Question

You are the manager of a restaurant for a fast-food franchise. Last month, the mean waiting time at the drive-through window for branches in your geographical region, as measured from the time a customer places an order until the time the customer receives the order, was 3.7 minutes. You select a random sample of 81 orders. The sample mean waiting time is 3.93 minutes, with a sample standard deviation of 0.9 minute. Complete parts (a) and (b) below. a. At the 0.10 level of significance, is there evidence that the population mean waiting time is different from 3.7 minutes? State the null and alternative hypotheses. Ho: u = 3.7 Hy: u 3.7 (Type integers or decimals.) Determine the test statistic. The test statistic is 2.30. (Round to two decimal places as needed.) Find the p-value. p-value = (Round to three decimal places as needed.)

Answer 1

Question 1 The null hypothesis to be tested is Ho: u = 3.7 against the alternative H : 4 +3.7 We have the sample size n=81, sample mean 7 = 3.93 and sample standard deviation s =0.9 The test statistic is X-u 3.93–3.7 Z = = 2.30 0.9 Jn 181 The p-value is P(Z >2.3) = 0.0214 S Normal Curve, mean = 0, SD = 1 Shaded Area = 0.0214 CO 02 0.1 90 23 — Since p-value is less than the significance level specified we will reject the null hypothesis

: Question 2 Here the null hypothesis to be tested is Ho: u 29.75 against the alternative H,: u< 9.75 We have the sample size n=50, sample mean = 9.01 and sample standard deviation s=5.5 The test statistic is -u 9.01-9.75 Z = -0.9514 5.5 Jn 50 The p-value is P(Z <-0.9514) = 0.1707 s Normal Curve, mean = 0, SD = 1 Shaded Area = 0.1707 03 02 0.1 09514 0 since the p-value is greater than the significance level we will accept the null hypothesis

Question 3 The null hypothesis to be tested is Ho:p=0.45 against the alternative H,: #0.45 73 We have sample size n=120. The sample proportion is 0.6083 120 p-p 0.6083-0.45 The test statistic is Z= = 3.48 p(1-p) 0.45(1-0.45) 120 P-value is P(Z>3.48)=0.001 (by approximating to 3 digits) since the p-value is less than the significance level specified, we will reject the null hypothesis. i.e. there is sufficient evidence that the proportion of unpaid interns are different from 0.45 n b. The null hypothesis to be tested is H, :p=0.45 against the alternative H:u #0.45 56 - We have sample size n=120. The sample proportion is ø ==0.4667 120 -p 0.4667 -0.45 The test statistic is Z- = 0.36 p(1-p) 0.45(1-0.45) 120 P-value is P(12>0.37)=0.711 (by approximating to 3 digits) since the p-value is greater than the significance level specified, we failed to reject the null hypothesis. i.e. there is not sufficient evidence that the proportion of unpaid interns are different from 0.45 n