Question

In a random sample of seven aerospace engineers, the sample mean monthly income is $6824 and the sample standard deviation is $340. Construct a 95% confidence interval for the population mean. Assume that the monthly incomes are normally distributed.

Answer 1

The 95% confidence interval for the population mean is ($6510, $7138).

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 = 7 - 1 = 6

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 6 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.95)/(2) = 0.975`. So we have T = 2.4469.

The margin of error is:

`M = T(s)/(√(n)) = 2.4469(340)/(√(7)) = 314`

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 6824 - 314 = $6510.

The upper end of the interval is the sample mean added to M. So it is 6824 + 314 = $7138.

The 95% confidence interval for the population mean is ($6510, $7138).