Question

Suppose you have calculated the width of the 90% confidence interval for customer the average waiting time as ± 10 minutes, based on 10 replications of your experiment. However, your boss isn’t satisfied with your choice of a 90% confidence level and wants you to run additional experiments to give him the average waiting time with a 99% confidence interval of ± 10 minutes. Approximately how many additional replications will be required? (HINT: You may assume that the sample variance doesn’t change significantly when you incorporate the data from the additional replications.)

Answer 1

Let X be the waiting time in minutes of any given customer.

Let s be the sample variation. The estimated value of the population variance is

The standard error of the mean for a sample of size n=10 is

The level of significance for 90% confidence interval is

The critical value of z is

which is same as

Using standard normal tables we get

The margin of error obtained for 90% confidence interval with n=10 sample is ± 10 minutes

The level of significance for 99% confidence interval is

The critical value of z is

which is same as

Using standard normal tables we get

Assuming that the sample variance remains the same, the estimate of the population standard deviation remains at

the estimate of the standard error of the mean for a sample of size n is

we want the margin of error obtained for 99% confidence interval with a sample of size n  to be  ± 10 minutes

Rounding this up, we need a sample of size n=25 to get the average waiting time with a 99% confidence interval of ± 10 minutes.

ans: We need to run 15 additional experiments to give him the average waiting time with a 99% confidence interval of ± 10 minutes.