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The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day. Assume that the population standard deviation is 2.3 gallons. The mean water usage per family was found to be 18.5 gallons per day for a sample of 717 families. Construct the 80% confidence interval for the mean usage of water. Round your answers to one decimal place.

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Answer:

The 80% confidence interval for the mean usage of water is between 18.4 and 18.6 gallons per day.

Explanation:

We have that to find our
\alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:


\alpha = (1-0.8)/(2) = 0.1

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
1-0.1 = 0.90, so
z = 1.28

Now, find the margin of error M as such


M = z*(\sigma)/(√(n))

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


M = 1.28*(2.3)/(√(717)) = 0.1

The lower end of the interval is the sample mean subtracted by M. So it is 18.5 - 0.1 = 18.4 gallons per day.

The upper end of the interval is the sample mean added to M. So it is 18.5 + 0.1 = 18.6 gallons per day.

The 80% confidence interval for the mean usage of water is between 18.4 and 18.6 gallons per day.

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