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

And the best answer for this case would be:

About $47.69

Step-by-step 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".

represent the sample mean

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:

  (1)

In order to calculate the critical value we need to find first the degrees of freedom, given by:

Since the Confidence is 0.90 or 90%, the value of and , 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

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

And the best answer for this case would be:

About $47.69

a 10 percent increase

answer: it's d

step-by-step explanation: