Question

The owner of the Gulf Stream Café wished to estimate the mean number of lunch customers per day. A sample of 40 days revealed a mean of 160 per day, with a standard deviation of 20 per day. Develop a 98% confidence interval for the mean number of customers per day.

Answer 1

It is given that a sample of 40 days revealed a mean of 160 per day, with a standard deviation of 20 per day.

 

Our interest is to construct 98 percent confidence interval for the mean number of customers per day.

We have, X̅ = 160, s = 20, and n = 40.

 

Confidence level, 1 – α = 98%

 

Degrees of freedom,

df = n – 1

     = 40 – 1

     = 39

 

Using a t Distribution table, for 98% confidence with 39 df, we get t = 2.426.

 

The 98% confidence interval for the true mean,

X ± t s/√n = {160 ± (2.426)20/√40}

                  = (160 ± 7.67)

                 = (160 – 7.67, 160 + 7.67)

                 = (152.33, 167.67)

 

Thus, the 98% confidence interval for the mean number of customers per day is between 152 and 168.

Thus, the 98% confidence interval for the mean number of customers per day is between 152 and 168.