Question

A very busy and fancy restaurant records the wait time for each customer that comes in. Below are 12 customers’ wait times (in minutes):

51        60        59        72        80        83        54        66        61        81        66        62

1.The manager wants to know the percentage of wait times that are greater than 55 minutes. According to the sample data, what would be a point estimate of the true percentage of wait times that are greater than 55 minutes?

2.Use this point estimate to construct and interpret a 90% confidence interval for p, the true proportion of wait times that are greater than 55 minutes

3.Conduct a hypothesis test to see if the true proportion of wait times over 70 minutes is less than 40%. What are the null and alternative hypotheses you will test?

4.What would be the test statistic for testing the hypotheses from (3).

5.What is the corresponding p-value?

6.Do you reject the null hypothesis based on a significance level of .10? What do you conclude about the true proportion of wait times over 70 minutes?

Answer 1

1. From the above data, we have 10 customers who wait more than 55 minutes.

So the required point estimate = 10/12 = 0.8333

2. 90% CI for p is:

p +- z0.05*√p*(1-p)/n

= 0.8333 +- 1.645*√0.8333*0.1667/12

= 0.8333 +- 0.1770

= (0.6563, 1)

(Since probability values can lie between 0 to 1 only, if confidence interval for the upper bound increases by 1 we bound it to 1)

3. To test the hypothesis of wait times over 70 mins <40%, we have,

H0: p= 0.4 vs H1: p<0.4

4. Proportion of wait times over 70 mins from sample above = 4/12 = 0.333

So, test statistic: 0.333-0.4/√0.4*0.6/12 = -0.474