Question

The top food snacks consumed by adults aged 18-54 are gum, chocolate candy, fresh fruit, potato chips, breath mints/candy, ice cream, nuts, cookies, bars, yogurt, and crackers. Out of a random sample of 29 men, 15 ranked fresh fruit in their top five snack choices. Out of a random sample of 38 women, 20 ranked fresh fruit in their top five snack choices. Is there a difference in the proportion of men and women who rank fresh fruit in their top five list of snacks? (a-1) Choose the appropriate hypotheses. Assume πΜ is the proportion of men and πw is the proportion of women. O a (a-2) State the decision rule for α = .01. (Round your answers to 3 decimal places. A negative value should be indicated by a minus sign.) Reject the null hypotheses if Zcalc b) Calculate the sample proportions. (Round your answers to 4 decimal places.) or Zcalc Sample proportion Women

Answer 1

Answer Date: 29/01/2019 a1) To test the hypothesis is that there is a difference in proportion of men and women who rank fresh fruit in their top five list of snakes at 1% significance level. The null and alternative hypothesis is The answer is d a-2 Thez-critical value is, From the standard normal table, the z-critical value for two-tailed at 99% confidence is 2.575 The z-critical value is + 2.575

n,n, 0.5172-0.5263 0.5224(1-0.5224) 29 38 0074 The value of test statistics is 0.074 The p-value is, From the standard normal distribution table, the area under to the left of-0.074 multiply by 2 and p-value is, p 200.4705) 0.9410 Thep-value is 10.9410.

Decision The z-test statistics which is -0.074 corresponds to sample statistics is not falls in the critical region which is -2.575, so the null hypothesis is not rejected at 1% level of significance. There is insufficient evidence to indicate that there is a difference in proportion of men and women who rank fresh fruit in their top five list of snakes. The result is not statistically significant. d) The assumption is approximate p, by a normal distribution is np, 210 and n (1-p,)210 with n 29 is, P 29 x15/29 -29 m(1-A)-298 (1-15/29) -14 The assumption is approximate P, by a normal distribution is np, 210and n(-P)210 with n 29 is satisfied