Question

A sample of 25 commuters in the Washington, D.C., area yielded the sample mean commute time of 27.97 minutes and sample standard deviation of 10.04 minutes. Construct and interpret a 99% confidence interval for the mean commute time of all commuters in Washington D.C. area.

Answer 1

Answer:

The 99% confidence interval for the mean commute time of all commuters in Washington D.C. area is between 22.7994 minutes and 33.1406 minutes. The interpretation is that we are 99% sure that the true mean ommute time of all commuters in Washington D.C. area is between 22.7994 minutes and 33.1406 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 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.

$M = 2.575*\frac{10.04}{\sqrt{25}} = 5.1706$

The lower end of the interval is the mean subtracted by M. So it is 27.97 - 5.1706 = 22.7994 minutes

The upper end of the interval is the mean added to M. So it is 6.4 + 0.3944 = 33.1406 minutes.

The 99% confidence interval for the mean commute time of all commuters in Washington D.C. area is between 22.7994 minutes and 33.1406 minutes. The interpretation is that we are 99% sure that the true mean ommute time of all commuters in Washington D.C. area is between 22.7994 minutes and 33.1406 minutes.