Question

In a study of factors affecting whether soldiers decide to reenlist, 320 subjects were measured for an index of satisfaction. The sample mean is 28.8 and the sample standard deviation is 7.3. Use the given sample data to construct the 98 percent confidence interval for the population mean.

Answer 1

The 98 percent confidence interval for the population mean is between 27.85 and 29.75 subjects.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.98)/(2) = 0.01`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.01 = 0.99`, so
`z = 2.325`

Now, find M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.325(7.3)/(√(320)) = 0.9488`

The lower end of the interval is the mean subtracted by M. So it is 28.8 - 0.9488 = 27.85 subjects.

The upper end of the interval is the mean added to M. So it is 28.8 + 0.9488 = 29.75 subjects.

The 98 percent confidence interval for the population mean is between 27.85 and 29.75 subjects.