I am working on the following question:
For a two-sided t-test to test $H_0: \mu = 6.5$ versus $H_1: \mu \neq 6.5$, based on a sample of size $n$, the test statistic value is $-2.16$ and the $p$-value is $0.026$. Then the p-value for the t-test to test $H_0: \mu = 6.5$ versus $H_1: \mu > 6.5$, based on the same sample, is
A. 0.974
B. 0.026
C. 0.013
D. 0.987
E. None of the Above
The answer to the question in the book is showing as 0.987 However, I don't understand how this is the correct answer. Given that the test statistic is 2.16, for a one sided test, the p-value would be given by the following:
$1-P(Z<2.16)$
From a normal distribution table, this comes as $1-0.9846$ which is $0.0154$
So shouldn't the answer be none of the above?
Thanks
The stated answer of D is indeed correct.
To see why, you have to look carefully at the relationship between the test statistic, which is $t = -2.16$, and the new alternative hypothesis, which is $H_1: \mu > 6.5$. I'll use this diagram from Tests of Hypotheses Based on a Single Sample by Devore and Berk as a guide:
The p-value given to you is one calculated with picture 3, i.e. a two-tailed test. This is saying that the total area captured in the sum of both tails beyond the test statistic is $0.026$ -- or in other words, that $0.013$ of the area is to the left of the stated test statistic of $t = -2.16$, and another $0.013$ of the area is to the right of its mirror image, $t = 2.16$.
Since the alternative hypothesis has the structure $H_1: \mu > 6.5$, that means your job is to shade an area to the right of something, as shown in situation 1 above. What's different, though, is that your test statistic is negative, meaning that the vertical bar should actually be drawn to the left of $t = 0$ on the curve. That means that when you shade an area, you're actually going to shade most of the area under the curve. We already established above that the amount of area to the left of the test statistic was $0.013$, which means that the area to the right of the test statistic is $1 - 0.013 = \fbox{0.987}$.
Why your answer isn't correct: First, note that you shouldn't consult a table of normal values at all; you should consult a table of $t$-distribution values, because this is a $t$-test and not a $z$-test. Also, after making that switch, you should still be computing $1 - P(T < \color{red}{\mathbf{-}}2.16)$ to capture the area to the right of $-2.16$.
Why answer C isn't correct: Because the stated alternative hypothesis is $H_1: \mu > 6.5$, not $H_1: \mu < 6.5$. Intuitively, this corresponds with expecting (before the experiment starts) for the sample to come out higher than the stated average (hence $H_1: \mu > 6.5$), but having the sample actually come out lower than the stated average (hence the negative test statistic).