At the time she was hired as a server at the Grumney Family Restaurant, Beth Bri
ID: 3155507 • Letter: A
Question
At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $85 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $4.50. Over the first 47 days she was employed at the restaurant, the mean daily amount of her tips was $86.06. At the 0.01 significance level, can Ms. Brigden conclude that her daily tips average more than $85? a. State the null hypothesis and the alternate hypothesis. H0: 85 ; H1: > 85 H0: >85 ; H1: = 85 H0: 85 ; H1: < 85 H0: = 85 ; H1: 85 b. State the decision rule. Reject H1 if z > 2.33 Reject H0 if z < 2.33 Reject H1 if z < 2.33 Reject H0 if z > 2.33 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
Explanation / Answer
a)
Formulating the null and alternative hypotheses,
Ho: u <= 85
Ha: u > 85 [ANSWER, A]
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b)
As we can see, this is a right tailed test.
Thus, getting the critical z, as alpha = 0.01 ,
alpha = 0.01
zcrit = + 2.33
Hence,
Reject Ho when z > 2.33. [ANSWER, D]
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C)
Getting the test statistic, as
X = sample mean = 86.06
uo = hypothesized mean = 85
n = sample size = 47
s = standard deviation = 4.5
Thus, z = (X - uo) * sqrt(n) / s = 1.614887528 [ANSWER]
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d)
As z<2.33, then we DO NO REJECT HO.
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e)
Also, the p value is, as this is right tailed,
p = 0.053167527 [ANSWER]
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