50 of 100 against 30 of 100
- Input
- x₁ = 50, n₁ = 100, x₂ = 30, n₂ = 100
- Expected output
- z = 2,887, p = 0,0039
p̄ = 0.4; odds ratio = 2.33.
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two proportion z test calculator
In the "z, 2 proportions" tab, enter the successes and the total of each group (for example, 56 conversions in 1000 visits and 78 in 1000). The calculator uses the pooled proportion under H0, computes z and the p-value, and shows the difference, the confidence interval, the odds ratio and the relative risk.
p̄ = 0.4; odds ratio = 2.33.
95% CI: −0.0439 to −0.0001.
Critical value −1.645.
95% CI of the difference: 0.0012 to 0.1252.
95% Wilson CI: 0.502 to 0.691.
The p-value is the probability of observing a statistic as extreme as your sample, or more, assuming H0 is true. It is not the probability that H0 is true. With p = 0.0794 and α = 0.05, the data are not "unlikely enough" to reject H0; with p = 0.0028, they are. Very small p-values appear in scientific notation: 2.5e-7 means 0.00000025.
In a two-sided test they are equivalent: z² = χ² without Yates correction, and the p-values coincide. The z test allows one-sided tails and gives the CI of the difference; the chi-square generalizes to more than two groups.
Yes, in the "z, 1 proportion" tab: enter successes, total and p₀. It uses the standard error under H0 and gives the Wilson interval, more reliable than Wald for proportions close to 0 or 1.
It depends on the minimum difference that matters and the desired power. With small samples only huge effects become significant. Use the sample size calculator to plan before collecting, and do not stop the test just because the p-value crossed 0.05.
10 values read (separator: comma).
10 values read (separator: comma).
Interpretation
There is not enough statistical evidence, at significance level α = 0.05, that the difference between the population means (group 1 − group 2) is different from 0 (observed difference: -1.58; p-value = 0.079394). Effect size: Cohen d = -0.8322 (large). Failing to reject H0 does not prove it true: the data simply do not provide enough evidence against it, which can also reflect a small sample.
| Group | Size (n) | Mean | s |
|---|---|---|---|
| Group 1 | 10 | 0.75 | 1.789 |
| Group 2 | 10 | 2.33 | 2.0022 |
Difference between means (group 1 − group 2): -1.58. Standard error: 0.8491.
F test of variances: F = 0.7983, p = 0.742720. No sign of different variances.
Levene (Brown-Forsythe): F = 0.2482, p = 0.624374.
SE = √(s₁²/n₁ + s₂²/n₂) = √(3.2006/10 + 4.009/10) = 0.8491t = ((x̄₁ − x̄₂) − δ₀) / SE = ((0.75 − 2.33) − 0) / 0.8491 = -1.8608df = (s₁²/n₁ + s₂²/n₂)² / [ (s₁²/n₁)²/(n₁−1) + (s₂²/n₂)²/(n₂−1) ] = 17.7765p = 2 · P(T ≥ |t|) = 2 · P(T ≥ 1.8608) = 0.079394tcrit = ±2.1028CI 95% = [-3.3655; 0.2055]Two independent samples, each approximately normal (or large). Welch does not require equal variances; the pooled t does. If the same subjects were measured twice, use the paired t.