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​

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

Now, we have to find z in the Z-table as such z has a p-value of .

That is z with a pvalue of , so Z = 2.575.

Now, find the margin of error M as such

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

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.

step-by-step explanation:

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this is false ^.^ lol