Question

The mean bill of 16 randomly selected customers in a restaurant is found to be $32, with a standard deviation of $3.50. Construct a 99% confidence interval for the mean bill of a customer.

Answer 1

Here population standard deviation is not given and we use sample standard deviation(s) instead of population standard deviation ( ). Also sample size is not sufficiently large( >= 30) so we need to assume that the sample comes from normal population. Then we can used one sample t confidence interval for population mean( )

Using minitab:

Step 1) Click on Stat>>>Basic Statistics >>1 sample t...

Select summarized data

Sample size: 16

Mean: 32

Standard deviation: 3.50

Step 2) Click on Option

Confidence level = 99

Alternative: Not equal

Then click on OK

Again Click on OK

Then we get the following output

From the above output, the 99% confidence interval of the population mean is  (29.422, 34.578).