Question 1062011: 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 $2.81. Over the first 42 days she was employed at the restaurant, the mean daily amount of her tips was $82.74. At the 0.10 significance level, can Ms. Brigden conclude that her daily tips average more than $80?
a. State the null hypothesis and the alternate hypothesis.
H0: μ ≥ 80 ; H1: μ < 80
H0: μ = 80 ; H1: μ ≠ 80
H0: μ >80 ; H1: μ = 80
H0: μ ≤ 80 ; H1: μ > 80
b. State the decision rule.
Reject H1 if z > 1.28
Reject H0 if z < 1.28
Reject H0 if z > 1.28
Reject H1 if z < 1.28
c. Compute the value of the test statistic. (Round your answer to 2 decimal places.)
Value of the test statistic
d. What is your decision regarding H0?
Do not reject H0
Reject H0
e. What is the p-value? (Round your answer to 4 decimal places.)
p-value
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! It's mu=80 vs. mu NE 80. "You can average $80/day in tips." That means it could be on either side or a two way test.
For the next part, reject Ho if |z|>1.28. That isn't one of the choices, so I would say reject Ho if z>1.28.
The test statistic is z=(82.74-80)/2.81/sqrt (42).
That is 2.74*sqrt(42)/2.81=6.32. Reject Ho.
This is highly significant at < 0.0001 level
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