In an effort to estimate the mean dollars spent per visit by customers of a food store, the manager has selected a random sample of 100 cash register receipts. The mean of these was $45.67 with a sample standard deviation equal to $12.30. Assuming that he wants to develop a 90 percent confidence interval estimate, the upper limit of the confidence interval estimate is:

The upper limit of the confidence interval estimate is $66.09

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 100 - 1 = 99

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of ). So we have T = 2.68

The margin of error is:

M = T*s = 1.6604*12.30 = 620.42.

In which s is the standard deviation of the sample.

The upper end of the interval is the sample mean added to M. So it is 45.67 + 20.42 = 66.09

The upper limit of the confidence interval estimate is $66.09