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 1.3 gallons. The mean water usage per family was found to be 15 gallons per day for a sample of 499 families. Construct the 99% confidence interval for the mean usage of water. Round your answers to one decimal place.

Answer 1

`15-2.58(1.3)/(√(499))=14.8`

`15+2.58(1.3)/(√(499))=15.2`

So on this case the 99% confidence interval would be given by (14.8;15.2)

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=15` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma =1.3` represent the population standard deviation

n=499 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

Since the Confidence is 0.99 or 99%, the value of
`\alpha=0.01` and
`\alpha/2 =0.005`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that
`z_(\alpha/2)=2.58`

Now we have everything in order to replace into formula (1):

`15-2.58(1.3)/(√(499))=14.8`

`15+2.58(1.3)/(√(499))=15.2`

So on this case the 99% confidence interval would be given by (14.8;15.2)