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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?

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  • $\begingroup$ Did the question mention anything about the variance being known/unknown? $\endgroup$
    – B.Liu
    Commented Apr 6, 2021 at 23:55
  • $\begingroup$ @B.Liu unfortunately no, that is the full question $\endgroup$
    – rabito
    Commented Apr 6, 2021 at 23:55
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    $\begingroup$ 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. $\endgroup$
    – Dave
    Commented Apr 6, 2021 at 23:59
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    $\begingroup$ 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. $\endgroup$
    – Firebug
    Commented Apr 7, 2021 at 0:07
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    $\begingroup$ 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. $\endgroup$
    – Firebug
    Commented Apr 7, 2021 at 0:12

1 Answer 1

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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)
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    $\begingroup$ 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? $\endgroup$
    – rabito
    Commented Apr 7, 2021 at 16:43
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    $\begingroup$ 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. $\endgroup$
    – BruceET
    Commented Apr 7, 2021 at 20:27

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