Question

A random sample of 200 was selected to estimate the average amount of time retired folks in Arizona listened to the radio during the day. The sample mean was 110 minutes, and the standard deviation was 30 minutes. The 95% confidence limits for the population mean listening time are _____ and _____.

Answer 1

The 95% confidence limits for the population mean listening time are 105.84 minutes and 114.16 minutes.

Explanation:

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

`\alpha = (1-0.95)/(2) = 0.025`

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
`1-0.025 = 0.975`, so
`z = 1.96`

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 = 1.96*(30)/(√(200)) = 4.16`

The lower end of the interval is the sample mean subtracted by M. So it is 110 - 4.16 = 105.84 minutes.

The upper end of the interval is the sample mean added to M. So it is 110 + 4.16 = 114.16 minutes.

The 95% confidence limits for the population mean listening time are 105.84 minutes and 114.16 minutes.