I am writing out a t-test by hand and I am confused about how to mathematically express the step of calculating / looking up the $p$-value of the t-statistic. This is what I have so far from the following example:

Suppose I take a random sample of size $n=36$ (assume CLT holds) from a population with mean and standard deviation $\mu=67$ and $\sigma = 7$. Also suppose the mean of the sample is $\bar{x} = 71$. Is there sufficient statistical evidence to prove that the mean of the population is actually greater than 67?

My solution thus far is: $$ H_0: \mu = 67 \\ H_A: \mu > 67 \\ $$

$$t = \frac{71 - 67}{\frac{7}{\sqrt{36}}} = \frac{7}{6} \approx 3.428571$$

I understand that the $p$-value can be calculated in R with 1 - pt(3.428571, df=35) resulting in 0.0007848787, but how would I express this mathematically? My best guess is something like:

$$ P(\bar{X} > t_{\text{df}=35}) \approx 0.0007848787 $$

...but I'm hesitant about this because I'm pretty sure $\bar{X}$ is on a different distribution than the t-statistic.

I suppose another way of asking this question is how to mathematically express pt(3.428571, df=35).

I've searched several tutorials for answers but because they tend to be geared more towards beginners they largely gloss over the mathematical expressions (especially for the $p$-value calculations).

This question is related in the sense that it addresses how to calculate the $p$-value for left-/right-tailed tests but it doesn't really answer my specific question of how to mathematically express the $p$-value in terms of the t-statistic.

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    $\begingroup$ What is your test statistic here? What is the definition of $p$-value? $\endgroup$ Commented Nov 13, 2022 at 2:29
  • $\begingroup$ The test statistic here is the ratio defined by the difference between the sample mean and the hypothesized population mean divided by the standard error (where the standard error is the sample standard deviation divided by the square root of the sample size). In LaTex, it is $\frac{\bar{x} - \mu}{s / \sqrt{n}}$. The $p$-value in this case is the probability of observing a sample mean of 71 or greater when the population mean is in fact 67. Does this answer your question? $\endgroup$
    – jglad
    Commented Nov 13, 2022 at 2:55
  • $\begingroup$ It should be $P_{\mu_0}(T \geq t_{\text{obs}})$, where $T = \sqrt{n}(\bar{X} - \mu_0)/s$, which has a $t_{n - 1}$ distribution under $H_0$. $\endgroup$
    – Zhanxiong
    Commented Nov 13, 2022 at 2:56
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    $\begingroup$ @jglad they were not questions. They were hints. And Zhanxiong wrote it explicitly. You were calculating $t$ but wrote $\bar x$ in writing the $p$-value. $\endgroup$ Commented Nov 13, 2022 at 3:04
  • $\begingroup$ Does this answer your question? Calculate p-value for one-sample T-test in SPSS: 1 - p/2? $\endgroup$
    – dipetkov
    Commented Nov 13, 2022 at 14:00

1 Answer 1


Depending on the assumptions one is willing to make, the problem can be stated mathematically as follows.

Case I

Let $X_1,\ldots,X_n$ an i.i.d. sample with $X_i\sim N(\mu, \sigma^2)$, with $\mu, \sigma^2$ both unknown finite parameters. Furthermore, let $X = n^{-1}\sum_{i=1}^n X_i$ and $S^2 = (n-1)^{-1}\sum_{i=1}^n (X_i-\bar X)^2$ be respectively the sample average and the sample variance and let us be interested in testing $H_0:\mu=\mu_0$ against $H_1:\mu>\mu_0$.

It can be proved that

$$ \frac{\sqrt{n}(\bar {X}-\mu)}{\sigma^2}\sim N(0,1), $$

$$ \frac{(n-1)S^2}{\sigma^2}\sim \chi_{n-1}^2, $$ and that $\bar X$ and $S^2$ are independent. Thus

$$ T_n = \frac{\frac{\sqrt{n}(\bar {X}-\mu)}{\sigma^2}}{\sqrt{\frac{\frac{(n-1)S^2}{\sigma^2}}{n-1}}} = \frac{\sqrt{n}(\bar {X}-\mu)}{\sqrt{S^2}}\sim t_{n-1}, $$ is the usual $t$ statistic, which follows a $t$-Student distribution with $n-1$ degrees for freedom.

A test of size $\alpha$ has rejection $$R_{\alpha} = \{X_1,\ldots,X_n: T_n \geq t_{n-1,1-\alpha}\}.$$

If we denote by $T_{n}^{obs}$, the observed $t$ statistics, the $p$-value (see here) is given by $$ \sup_{\mu\leq \mu_{0}} P_\theta(T_n \geq T_{n}^{obs}) = P_{\mu_0}(t_{n-1}\geq T_{n}^{obs}). $$

Case II

If $\sigma^2$ is known then there is no need to estimate it and the statistic to be used is

$$ Z_n = \frac{\sqrt{n}(\bar X-\mu)}{\sigma}\sim N(0,1). ,$$

The test of size $\alpha$ is to reject $H_0$ if $Z_n\geq z_{1-\alpha}$ and the $p$-value is

$$ \sup_{\mu\leq \mu_{0}} P_\theta(Z_n \geq Z_{n}^{obs}) = P_{\mu_0}(Z\geq Z_{n}^{obs}). $$

Case III

Lastly, if we don't wish to assume the normality of the sample but only that the sample is iid from a population with mean $\mu$ and variance $\sigma^2$, both unknown, then we can use the statistic

$$ Z_n^{s} = \frac{\sqrt{n}(\bar X-\mu)}{S^2}, $$ which by CLT converges to $N(0,1)$; the test and the $p$-value are computed as in Case II.

Case IV

Again, we do not make the normality assumption but only that the sample is iid from a population with mean $\mu$ and variance $\sigma^2$, with $\sigma^2$ known. The test statistic is as in Case II, and its distribution $N(0,1)$ holds only in the limit, for large $n$.

Example. Your example seems to follow in Case IV, with $n=36$, $\bar x = 71$, $\mu_0=67$ and $\sigma^2 = 7^2$. Thus the observed test statistic is

$$ \frac{\sqrt{36}(71-67)}{7}=3.428571 $$

and $p$-value = $P(Z\geq 3.428571) = 0.000303$.

Nevertheless, to have less conservative results, in Case III and Case IV, some statisticians prefer to use the $t_{n-1}$ distribution in place of the limiting $N(0,1)$ distribution.

  • $\begingroup$ Thank you for the excellent and thorough answer! This helps a lot. If you don't mind me asking a follow-up question--let's assume that we are indeed using the $t_{n-1}$ distribution instead of the $N(0,1)$ distribution as you mention some statisticians prefer. In the given example (and following the assumptions of Case IV), does it make sense to express the $p$-value as $P(t_{35} \geq 3.428571 ) = 0.00078$, similar to the expression in Case I? $\endgroup$
    – jglad
    Commented Nov 13, 2022 at 22:44
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    $\begingroup$ Glad to see it helps. Sure. Indeed, you will typically get a higher $p$-value (and wider CIs) since the $t$ distribution has heavier tails than the standard normal. However, this will help you prevent your test from having higher $\alpha$ (or the CI to have confidence than 1-$\alpha$ lower) when you are not too far from the limiting normal distribution. $\endgroup$
    – utobi
    Commented Nov 13, 2022 at 22:54

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