Question

Four hundred randomly selected working adults in a certain state, including those who worked at home, were asked the distance from their home to their workplace. The average distance was x-bar equal to 8.84 miles. Assume that the standard deviation of the population of all working adults in that state is 2.70 miles. Construct a 99% confidence interval for the mean distance from home to work for all residents of this state. Droble​

Answer 1

Answer:

The 99% confidence interval for the mean distance from home to work for all residents of this state is between 8.49 and 9.19 miles.

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 Z-table as such z has a p-value of $1 - \alpha$.

That 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.

$M = 2.575\frac{2.7}{\sqrt{400}} = 0.35$

The lower end of the interval is the sample mean subtracted by M. So it is 8.84 - 0.35 = 8.49 miles.

The upper end of the interval is the sample mean added to M. So it is 8.84 + 0.35 = 9.19 miles.

The 99% confidence interval for the mean distance from home to work for all residents of this state is between 8.49 and 9.19 miles.