A random sample of 128 lunch customers was selected at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 128.2 minutes and the sample standard deviation equals 24.9 minutes. The objective is to create a 99% confidence interval for the mean amount of time that customers stay. What is the confidence interval?

Answer: (122.52, 133.88)

Step-by-step explanation:

We are to construct a 99% confidence interval for population mean time.

From our question, we have the following parameters

Sample size (n) = 128

Sample mean (x) = 128.2

Sample standard deviation (s) = 24.9

To construct a 99% confidence interval, we use the formulae below.

u = x + Zα/2 * s/√n ... For upper limit

u = x - Zα/2 * s/√n ... For lower limit

We are using a z score for our critical value (Zα/2) and that's because our sample size is greater than 30 (n = 128), even though we have our sample standard deviation.

The value of Zα/2 is gotten using a z distribution table and has a value of 2.58

For lower limit, we have that

u = 128.2 - 2.58 / (24.9/√128)

u = 128.2 - 2.58 (2.2)

u = 128.2 - 5.678

u = 122.52.

For upper limit, we have that

u = 128.2 + 2.58 / (24.9/√128)

u = 128.2 + 2.58 (2.2)

u = 128.2 + 5.678

u = 133.88

Hence the 99% confidence interval for the mean amount of time customer stays is (122.52, 133.88)