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.

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.

Answer 2

To determine if the average tip Maureen received last year was more than $4.75, we can perform a hypothesis test using the given information.

Here are the steps to conduct the hypothesis test:

Step 1: State the null and alternative hypotheses:
The null hypothesis (H0): Average tip received by Maureen last year is $4.75.
The alternative hypothesis (Ha): Average tip received by Maureen last year is more than $4.75.

Step 2: Determine the significance level:
In this case, the significance level is given as 1%, which means the level of significance (alpha) is 0.01.

Step 3: Calculate the test statistic:
We can use the t-test statistic formula since the population standard deviation is unknown:

t = (X - μ) / (s / sqrt(n))

where:
X = Sample mean ($5.25)
μ = Population mean ($4.75)
s = Sample standard deviation ($1.15)
n = Sample size (52)

Plugging in the given values, we get:
t = (5.25 - 4.75) / (1.15 / sqrt(52))

Step 4: Determine the critical value:
Since we are testing whether the average tip is higher than $4.75, it is a one-tailed test in the right direction. At a 1% significance level, using a t-distribution with (n-1) degrees of freedom (51), we can find the critical value.

The critical value can be found using statistical tables or software. For a one-tailed test at a 1% level of significance with 51 degrees of freedom, the critical value is approximately 2.685.

Step 5: Compare the test statistic with the critical value:
If the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, if the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis.

Step 6: Make a conclusion:
If we reject the null hypothesis, it suggests that the average tip received by Maureen last year was more than $4.75. If we fail to reject the null hypothesis, it means there is not enough evidence to support the claim that the average tip was more than $4.75.

Now, you can compute the test statistic using the formula and compare it with the critical value to draw a conclusion about Maureen's claim.