At the time she was hired as a server at the Grumney Family Restaurant, Beth Bri
ID: 3022615 • Letter: A
Question
At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $78 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $4.20. Over the first 40 days she was employed at the restaurant, the mean daily amount of her tips was $79.24. At the 0.02 significance level, can Ms. Brigden conclude that her daily tips average more than $78?
a) State the null hypothesis and the alternate hypothesis.
H0: = 78 ; H1: 78
b) State the decision rule.
Reject H0 if z < 2.05
c) Compute the value of the test statistic.
d) What is your decision regarding H0?
Do not reject H0
e) What is the p-value?
H0: 78 ; H1: > 78 H0: >78 ; H1: = 78 H0: 78 ; H1: < 78H0: = 78 ; H1: 78
b) State the decision rule.
Reject H1 if z < 2.05 Reject H1 if z > 2.05 Reject H0 if z > 2.05Reject H0 if z < 2.05
c) Compute the value of the test statistic.
d) What is your decision regarding H0?
Reject H0Do not reject H0
e) What is the p-value?
Explanation / Answer
a)
Formulating the null and alternative hypotheses,
Ho: u <= 78
Ha: u > 78 [ANSWER, A]
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B)
As we can see, this is a right tailed test.
Thus, getting the critical z, as alpha = 0.02 ,
alpha = 0.02
zcrit = + 2.053748911
Hence, reject Ho when z > 2.05. [ANSWER, C]
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c)
Getting the test statistic, as
X = sample mean = 79.24
uo = hypothesized mean = 78
n = sample size = 40
s = standard deviation = 4.2
Thus, z = (X - uo) * sqrt(n) / s = 1.867249666 [ANSWER, TEST STATISTIC]
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D)
Comparing z < 2.05, we DO NOT REJECT THE NULL HYPOTHESIS. [ANSWER]
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E)
As this is a right tailed test, by table/technology
p = 0.030933365 [ANSWER]
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