P-Value Visualizer

The simulator has one slider and two rows of buttons:

The plot is the null distribution: a standard normal curve from −4 to 4. The P-value region is shaded in coral, and a dashed coral line marks z (plus −z in two-sided mode, where both tails are shaded). Faint dashed lines mark ±1.645, ±1.96 and ±2.576, the critical values for common levels.

The results box lists the Test statistic, the P-value with how it was found (P(Z ≤ z), P(Z ≥ z) or 2·P(Z ≥ |z|)) to four decimals, or "< 0.0001" when tiny, and α. A verdict bar then reads either REJECT H0 (statistically significant) or FAIL TO REJECT H0 (not significant).

The verdict compares the unrounded P-value with α. At z = 1.96 two-sided, the box shows 0.0500 but the verdict says reject, because the unrounded value is 0.049996. On real work, report the P-value and compare it with α yourself.

Slide z toward 0 and the shaded area grows. At z = 0, the right-tailed P-value is 0.5000 and the two-sided P-value is 1.0000: a statistic right at the center of the null distribution is no evidence against $H_0$ at all.

Try z = 1.96 with Right, then with Two-sided: the P-value doubles from 0.0250 to 0.0500. The same statistic is less convincing when the alternative did not predict a direction.

For a one-sample z-test for a proportion with null hypothesis $H_0\colon p = p_0$:

This is the formula sheet's standardized test statistic, (statistic − parameter)/(standard error of the statistic). The standard error uses $p_0$, because the test assumes $H_0$ is true.

The P-value is the probability, assuming $H_0$ is true, of getting a statistic at least as extreme as the one observed, in the direction of $H_a$. If P-value $< \alpha$, reject $H_0$: there is convincing evidence for $H_a$. If P-value $\ge \alpha$, fail to reject $H_0$: there is not convincing evidence for $H_a$.

Problem: A phone company says 60% of its customers would recommend it. A manager suspects the true proportion is higher. In a random sample of 200 customers, 132 would recommend it. Test at $\alpha = 0.05$.

Step 1: Hypotheses. $H_0\colon p = 0.60$ and $H_a\colon p > 0.60$, where $p$ is the proportion of all the company's customers who would recommend it.

Step 2: Conditions. Random sample; 200 is less than 10% of all customers; $np_0 = 200(0.60) = 120$ and $n(1-p_0) = 200(0.40) = 80$, both at least 10.

Step 3: Test statistic. $\hat{p} = 132/200 = 0.66$. $\sigma_{\hat{p}} = \sqrt{0.60(0.40)/200} \approx 0.0346$, so $z = (0.66 - 0.60)/0.0346 \approx 1.73$.

Step 4: P-value. $P(Z \ge 1.73) \approx 0.0418$. If 60% of customers really would recommend the company, there is about a 4.2% chance that a random sample of 200 gives a $\hat{p}$ of 0.66 or higher.

Step 5: Conclude. Because $0.0418 < 0.05$, reject $H_0$. There is convincing evidence that more than 60% of the company's customers would recommend it.

Check it in the simulator: set z to 1.73 with Right and α = 0.05. The P-value reads 0.0418 and the verdict is REJECT. Click α = 0.01: the same P-value now gives FAIL TO REJECT. Click Two-sided: the P-value doubles to 0.0836, which would not be significant at 0.05.

P-values are topic 3.6, part of significance tests for a population proportion. Free-response questions regularly ask you to carry out a full test, or to interpret a given P-value in context, which means stating the assumption that $H_0$ is true, the observed result, and the direction of $H_a$.