Question

A city planner wants to estimate the average monthly residential water usage in the city. He selected a random sample of 100 households from the city, which gave a mean water usage of 4500 gallons over a 1-month period. Based on earlier data, the population standard deviation of the monthly residential water usage in this city is 600 gallons. Make a 90% confidence interval for the average monthly residential water usage for all households in this city

Answer 1

The 90% confidence interval for the average monthly residential water usage for all households in this city is between 4401.3 gallons and 4598.7 gallons.

Explanation:

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

`\alpha = (1-0.9)/(2) = 0.05`

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
`1-0.05 = 0.95`, so
`z = 1.645`

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 = 1.645*(600)/(√(100)) = 98.7`

The lower end of the interval is the sample mean subtracted by M. So it is 4500 - 98.7 = 4401.3 gallons

The upper end of the interval is the sample mean added to M. So it is 4500 + 98.7 = 4598.7 gallons

The 90% confidence interval for the average monthly residential water usage for all households in this city is between 4401.3 gallons and 4598.7 gallons.