An economist wondered if people who go grocery shopping on weekdays go more or less often on Fridays than any other day. She figured that if it were truly random, 20% of these shoppers would go grocery shopping on Fridays. She randomly sampled 75 consumers who go grocery shopping on weekdays and asked them on which day they shop most frequently. Of those sampled, 24 indicated that they shop on Fridays more often than other days.The economist conducts a one-proportion hypothesis test at the 1% significance level, to test whether the true proportion of weekday grocery shoppers who go most frequently on Fridays is different from 20%.(a) H0:p=0.2; Ha:p≠0.2, which is a two-tailed test.(b) Use Excel to test whether the true proportion of weekday grocery shoppers who go most frequently on Fridays is different from 20%. Identify the test statistic, z, and p-value from the Excel output, rounding to three decimal places.Provide your answer below:test statistic = p-value =

Answer:There is not enough statistical evidence to state that the true proportion of grocery shoppers from Monday to Friday that goes most frequently on Fridays is different from 20%. The p - value is 0.0130.Step-by-step explanation:To solve this problem, we run a hypothesis test about the population proportion. Proportion in the null hypothesis (p0) = 0.2 Sample size (n) = 75 Sample proportion (sp) = 24/75 = 0.32 Significance level = 0.01 H0: p = 0.2 Ha: p [tex]\neq[/tex] 0.2 Test statistic = (sp - p0) / sqrt ((p0) (1-p0) / n) Left critical Z value (for 0.01) = -2.5758 Right critical Z value (for 0.01) = 2.5758 Calculated statistic = (0.32 - 0.2) / sqrt ((0.32) (1-0.32) / 75) = 2.2278 Since, -2.5758 < Test statistic < 2.5758, the null hypothesis cannot be rejected. There is not enough statistical evidence to state that the true proportion of grocery shoppers from Monday to Friday that goes most frequently on Fridays is different from 20%. The p - value is 0.0130.