A p-value measures the probability of observing a test statistic as extreme as, or more extreme than, the test statistic actually observed, given that the null hypothesis is true.

\[ p = P\left(|T| \ge |t_{\text{obs}}| \mid H_0\right) \]

If the p-value is small, it means the observed result would be unlikely if the null hypothesis were true.

A common rule is:

• If \(p \le 0.05\): reject \(H_0\) (there is evidence against the null hypothesis)
• If \(p > 0.05\): fail to reject \(H_0\) (there is not strong evidence against it)

1. State a null hypothesis
   
2. Choose a test statistic
   
3. Find its distribution under \(H_0\)
   
4. Compare observed value to that distribution

\[ H_0: \mu = 70 \]

\[ H_a: \mu \neq 70 \]

\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]

\[ t \sim t_{n-1} \]

We analyze exam scores for 40 students.

Sample mean: 71

Test statistic: 2.048

p-value: 0.047

curve_data <- data.frame(t = seq(-4, 4, length.out = 400))
curve_data$density <- dt(curve_data$t, df = n - 1)

tail_left <- subset(curve_data, t <= -abs(t_stat))
tail_right <- subset(curve_data, t >= abs(t_stat))

The curve shows the distribution of the test statistic assuming the null hypothesis is true.

The dashed lines mark the observed test statistic.

The shaded red areas represent values as extreme or more extreme than the observed value.

The total shaded area is the p-value.

t.test(scores, mu = 70)

## 
##  One Sample t-test
## 
## data:  scores
## t = 2.0478, df = 39, p-value = 0.04735
## alternative hypothesis: true mean is not equal to 70
## 95 percent confidence interval:
##  70.01227 71.98773
## sample estimates:
## mean of x 
##        71

The p-value is 0.047.

This provides evidence against the null hypothesis.

Smaller p-values indicate stronger evidence against \(H_0\).