Problem 8, 13, 53, 54, Chapter 10

Problem 8problem 13problem 53problem 54ch 10 Problem 8at The Time She

At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $80 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $9.95. Over the first 35 days she was employed at the restaurant, the mean daily amount of her tips was $84.85. At the .01 significance level, can Ms. Brigden conclude that her daily tips average more than $80?

a) State the null hypothesis and alternative hypothesis

b) State the decision rule

c) Compute the value of the test statistic

d) Determine your decision regarding H0

e) Find the p-value and interpret it

Paper For Above instruction

The scenario presents a hypothesis testing problem regarding whether Beth Brigden's average daily tips exceed a certain benchmark. To investigate this, statistical hypothesis testing is essential, specifically a one-sample z-test for the population mean. The null hypothesis (H0) assumes no increase over the stated average, while the alternative hypothesis (H1) assesses whether her tips are statistically significantly higher than $80.

Hypotheses Formulation

The null hypothesis (H0) posits that the true mean of Brigden's daily tips is $80, the value she was initially told:

H0: μ = $80

The alternative hypothesis (H1) suggests that her average tips are greater than $80:

H1: μ > $80

Decision Rule

The significance level (α) is given as 0.01. Since this is a one-tailed test (testing if the mean is greater than $80), the critical z-value corresponding to an upper-tail probability of 0.01 must be used. Consulting standard normal distribution tables, the critical z-value is approximately 2.33.

The decision rule is: reject H0 if the calculated z-statistic > 2.33. Otherwise, do not reject H0.

Calculation of Test Statistic

The test statistic for a z-test for the mean when the population standard deviation is known is given by:

z = (x̄ - μ0) / (σ / √n)

where:

- x̄ (sample mean) = $84.85

- μ0 (hypothesized mean) = $80

- σ (population standard deviation) = $9.95

- n (sample size) = 35

Calculating:

z = (84.85 - 80) / (9.95 / √35)


= 4.85 / (9.95 / 5.916)


= 4.85 / 1.684


≈ 2.88

Decision Based on the Test Statistic

The calculated z-value of approximately 2.88 exceeds the critical value of 2.33. Therefore, we reject the null hypothesis at the 0.01 significance level, indicating there is statistically significant evidence that Beth Brigden's average daily tips are greater than $80.

P-Value Calculation and Interpretation

The p-value corresponds to the probability of observing a z-value as extreme as the calculated value, assuming H0 is true. For z = 2.88, the p-value from standard normal distribution tables or software is approximately 0.0019.

This p-value is less than the significance level (0.01), reaffirming the decision to reject H0. It implies that there is less than a 0.19% chance of observing such an increase in tips if the true mean were $80, providing strong evidence that her mean tips are indeed higher than $80.

Conclusion

Based on the hypothesis test, the evidence at the 1% significance level supports the conclusion that Beth Brigden's average daily tips exceed $80. This indicates her performance is statistically significantly better than the initial expectation.

References

  • Larsen, R. J., & Marx, M. L. (2012). An Introduction to Mathematical Statistics and Its Applications (4th ed.). Pearson.
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