In: Statistics and Probability
statistics and probability questions, solve clearly, show steps and in 30 minutes for thumbs up vote
A government report on remuneration for university
administrators in Canada lists average salaries for various
executive roles. The total annual salary for university presidents
is listed as $500,000. To test this reported figure, at LOC = 95%,
a random sample of 16 Canadian university presidents’ total annual
salaries was collected. The raw data is below (in $/year):
618,000, 464,000, 686,000, 500,000, 746,000, 704,000, 596,000,
629,000,
523,000, 756,000, 561,000, 608,000, 691,000, 663,000, 529,000,
442,000.
a) Find the mean and standard deviation using a spreadsheet.
b) Assess the government report’s claim, using the critical-value
method.
c) Use the p-value method to determine if there are any
commonly-used LOC values for which the conclusion would be opposite
to your answer from Part (a).
a)
b) Now we will conduct T test for testing the mean.
Since the null hypothesis is rejected we have evidence that government report’s claim that the total annual salary for university presidents is different than $500,000
c) Using the P-value approach: The p-value is p = 0.0005, and since p = 0.0005<0.05, it is concluded that the null hypothesis is rejected.
Commonly the least accepted LOC is 99% that is .
Hence there are none commonly used LOC values for which the conclusion would be opposite to your answer.
For Answer to be opposite the