Question 1176852: A university claims that the average tertiary entry (TE) score of applicants to its business studies
program has increased during the past three years. Three years ago, the mean and the standard deviation of TE scores of the university’s business studies program applicants were 920 and 20, respectively.
In a sample of 36 of this year's applicants for the program, the mean TE score was 925. At the 5% level of significance, can we conclude that the university’s claim is true? (Assume
that the standard deviation is unchanged.)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! population mean is assumed to be 920 with standard deviation equal to 20.
ample size is 36.
standard error = population standard deviation divided by square root of sample size = 20 / sqrt(36) = 20/6 = 3 and 1/3 = 10/3.
z-score = (x - m) / s
x = mean of the sample.
m = assumed mean of the population.
s = standard error
z-score = (925 - 920) / (10/3) = 1.5
one tailed alpha on the right of the normal distribution curve is 5% / 100 = .05
area to the right of z-score of 1.5 = .0668072287.
since the right tailed alpha of the sample z-score is greater than .05, the results are not considered significant and the assumption that the mean has increased is rejected.
the explanation is that the increase is not significant enough to determine that it was not due to random variation in the mean scores of samples of 36 elements each.
the determination is based on the assumption that the critical alpha is 5%.
any sample alpha higher than 5% is not considered significant.
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