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.

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

Step-by-step explanation:

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

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

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

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.

i believe the correct answer is b.

it would rotate about 180 degrees.