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​

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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 [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.99}{2} = 0.005[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.005 = 0.995[/tex], so Z = 2.575.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 2.575\frac{2.7}{\sqrt{400}} = 0.35[/tex]

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.