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 64 orders. The sample mean waiting time is 3.57 minutes, with a sample standard deviation of 0.8 minute.

A) At a 95% confidence level, is there evidence that the population mean waiting time is different from 3.7 minutes?

B) Because the sample size is 64, do you need to be concerned about the shape of the population distribution when conducting the t- test in (a)? Explain.

Answer 1

A) To test at a 95% confidence level or at a 5% significance level the population mean waiting time is different from 3.7 minutes,

Let u denotes the population mean. To test null hypothesis Ho u 3.7 against alternative hypothesis H1 u3.7 Let denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here 3.5700, s2 0.6400, s = 0.8000, n 64 The test statistic can be written as Vn (3.7) S which under Ho follows a t distribution with n -1 df. We reject Ho at 0.050 level of significance if P-value < 0.050 or if |tobs> to.025.n-1 Now The value of the test statistic = -1.300000 associated degrees of freedom 63 The critical value =1.998341 P((tn-1tobs) = 2 * P(t63 <-1.300000) 2 * 0.099169 = 0.198337 Р- value — Since p-value > 0.050 and tobsteritical = 1.998341, so We fail to reject Ho at 0.050 level of significance Hence, we can conclude that the population mean is not different from 3.7

b) Since sample size is large enough ( > 30), we can approximate the sampling distribution of sample mean follows approximately normal. So, when standard deviation of sample mean is not known, we can use t test.