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

Answer:

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

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:

$\alpha = \frac{1-0.8}{2} = 0.1$

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

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

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

$M = 1.28*\frac{2.3}{\sqrt{717}} = 0.1$

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.