In an examination of purchasing patterns of shoppers, a sample of 16 shoppers revealed that they spent, on average, $54 per hour of shopping. Based on previous years, the population standard deviation is thought to be $21 per hour of shopping. Assuming that the amount spent per hour of shopping is normally distributed, find a 90% confidence interval for the mean amount.

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=54[/tex] represent the sample mean

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

[tex]\sigma=21[/tex] represent the sample standard deviation

n=16 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 0.90 or 90%, the value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]

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

[tex]54-1.64\frac{21}{\sqrt{16}}=45.39[/tex]

[tex]54 +1.64\frac{21}{\sqrt{16}}=62.61[/tex]

So on this case the 90% confidence interval would be given by (45.39;62.61)