Question

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.

Answer 1

`54-1.64(21)/(√(16))=45.39`

`54 +1.64(21)/(√(16))=62.61`

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

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=54` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma=21` 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:

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

Since the Confidence is 0.90 or 90%, the value of
`\alpha=1-0.9=0.1` and
`\alpha/2 =0.05`, 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
`z_(\alpha/2)=1.64`

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

`54-1.64(21)/(√(16))=45.39`

`54 +1.64(21)/(√(16))=62.61`

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