Question

According to Nielsen//NetRatings, the average visitor to the American Greetings Website spends 11.85 minutes at the site. Assuming this finding to be based on a random sample of 20 visitors to the site, a POPULATION standard deviation of 3.0 minutes, and a population of visiting times that is approximately normally distributed, a 99% confidence interval for the population mean is closest to:________.

Answer 1

Answer:

The 99% confidence interval for the population mean is between 10.12 minutes and 13.58 minutes.

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:

$\alpha = \frac{1-0.99}{2} = 0.005$

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 $1-0.005 = 0.995$, so $z = 2.575$

Now, find the margin of error 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.

$M = 2.575*\frac{3}{\sqrt{20}} = 1.73$

The lower end of the interval is the sample mean subtracted by M. So it is 11.85 - 1.73 = 10.12 minutes

The upper end of the interval is the sample mean added to M. So it is 11.85 + 1.73 = 13.58 minutes

The 99% confidence interval for the population mean is between 10.12 minutes and 13.58 minutes.