P-Values

A p-value is the probability of getting a sample result at least as extreme as the one you observed, assuming the null hypothesis is true. A small p-value means your data would be rare if H₀ were true, so it counts as evidence against H₀. If the p-value is at or below the significance level α, you reject H₀.

Before this lesson: Hypothesis Testing and Normal Probability Calculations

What a p-value measures

Every test starts by assuming H₀ is true. The p-value then asks: in that world, how often would random sampling alone give a result this far from the H₀ value, or farther?

“Extreme” means in the direction of the alternative hypothesis Hₐ. For Hₐ with >, it is the right tail. For <, it is the left tail. For ≠, it is both tails, so you double one tail.

The smaller the p-value, the stronger the evidence against H₀. A p-value is a probability about the data, computed by assuming H₀. It is not the probability that H₀ is true.

A standard normal curve on a z axis from −3 to 3. The areas to the left of z = −2.1 and to the right of z = 2.1 are shaded orange, each labeled 0.0179. Dashed lines mark z = −2.1 and z = 2.1. Together the two tails make the p-value, about 0.0358.
For a two-sided test with z = −2.1, both tails count as “at least as extreme,” so the p-value is 0.0179 + 0.0179 ≈ 0.0358.

How to find and use a p-value

Alternativep-value for a z statisticTI-84
H_a: p \gt p_0area to the right of znormalcdf(z, 10^99)
H_a: p \lt p_0area to the left of znormalcdf(−10^99, z)
H_a: p \ne p_02 × the tail area beyond z2 · normalcdf(|z|, 10^99)

For a t statistic, use tcdf(lower, upper, df) the same way (see t-distribution).

\text{p-value} \le \alpha \Rightarrow \text{reject } H_0 \qquad \text{p-value} \gt \alpha \Rightarrow \text{fail to reject } H_0

How to interpret it: “Assuming H₀ is true, there is a [p-value] probability of getting a sample result as extreme as ours, or more extreme, by chance alone.”

P-value examples

Example 1: A right-tailed p-value

A test of H₀: p = 0.30 against Hₐ: p > 0.30 gives z = 1.5. Find the p-value and make a decision at α = 0.05.

P(Z \ge 1.5) = \text{normalcdf}(1.5, 10^{99})

Find the area to the right of z = 1.5.

\approx 0.0668

Read the calculator result.

0.0668 \gt 0.05

Compare the p-value with α.

Answer: the p-value is about 0.0668, which is greater than 0.05, so fail to reject H₀. The data don’t give convincing evidence that p > 0.30.

Example 2: A two-sided p-value

A test of H₀: p = 0.5 against Hₐ: p ≠ 0.5 gives z = −2.1. Find the p-value. What is the decision at α = 0.05? At α = 0.01?

P(Z \le -2.1) = \text{normalcdf}(-10^{99}, -2.1)

Find the left-tail area.

\approx 0.0179

Read the calculator result.

2 \cdot 0.0179 = 0.0358

Double it, since both tails count.

0.0358 \lt 0.05

Compare with α = 0.05.

0.0358 \gt 0.01

Compare with α = 0.01.

Answer: the p-value is about 0.0358. Reject H₀ at α = 0.05, but fail to reject at α = 0.01. The same data can lead to different decisions, which is why α is chosen before the test.

Example 3: A p-value from a t statistic

A one-sample t-test of Hₐ: μ > μ₀ gives t = 2.4 with df = 14. Find the p-value.

P(t \ge 2.4) = \text{tcdf}(2.4, 10^{99}, 14)

Find the right-tail area with df = 14.

\approx 0.0154

Read the calculator result.

0.0154 \lt 0.05

Compare with α = 0.05.

Answer: the p-value is about 0.0154, so reject H₀ at α = 0.05. A t statistic this large would happen only about 1.5% of the time if H₀ were true.

Example 4: Interpret a p-value (word problem)

A school says 25% of its students bike to school. A student council survey of 200 randomly chosen students finds 62 who bike. Test H₀: p = 0.25 against Hₐ: p > 0.25, and interpret the p-value.

\hat{p} = \dfrac{62}{200} = 0.31

Divide bikers by n.

0.25 \cdot 0.75 = 0.1875

Multiply p₀(1 − p₀).

\dfrac{0.1875}{200} = 0.0009375

Divide by n.

\text{SE} = \sqrt{0.0009375} \approx 0.03062

Take the square root (see square roots).

0.31 - 0.25 = 0.06

Subtract p₀ from p̂.

z = \dfrac{0.06}{0.03062} \approx 1.960

Divide by SE.

P(Z \ge 1.960) = \text{normalcdf}(1.960, 10^{99})

Find the right-tail area.

\approx 0.025

Read the calculator result.

Answer: the p-value is about 0.025. Interpretation: if 25% of all students at the school bike, there is about a 2.5% chance that a random sample of 200 would show 31% or more bikers. That is rare enough to reject H₀ at α = 0.05.

Example 5: Significant but small

An online store tests whether a new button changes the click rate from p = 0.5. With 40,000 visitors, p̂ = 0.505. Find the two-sided p-value. Does the result matter in practice?

0.5 \cdot 0.5 = 0.25

Multiply p₀(1 − p₀).

\dfrac{0.25}{40000} = 0.00000625

Divide by n.

\text{SE} = \sqrt{0.00000625} = 0.0025

Take the square root.

z = \dfrac{0.505 - 0.5}{0.0025} = \dfrac{0.005}{0.0025}

Subtract p₀ from p̂.

= 2

Divide.

P(Z \ge 2) \approx 0.02275

Find one tail with normalcdf(2, 10^99).

2 \cdot 0.02275 = 0.0455

Double it for a two-sided test.

Answer: the p-value is about 0.0455, so the result is statistically significant at α = 0.05. But the change is only half a percentage point. A huge sample can make a tiny difference significant, so always ask whether the size of the effect matters.

Common mistakes

MistakeExample of the errorFix
Calling the p-value the chance H₀ is true“p = 0.03 means a 3% chance H₀ is true”It is the chance of data this extreme, assuming H₀ is true.
Forgetting to double for a two-sided testReporting 0.0179 for Hₐ: p ≠ 0.5Two-sided p-value = 2 × one tail = 0.0358.
Using the wrong tailRight-tail area for Hₐ: μ < μ₀The tail follows the direction of Hₐ.
Treating “significant” as “important”A 0.5-point change with n = 40,000Check the size of the effect too.
Saying a large p-value proves H₀“p = 0.40, so the claim is true”Fail to reject H₀; the evidence is just not strong.

Related methods

Comparing p-values with confidence intervals

A p-value says whether the data are consistent with one specific value. A confidence interval shows all the plausible values at once (see confidence intervals). Reporting both gives the clearest picture.

Where p-values come from

Each test has its own test statistic and distribution. The next lesson, One-Proportion z-Test, puts every step together for a single proportion.

Practice problems

  1. A right-tailed z-test gives z = 1.2. Find the p-value.
  2. A left-tailed z-test gives z = −2.5. Find the p-value.
  3. A two-sided z-test gives z = 1.8. Find the p-value.
  4. A test gives p-value = 0.04. What is the decision at α = 0.05? At α = 0.01?
  5. A left-tailed t-test gives t = −1.9 with df = 20. Find the p-value.
  6. A test of whether a coin favors heads gives a p-value of 0.12. Interpret it in context.
  7. True or false: a p-value of 0.03 means there is a 3% chance that H₀ is true.
  8. A test gives p-value = 0.027. For which significance levels would you reject H₀?
  9. A two-sided t-test gives t = 2.6 with df = 11. Find the p-value.
  10. Which gives stronger evidence against H₀: a p-value of 0.002 or 0.04?
Show answers

1.

P(Z \ge 1.2) = \text{normalcdf}(1.2, 10^{99})

Right-tail area.

\approx 0.1151

Read the calculator result.

Answer: about 0.1151

2.

P(Z \le -2.5) = \text{normalcdf}(-10^{99}, -2.5)

Left-tail area.

\approx 0.0062

Read the calculator result.

Answer: about 0.0062

3.

P(Z \ge 1.8) = \text{normalcdf}(1.8, 10^{99})

One tail beyond z = 1.8.

\approx 0.0359

Read the calculator result.

2 \cdot 0.0359 = 0.0718

Double it for a two-sided test.

Answer: about 0.072

4.

0.04 \lt 0.05

Compare with α = 0.05.

0.04 \gt 0.01

Compare with α = 0.01.

Answer: Reject H₀ at α = 0.05; fail to reject at α = 0.01.

5.

P(t \le -1.9) = \text{tcdf}(-10^{99}, -1.9, 20)

Left-tail area with df = 20.

\approx 0.036

Read the calculator result.

Answer: about 0.036

6.

The p-value is computed assuming the coin is fair.

Answer: If the coin is fair, there is a 12% chance of getting at least this many heads by chance alone. That is not unusual, so the data don’t show the coin favors heads.

7.

A p-value is computed by assuming H₀ is true, so it can’t be the probability that H₀ is true.

Answer: False. It is the probability of data this extreme if H₀ is true.

8.

Reject H₀ whenever p-value ≤ α.

Answer: any α of 0.027 or more, such as 0.05 or 0.10 (but not 0.01)

9.

P(t \ge 2.6) = \text{tcdf}(2.6, 10^{99}, 11)

One tail with df = 11.

\approx 0.01235

Read the calculator result.

2 \cdot 0.01235 = 0.0247

Double it for a two-sided test.

Answer: about 0.0247

10.

0.002 \lt 0.04

Compare the p-values.

Answer: 0.002. The smaller the p-value, the stronger the evidence against H₀.

P-value FAQ

What does a p-value tell you?

It tells you how likely a result as extreme as yours would be if H₀ were true. A small p-value means the data are hard to explain by chance alone.

What is a good p-value?

There is no single “good” value. A p-value at or below your significance level, usually 0.05, counts as statistically significant. Smaller p-values give stronger evidence against H₀.

How do you find a p-value for a two-tailed test?

Find the tail area beyond your test statistic on one side, then multiply by 2. On a TI-84, use 2 · normalcdf(|z|, 10^99) for a z statistic.

Is a p-value of 0.05 significant?

At α = 0.05, yes: a p-value equal to α leads you to reject H₀. Many researchers also report the exact p-value so readers can judge the strength of the evidence.

Summary

  • The p-value is the probability, assuming H₀ is true, of a result at least as extreme as the one observed.
  • Use the tail that matches Hₐ; double one tail for a two-sided test.
  • If p-value ≤ α, reject H₀; if p-value > α, fail to reject H₀.
  • Smaller p-values mean stronger evidence against H₀.
  • Statistical significance is not the same as practical importance.

Previous lesson: Hypothesis Testing

Next lesson: One-Proportion z-Test

Back to the Statistics Index