Difference between two means (large samples)

The method

For large independent samples:

z = ((x̄ - ȳ) - (μx - μy)) / sqrt(sx²/nx + sy²/ny) approximately N(0, 1).

Two approximations are made: X̄ and Ȳ are approximately normal (Central Limit Theorem, because the samples are large), and σx² ≈ sx², σy² ≈ sy² (good estimates, because the samples are large).

From summary statistics

Group A: n = 50, Σx = 1240, Σx² = 31580, so x̄ = 24.8 and sx² = (31580 - 1240²/50)/49 = 16.8980.

Group B: n = 60, Σy = 1404, Σy² = 33560, so ȳ = 23.4 and sy² = 11.9729.

Standard error = sqrt(16.8980/50 + 11.9729/60) = 0.7331; z = 1.4/0.7331 = 1.9096.

Reading the result carefully

z = 1.910. For H1: μA ≠ μB at 5% (critical values ±1.96) this is not significant; for H1: μA > μB at 5% (1.6449) it is significant. The alternative hypothesis must be decided from the question's wording before you calculate.

Assumptions the mark scheme credits: independent random samples; samples large enough for the Central Limit Theorem; sample variances used as the population variances.

Worked example

Using the summary statistics for groups A and B, find a 95% confidence interval for μA - μB.

1. x̄ - ȳ = 1.4
2. Standard error = 0.7331
3. 1.4 ± 1.96 × 0.7331

Answer: (-0.037, 2.837)

Check yourself

1. Group A: n = 50, Σx = 1240, Σx² = 31580. Find an unbiased estimate of the population variance. Give your answer to 3 significant figures.
   Show answer

   16.9. (31580 - 1240²/50)/49 = 16.8980.

2. Group B: n = 60, Σy = 1404, Σy² = 33560. Find an unbiased estimate of the population variance. Give your answer to 3 significant figures.
   Show answer

   12.0. (33560 - 1404²/60)/59 = 11.9729.

3. Group A: n = 50, x̄ = 24.8, sx² = 16.898. Group B: n = 60, ȳ = 23.4, sy² = 11.973. Calculate the test statistic for H0: μA = μB. Give your answer to 3 significant figures.
   Show answer

   1.91. 1.4/sqrt(16.898/50 + 11.973/60) = 1.9096.

4. With the same data, find the upper limit of a 95% confidence interval for μA - μB. Give your answer to 3 significant figures.
   Show answer

   2.84. 1.4 + 1.96 × 0.7331 = 2.8370.

5. The test statistic is 1.910. The question asks whether the two population means are different, at the 5% level. What is the outcome?
   • Do not reject H0: there is insufficient evidence that the means differ
   • Reject H0: the means differ
   • Reject H0, because z exceeds 1.6449
   • Do not reject H0, because z is positive
   Show answer

   Do not reject H0: there is insufficient evidence that the means differ. 'Different' is two-tailed, so compare with 1.96; 1.910 < 1.96.

6. Why can sx² and sy² be used in place of the population variances here?
   • The populations are normal
   • The samples are the same size
   • The samples are large, so the sample variances are good estimates of the population variances
   • The sample means are close together
   Show answer

   The samples are large, so the sample variances are good estimates of the population variances. Large samples make s² reliable; the CLT also makes the sample means approximately normal.

7. Which statement correctly justifies the use of the normal distribution for X̄ - Ȳ when the population distributions are unknown?
   • The samples are large, so by the Central Limit Theorem X̄ and Ȳ are approximately normal
   • The samples are independent, so X̄ - Ȳ is normal
   • The variances are known, so X̄ - Ȳ is normal
   • X̄ - Ȳ is always normal
   Show answer

   The samples are large, so by the Central Limit Theorem X̄ and Ȳ are approximately normal. Examiners require a reference to large samples and the Central Limit Theorem.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit S3
• Pearson Edexcel IAL Statistics S3 (WST03) mark scheme, June 2024
• Pearson Edexcel IAL Statistics S3 (WST03) examiners' report, June 2025

Last updated 8 November 2026. Spotted a mistake? Tell us.