Two-Proportion z-Test

A two-proportion z-test checks whether two population proportions differ. You compare the sample proportions p̂₁ and p̂₂ using z = (p̂₁ − p̂₂) ÷ √(p̂c(1 − p̂c)(1/n₁ + 1/n₂)), where p̂c is the pooled proportion of successes in both samples. A small p-value is evidence that p₁ and p₂ are different.

Before this lesson: One-Proportion z-Test, Sampling Distributions for Differences and Type I and Type II Errors

How the test works

Suppose a school sends text reminders to some families and not to others. More families return a form in the text group. Is that gap real, or could chance alone make a gap that big?

The null hypothesis says there is no difference: H₀: p₁ = p₂, or p₁ − p₂ = 0. If H₀ is true, both groups share one success rate. Your best estimate of that shared rate is the pooled proportion: all the successes divided by all the trials.

The z statistic says how many standard errors the gap p̂₁ − p̂₂ is from 0. The standard error uses the pooled proportion, because the test assumes H₀ is true.

Two-proportion z-test formula

\hat{p}_c = \dfrac{x_1 + x_2}{n_1 + n_2}

z = \dfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}_c(1 - \hat{p}_c)\left(\dfrac{1}{n_1} + \dfrac{1}{n_2}\right)}}

• x₁ and x₂ are the numbers of successes, and n₁ and n₂ are the sample sizes.
• H₀: p₁ = p₂. Hₐ is p₁ > p₂, p₁ < p₂ or p₁ ≠ p₂.
• The p-value is the tail area beyond z that matches Hₐ (both tails for ≠).
• On a TI-84: STAT, TESTS, 6:2-PropZTest. Enter x₁, n₁, x₂, n₂ and the direction of Hₐ.

Conditions

• Random: two independent random samples, or two groups from a randomized experiment.
• 10%: when sampling without replacement, each sample is at most 10% of its population.
• Large Counts: n₁p̂c, n₁(1 − p̂c), n₂p̂c and n₂(1 − p̂c) are all at least 10.

Two-proportion z-test examples

Example 1: Find the pooled proportion

In one class, 18 of 30 students have a part-time job. In another class, 12 of 30 do. Find each sample proportion and the pooled proportion.

\hat{p}_1 = \dfrac{18}{30} = 0.6

\hat{p}_2 = \dfrac{12}{30} = 0.4

\hat{p}_c = \dfrac{18 + 12}{30 + 30}

= \dfrac{30}{60} = 0.5

Answer: p̂₁ = 0.6, p̂₂ = 0.4 and p̂c = 0.5. The pooled proportion always falls between the two sample proportions.

Example 2: Text reminders (word problem)

A school randomly assigns 480 families to two groups of 240. One group gets a text reminder about a form, and 156 return it. The other group gets no text, and 126 return it. Do text reminders raise the return rate? Use α = 0.05.

State: H₀: p₁ = p₂ and Hₐ: p₁ > p₂, where p₁ and p₂ are the true return rates with and without a text. Plan: two-proportion z-test. The groups come from random assignment, and the counts below are at least 10.

\hat{p}_1 = \dfrac{156}{240} = 0.65

\hat{p}_2 = \dfrac{126}{240} = 0.525

\hat{p}_c = \dfrac{156 + 126}{240 + 240} = \dfrac{282}{480} = 0.5875

240 \cdot 0.5875 = 141

240 \cdot 0.4125 = 99

0.5875 \cdot 0.4125 \approx 0.2423

\dfrac{1}{240} + \dfrac{1}{240} \approx 0.008333

0.2423 \cdot 0.008333 \approx 0.002019

\text{SE} = \sqrt{0.002019} \approx 0.04493

0.65 - 0.525 = 0.125

z = \dfrac{0.125}{0.04493} \approx 2.782

P(Z \ge 2.782) = \text{normalcdf}(2.782, 10^{99})

\approx 0.0027

0.0027 \lt 0.05

Conclude: the p-value is about 0.0027, which is less than 0.05, so reject H₀. There is convincing evidence that text reminders raise the return rate. Because this was a randomized experiment, the texts likely caused the increase.

Example 3: A two-sided test

A survey takes random samples of students at two large high schools. At School A, 54 of 300 walk to school. At School B, 45 of 200 walk. Is there a difference in the proportions who walk? Use α = 0.05.

\hat{p}_1 = \dfrac{54}{300} = 0.18

\hat{p}_2 = \dfrac{45}{200} = 0.225

\hat{p}_c = \dfrac{54 + 45}{300 + 200} = \dfrac{99}{500} = 0.198

200 \cdot 0.198 = 39.6

0.198 \cdot 0.802 \approx 0.1588

\dfrac{1}{300} + \dfrac{1}{200} \approx 0.008333

0.1588 \cdot 0.008333 \approx 0.001323

\text{SE} = \sqrt{0.001323} \approx 0.03637

0.18 - 0.225 = -0.045

z = \dfrac{-0.045}{0.03637} \approx -1.237

P(Z \le -1.237) = \text{normalcdf}(-10^{99}, -1.237)

\approx 0.1080

2 \cdot 0.1080 = 0.2160

0.2160 \gt 0.05

Answer: with H₀: p₁ = p₂ and Hₐ: p₁ ≠ p₂, the p-value is about 0.216, so fail to reject H₀. The data don’t give convincing evidence that the walking rates at the two schools differ.

Example 4: When the test can’t be used

A small study finds that 3 of 40 people on Diet A and 5 of 50 people on Diet B get a headache. Can you use a two-proportion z-test?

\hat{p}_c = \dfrac{3 + 5}{40 + 50} = \dfrac{8}{90} \approx 0.0889

40 \cdot 0.0889 \approx 3.56

3.56 \lt 10

Answer: No. The expected count of headaches in the first group is only about 3.56, so the Large Counts condition fails and the normal model is not reliable.

Example 5: Matching a test with an interval

Use the walking data from Example 3. Build a 95% confidence interval for p₁ − p₂. Does it agree with the test?

0.18 \cdot 0.82 = 0.1476

\dfrac{0.1476}{300} = 0.000492

0.225 \cdot 0.775 = 0.174375

\dfrac{0.174375}{200} = 0.000871875

0.000492 + 0.000871875 = 0.001363875

\text{SE} = \sqrt{0.001363875} \approx 0.03693

1.96 \cdot 0.03693 \approx 0.0724

-0.045 - 0.0724 = -0.1174

-0.045 + 0.0724 = 0.0274

Answer: about (−0.117, 0.027). The interval contains 0, so “no difference” is plausible. That matches the two-sided test, which failed to reject H₀. This interval is the confidence interval for a difference in proportions.

Common mistakes

Related methods

Estimating the size of the difference

A test answers “is there a difference?” To estimate how big the difference is, use the confidence interval for a difference in proportions, which uses p̂₁ and p̂₂ separately, not p̂c.

More than two groups

To compare proportions across three or more groups, use a chi-square test. With exactly two groups, it gives the same p-value as a two-sided two-proportion z-test.

Practice problems

1. In one sample, 30 of 80 people say yes. In another, 42 of 120 say yes. Find the pooled proportion.
2. Use Problem 1. Find the standard error for a two-proportion z-test.
3. Use Problems 1 and 2. Find z.
4. Use Problem 3. Find the p-value for Hₐ: p₁ ≠ p₂.
5. Samples of 50 and 60 give a pooled proportion of 0.1. Is the Large Counts condition met?
6. Write H₀ and Hₐ to test whether the proportion of left-handed students differs between boys and girls.
7. A website shows Ad A to 800 random visitors and Ad B to 800 others. Ad A gets 64 clicks and Ad B gets 40. Test whether Ad A has a higher click rate. Use α = 0.05.
8. A two-proportion z-test comparing two towns’ recycling rates gives a p-value of 0.21. Write the conclusion at α = 0.05.
9. Why does the test use the pooled proportion in the standard error?
10. In a randomized experiment, a two-proportion z-test of a new study app gives a p-value of 0.003. Can you conclude the app causes higher pass rates at α = 0.01?

\hat{p}_c = \dfrac{30 + 42}{80 + 120} = \dfrac{72}{200}

= 0.36

0.36 \cdot 0.64 = 0.2304

\dfrac{1}{80} + \dfrac{1}{120} \approx 0.02083

0.2304 \cdot 0.02083 \approx 0.0048

\text{SE} = \sqrt{0.0048} \approx 0.06928

\hat{p}_1 = \dfrac{30}{80} = 0.375

\hat{p}_2 = \dfrac{42}{120} = 0.35

0.375 - 0.35 = 0.025

z = \dfrac{0.025}{0.06928} \approx 0.361

P(Z \ge 0.361) = \text{normalcdf}(0.361, 10^{99}) \approx 0.3590

2 \cdot 0.3590 = 0.718

50 \cdot 0.1 = 5

5 \lt 10

\hat{p}_A = \dfrac{64}{800} = 0.08

\hat{p}_B = \dfrac{40}{800} = 0.05

\hat{p}_c = \dfrac{64 + 40}{800 + 800} = \dfrac{104}{1600} = 0.065

800 \cdot 0.065 = 52

0.065 \cdot 0.935 = 0.060775

\dfrac{1}{800} + \dfrac{1}{800} = 0.0025

0.060775 \cdot 0.0025 \approx 0.0001519

\text{SE} = \sqrt{0.0001519} \approx 0.01232

0.08 - 0.05 = 0.03

z = \dfrac{0.03}{0.01232} \approx 2.435

P(Z \ge 2.435) = \text{normalcdf}(2.435, 10^{99}) \approx 0.0074

0.0074 \lt 0.05

0.21 \gt 0.05

0.003 \lt 0.01

Two-proportion z-test FAQ

What is the formula for a two-proportion z-test?

The test statistic is z = (p̂₁ − p̂₂) ÷ √(p̂c(1 − p̂c)(1/n₁ + 1/n₂)), where p̂c is the pooled proportion. The p-value is the matching tail area under the standard normal curve.

What is the pooled proportion?

The pooled proportion is the total number of successes in both samples divided by the total sample size: p̂c = (x₁ + x₂)/(n₁ + n₂). It estimates the shared rate that H₀ assumes.

How do you do a 2-PropZTest on a TI-84?

Press STAT, choose TESTS, then 6:2-PropZTest. Enter x₁, n₁, x₂, n₂ and the direction of Hₐ, then choose Calculate. The calculator shows z, the p-value and p̂c.

When do you pool in a two-proportion test?

Pool only in the significance test, because it assumes p₁ = p₂. A confidence interval for p₁ − p₂ doesn’t assume the proportions are equal, so it uses p̂₁ and p̂₂ separately.

Summary

• Use a two-proportion z-test to check H₀: p₁ = p₂.
• Pool the data: p̂c = (x₁ + x₂)/(n₁ + n₂).
• z = (p̂₁ − p̂₂) ÷ √(p̂c(1 − p̂c)(1/n₁ + 1/n₂)).
• Check Random, 10% and Large Counts with p̂c for both groups.
• Find the p-value that matches Hₐ, compare it with α and conclude in context. Only a randomized experiment supports cause and effect.

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