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.

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".

[tex]\bar X=15[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma =1.3[/tex] 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:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], 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 [tex]z_{\alpha/2}=2.58[/tex]

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

[tex]15-2.58\frac{1.3}{\sqrt{499}}=14.8[/tex]

[tex]15+2.58\frac{1.3}{\sqrt{499}}=15.2[/tex]

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