufficient evidence to support the claim that the percentage of women preferring white wine is 25% higher than men.