A full-time PhD students received an average salary of $12, 837 according to U.S
ID: 3238185 • Letter: A
Question
A full-time PhD students received an average salary of $12, 837 according to U.S. Department of Education. The dean of graduate studies at a large state University feels that PhD. Students in his state cam more than this. He surveys 44 randomly selected students and finds their average salary is $14, 445 with a standard deviation of S150. With a alpha = 0.05, is the dean correct? A manufacturer states that the average lifetime of its television sets is more than 84 months. The standard deviation of the population is 10 months. One hundred sets are randomly selected and tested. The average lifetime of the sample is 85.1 months. Test the claim that the average lifetime of the sets is more than 84 months, and find the P-value. On the basis of the P-value, should the null hypothesis be rejected at a alpha = 0.01? (Hypothesis testing)Explanation / Answer
(6a)
Data:
n = 44
= 12837
s = 150
x-bar = 14445
Hypotheses:
Ho: 12837
Ha: > 12837
Decision Rule:
= 0.05
Degrees of freedom = 44 - 1 = 43
Critical t- score = 1.681070704
Reject Ho if t > 1.681070704
Test Statistic:
SE = s/n = 150/44 = 22.61335084
t = (x-bar - )/SE = (14445 - 12837)/22.6133508433323 = 71.10843551
p- value = 1.55347E-46
Decision (in terms of the hypotheses):
Since 71.10843551 > 1.681070704 we reject Ho and accept Ha
Conclusion (in terms of the problem):
There is sufficient evidence that > 12837. The dean is right.
(6b)
Data:
n = 100
= 84
s = 10
x-bar = 85.1
Hypotheses:
Ho: 84
Ha: > 84
Decision Rule:
= 0.01
Critical z- score = 2.326347874
Reject Ho if z > 2.326347874
Test Statistic:
SE = s/n = 10/100 = 1
z = (x-bar - )/SE = (85.1 - 84)/1 = 1.1
p- value = 0.135666061
Decision (in terms of the hypotheses):
Since 1.1 < 2.326347874 we fail to reject Ho
Conclusion (in terms of the problem):
There is no sufficient evidence that > 84. The claim is not valid.
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