# Finding the P-value from T-test

Suppose $$X_1,…,X_n$$ are modeled as normally distributed with mean $$μ$$, and have T statistic $$1.4$$.

What is the p-value for testing $$H_0:μ=0$$ against $$H_a:μ<0$$?

How can I find the p-value if I am not given degrees of freedom?

• Did the question mention anything about the variance being known/unknown? Commented Apr 6, 2021 at 23:55
• @B.Liu unfortunately no, that is the full question Commented Apr 6, 2021 at 23:55
• That the test statistic is stated to be a t-stat and not a z-stat tells me that the variance is unknown. The best that I can think to do is to give an asymptotic p-value, assuming the sample size goes to infinity and the null distribution converges to standard normal.
– Dave
Commented Apr 6, 2021 at 23:59
• I might've misunderstand the question, but the degrees of freedom is given in terms of the sample size. Having the T-statistic and the degrees of freedom is sufficient to deduce a p-value. Commented Apr 7, 2021 at 0:07
• Of course it is: it's $n$. You can just write the p-value in terms of the t-distribution cdf. My former statistics professors would probably have meant something akin to that in a question written similarly to the one you posted here. Commented Apr 7, 2021 at 0:12

To begin with required structure, suppose I have normal data in vector x with summary statistics as shown:

summary(x);  length(x);  sd(x)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
-2.6342 -1.0172 -0.6990 -0.6650  0.2648  0.4705
[1] 20           # sample size
[1] 0.9254546    # sample SD


A t test of $$H_0: \mu=0$$ against $$H_a: \mu < 0$$ will reject for a sufficiently small (negative) value of the t statistic. Results from t.test in R are as shown below: $$T = -3.2135$$ is sufficiently small to give P-value $$0.002287,$$ so we reject $$H_0$$ at the 5% level.

t.test(x, mu=0, alt="less")

One Sample t-test

data:  x
t = -3.2135, df = 19, p-value = 0.002287
alternative hypothesis: true mean is less than 0
95 percent confidence interval:
-Inf -0.3071727
sample estimates:
mean of x
-0.6649959


The P-value is the probability under $$H_0$$ (used to compute $$T)$$ that $$T \le -3.2135.$$ Typically, one cannot find exact P-values from a printed t table because not enough percentage points are given. But we can use R to show how the exact P-value can be computed. In R, pt is a CDF of the t distribution matching the given DF. Except for minor rounding, the result is as reported in the output of t.test.

pt(-3.2135, 19)
[1] 0.002286805


For a left-tailed test as above, it makes little sense to ask for the P-value matching $$T = 1.4$$ because we are clearly not going to reject $$H_0$$ in favor of $$H_a$$ based on a positive t statistic. The P-value for a test against $$H_a: \mu > 0$$ based on $$T =1.4$$ is the probability under $$H_0$$ that $$T \ge 1.4,$$ which is #0.0888;$we could reject at the 10% level, but not at the 5% level. 1 - pt(1.4, 19) [1] 0.08881538  You are correct that you need to know the degrees of freedom in order to find the P-value. However, for large sample sizes you don't need to know the exact DF in order to get an approximate P-value. Suppose $$n = 50, 100, 200,$$ so that degrees of freedom are $$\nu = 49, 99, 199.$$ If we are testing $$H_0: \mu = 0$$ against $$H_a: \mu < 0,$$ and $$T = -3.2135,$$ then the P-values would all be near $$0.003,$$ leading to rejection at the 1% level (and below). dt(-3.2135, 49); dt(-3.2135, 99); dt(-3.2135, 199) [1] 0.003329379 [1] 0.002787947 [1] 0.00253066  Note: The sample x used in the example above was sampled in R with the following code: set.seed(2021) x = rnorm(20, -1 , .85)  • Thanks this is very helpful. It makes sense that to find$\mu > 0$or$\mu < 0$to use the pt command; however, if you are trying to find p-value that$\mu \neq 0$is this just 1 then? Commented Apr 7, 2021 at 16:43 • For a 2-sided test, find total probability in two tails beyond$\pm T.$Example: If$T = -3.2135,\$ then use code 2*pt(-3.2135, 19). // When using t.test, omit alt parameter because two-sided is the default. Then the P-value in output is suitably doubled. Commented Apr 7, 2021 at 20:27