Question

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.

Answer 1

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.