Question

A study was done to determine the average number of homes that a homeowner owns in his or her lifetime. For the 50 homeowners surveyed, the sample average was 5.1 and the sample standard deviation was 3.8. Calculate the 95% confidence interval for the true average number of homes that a person owns in his or her lifetime.

Answer 1

The 95% confidence interval for the true average number of homes that a person owns in his or her lifetime is (4,6.2).

Explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

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

df = 50 - 1 = 49

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 49 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.95)/(2) = 0.975`. So we have T = 2.0096

The margin of error is:

`M = T(s)/(√(n)) = 2.0096(3.8)/(√(50)) = 1.1`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 5.1 - 1.1 = 4

The upper end of the interval is the sample mean added to M. So it is 5.1 + 1.1 = 6.2.

The 95% confidence interval for the true average number of homes that a person owns in his or her lifetime is (4,6.2).