Question

The time spent, in hours, of teenagers on social media per year are normally distributed with a population standard deviation of 442 hours and an unknown population mean. If a random sample of 24 teenagers is taken and results in a sample mean of 1330 hours, find a 99% confidence interval for the population mean. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 You may use a calculator or the common z values above. Round your answer to TWO decimal places.

Answer 1

The 99% confidence interval for the population mean is between 1087.59 hours and 1572.41 hours.

Explanation:

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

`\alpha = (1-0.99)/(2) = 2.576`

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
`1-0.05 = 0.95`, so
`z = 1.645`

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 = 2.576*(442)/(√(24)) = 232.41`

The lower end of the interval is the sample mean subtracted by M. So it is 1330 - 242.41 = 1087.59 hours.

The upper end of the interval is the sample mean added to M. So it is 1330 + 242.41 = 1572.41 hours.

The 99% confidence interval for the population mean is between 1087.59 hours and 1572.41 hours.