Question

The manager of a grocery store has selected a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3.0 minutes. It is known that the standard deviation of the checkout time is 1 minute. The 95% confidence interval for the average checkout time for all customers is ________?

Answer 1

The 95% confidence interval would be given by (2.804;3.196)

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=3.0` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma=1` represent the population standard deviation

n=100 represent the sample size

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 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that
`z_(\alpha/2)=1.96`

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

`3.0-1.96(1)/(√(100))=2.804`

`3.0+1.96(1)/(√(10))=3.196`

So on this case the 95% confidence interval would be given by (2.804;3.196)