Question

Maureen is a cocktail hostess in a very exclusive private club. The IRS is auditing her tax return. Maureen claims that her average tip last year was $4.75. To support this claim, she sent the IRS a random sample of 52 credit card receipts showing her bar tips. When the IRS got the receipts, they computed the sample average and found it to be X = $5.25 with sample deviations s = $1.15. Do these receipts indicate that the average tip Maureen received last year was more than $4.75? Use a 1% level of signicance.

Answer 1

Use a one-sample z-test.

z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

With your data:
z = (5.25 - 4.75)/(1.15/√52) = ?

Finish the calculation.

Check a z-table at .01 level of significance for a one-tailed test.
If the z-test statistic exceeds the critical value from the z-table, reject the null. If the z-test statistic does not exceed the critical value from the z-table, do not reject the null. Draw your conclusions from there.

I hope this will help get you started.