Question

In an effort to estimate the mean dollars spent per visit by customers of a food store, the manager has selected a random sample of 100 cash register receipts. The mean of these was $45.67 with a sample standard deviation equal to $12.30. Assuming that he wants to develop a 90 percent confidence interval estimate, the upper limit of the confidence interval estimate is: Question 47 options: about $47.69 about $2.02 approximately $65.90 None of the above

Answer 1

`45.67-1.66(12.30)/(√(100))=43.63`

`45.67+1.66(12.30)/(√(100))=47.69`

And the best answer for this case would be:

About $47.69

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=45.67` represent the sample mean

`\mu` population mean (variable of interest)

s=12.3 represent the sample standard deviation

n=100 represent the sample size

Solution to the problem

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

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=100-1=99`

Since the Confidence is 0.90 or 90%, the value of
`\alpha=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: "=-T.INV(0.05,99)".And we see that
`t_(\alpha/2)=1.66`

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

`45.67-1.66(12.30)/(√(100))=43.63`

`45.67+1.66(12.30)/(√(100))=47.69`

And the best answer for this case would be:

About $47.69