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

Answer:

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 $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 2.576$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 2.576*\frac{442}{\sqrt{24}} = 232.41$

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.