P-Value

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P-Value: meaning, method and solved example (AP Statistics) – Maths PDF


In short: P-value is the probability of seeing a sample result at least as far from the null claim as the one observed, calculated on the assumption that the null hypothesis is true.

AP StatisticsStatisticsAP Statistics Unit 3

What is P-Value?

A p-value puts a number on surprise. Imagine the null hypothesis is correct and the study is repeated endlessly; the p-value tells you what fraction of those repeats would land as far from \(H_0\) as your actual sample did.

When that fraction is tiny, your sample would be a rare event under \(H_0\), and the null becomes hard to believe. When it is large, the sample looks ordinary. Keep in mind what it is not: it is neither the chance that \(H_0\) is true nor the chance that results came from luck alone. It is a probability about the data, given \(H_0\).

How to spot it

  • Computer output or a table prints a number labeled P beside a test statistic.
  • The question asks you to interpret that number in context, in one sentence.
  • A decision is required by weighing it against a significance level.
  • A normal or t curve is shaded in one or both tails beyond the observed statistic.

The method

  1. Compute the test statistic (a \(z\) or \(t\) value) from the sample.
  2. Find the area in the tail beyond that value, in the direction of \(H_a\); for a two-sided test, double it.
  3. Interpret it in a sentence: “If \(H_0\) were true, the probability of a result this extreme is …”.
  4. Compare with \(\alpha\) to decide whether to reject \(H_0\).
02.19-2.190.01430.0143two-sided p-value = 0.0285z = (0.60 – 0.50) / 0.0456 = 2.19
Each shaded tail holds 0.0143 of the area under the curve; together they make 0.0285.

P-Value: solved example

Problem

A game developer says a loot box yields a rare item half the time. A player opens 120 boxes and gets the rare item 72 times (60%). Run a two-sided test of \(H_0: p = 0.50\) and interpret the result.

Solution

Under \(H_0\) the standard error is \(\sqrt{\frac{0.5 \times 0.5}{120}} \approx 0.0456\), so \(z = \frac{0.60 – 0.50}{0.0456} \approx 2.19\).

One tail beyond 2.19 has area \(1 – 0.9857 = 0.0143\). Doubling for both directions gives 0.0285.

Meaning: if the true rate really were 50%, only about 2.85% of 120-box samples would deviate this much, high or low.

Try it

Question 1

A test reports 0.03. Is there a 3% chance that \(H_0\) is true?

Show the answer

No. That figure is the chance of data this extreme if \(H_0\) were true, not a probability for \(H_0\) itself.

Question 2

A test reports 0.30. What does it suggest about the data?

Show the answer

A result like this would show up about 30% of the time by chance under \(H_0\), so the data offer little evidence against it.

P-Value vs. Significance Level

The significance level \(\alpha\) is a cutoff chosen before the study. The p-value is computed from the data after the sample is collected. The decision rule compares them: reject \(H_0\) when the p-value is smaller than \(\alpha\).

Significance Level (Alpha)

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