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

Answer:

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.

Step-by-step explanation:

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

$\alpha = \frac{1-0.9}{2} = 0.05$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 1.645*\frac{600}{\sqrt{100}} = 98.7$

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.