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In my notes, I am looking at a hypothesis testing of a test statistic distributed under the t-distribution because the population SD isn't known. The t score was computed for the test statistic, $\mu$, to be $t = -1.31$.

Then it says that if we want to test whether $\mu < \mu_0$ where $\mu_0$ is the value of our test statistic under the null hypothesis, we'd need to compute the area of the t distribution curve to the left of $t = -1.31$ to get our p value. This makes sense. Then it says if we want to test whether $\mu \neq \mu_0$, then we compute the area of the t distribution to the left of $t = -1.31$ and to the right of $t = +1.31$ (or just double the area to the left of $t = -1.31$). This also makes sense. Then it says that if we want to test whether $\mu > \mu_0$, we compute the area to the right of $t = -1.31$. My question is, should it be to the right of $t = -1.31$ or to the right of $t = 1.31$? The former makes sense mathematically, but the p-value is obviously going to be more than 0.5 in this case, so what is the point?

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    $\begingroup$ If $T = -1.31,$ then obviously $\bar X < \mu_0$ so you have no evidence that $\mu > \mu_0$ and it makes sense to have a large P-value. $\endgroup$
    – BruceET
    May 5, 2021 at 17:58

1 Answer 1

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Suppose I'm testing $H_0: \mu = 55$ vs. $H_a: \mu > 55$ and I happen to have $n = 20$ observations from $\mathsf{Norm}(50, 7).$ Then I might get $\bar X = 52.76, T = -1.32,$ and P-value $0.898,$ so I correctly fail to Reject $H_0.$

Example in R:

set.seed(2021)
t.test(rnorm(20, 50, 7), mu = 55, alt = "g")

        One Sample t-test

data:  rnorm(20, 50, 7)
t = -1.3151, df = 19, p-value = 0.8979
alternative hypothesis: true mean is greater than 55
95 percent confidence interval:
 49.81208      Inf
sample estimates:
mean of x 
52.75886 

1 - pt(-1.3151, 19)
[1] 0.897936
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  • $\begingroup$ For this particular example, on the alternative hypothesis, if you changed $\mu > 55$ to $\mu > c$ where $c > 55$, then you'd still get the same p value right? What if $c < 55$, it seems you'd still get the same p value? $\endgroup$
    – David
    May 6, 2021 at 15:03
  • $\begingroup$ Or in other words, I'm confused about the purpose of $c$ here. It seems moot? $\endgroup$
    – David
    May 6, 2021 at 15:04

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