problem.
8.2.7. At a certain university, an English proficiency
test must be passed before undergraduates can receive their
degrees. Some students have been known to take the test twice
before passing it. A random sample of 25 such students was taken,
and the number of “comma errors” was counted on the first and
second tests. The average difference on the two tests was a
decrease of 2.4 errors. The standard deviation was 6.0.
If a college administrator wants to test to show that there was
no improvement, what are the null and alternative hypotheses?
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the first equation we'd reject the null hypothesis because it is
less than .05, but in the other we wouldn't reject the null
hypothesis. In my textbook, the table shows that t .05, 24 = 1.711.
Since t (2.00) is greater than 1.711, we reject null correct? In my
textbook, the correct answer is to reject the null.
Ho: μd = 0
Ha: pd > 0
to = 2.00
0.025 < 1/2 * pualue < 0.05