4.2 - Comparing means with a normal distribution | Cambridge A-Level Further Mathematics

A normal test can compare independent population means when variances are known, or approximately when samples are sufficiently large. You will calculate the standard error of a difference, choose the correct tail, and distinguish these situations from pooled and paired t-tests.

Variance of a difference

Why does subtracting two independent sample means add their uncertainty? Either mean can move away from its population value, so both sources of variation contribute to the difference.

For independent random samples, $\operatorname{Var}(\bar X)=\sigma_X^2/n_X$ and $\operatorname{Var}(\bar Y)=\sigma_Y^2/n_Y$ . Independence makes the covariance zero. Thus the minus sign in the difference disappears on squaring:

\operatorname{Var}(\bar X-\bar Y) =\frac{\sigma_X^2}{n_X}+\frac{\sigma_Y^2}{n_Y}.

The standard error is the square root of this variance. Do not subtract variances or add standard deviations.

Standard error of independent means

With known population variances,

\operatorname{SE}(\bar X-\bar Y) =\sqrt{\frac{\sigma_X^2}{n_X}+\frac{\sigma_Y^2}{n_Y}}.

With unknown variances and two sufficiently large samples, use the estimate

\widehat{\operatorname{SE}}(\bar X-\bar Y) =\sqrt{\frac{s_X^2}{n_X}+\frac{s_Y^2}{n_Y}},

where both sample variances are unbiased estimates. These difference standard errors are not printed in the formula booklet; the sample-mean variance and unbiased variance formulae are provided.

Normal populations with known variances give an exact normal difference for any sample sizes. For non-normal populations with finite variances, sufficiently large independent samples allow a normal approximation through the central limit theorem. The required size depends on the population shapes: no single cutoff guarantees accuracy for every distribution. Strong skewness or extreme outliers can require much larger samples. Large-sample questions will give or justify the normal approximation.

Worked example 1: known variances

Question. Independent samples have sizes $40$ and $50$ . Their population variances are $36$ and $25$ , respectively. Calculate the variance and standard error of the difference of sample means.

Plan. Add the two sample-mean variances, then take a square root.

\operatorname{Var}(\bar X-\bar Y) =\frac{36}{40}+\frac{25}{50}=1.4.

\operatorname{SE}(\bar X-\bar Y)=\sqrt{1.4}=1.183215\ldots

Answer. Variance $1.4$ , standard error $1.18$ to 3 significant figures.

Check. The separate standard errors are $\sqrt{0.9}$ and $\sqrt{0.5}$ . Their squared values add to $1.4$ ; the combined standard error exceeds either component but is less than their sum.

Now change only the variance information: the same numerical values must be estimated from the two large samples. The standard error becomes an estimate rather than a known value.

Worked example 2: large-sample estimates

Question. Independent samples have $n_X=40$ , $\sum(x-\bar x)^2=1404$ , and $n_Y=50$ , $\sum(y-\bar y)^2=1225$ . A normal approximation is appropriate. Estimate the standard error of the difference of sample means.

Plan. Divide each centred sum by its own $n-1$ , then combine the estimated sample-mean variances.

s_X^2=\frac{1404}{39}=36, \qquad s_Y^2=\frac{1225}{49}=25.

\widehat{\operatorname{SE}}(\bar X-\bar Y) =\sqrt{\frac{36}{40}+\frac{25}{50}}=\sqrt{1.4}.

Answer. The estimated standard error is $1.18$ to 3 significant figures.

Check. The two estimated variances equal the known variances in Worked example 1, so the numerical answer agrees. No equal-population-variance assumption was used: the two contributions remain separate.

Known unequal variances and large-sample estimated unequal variances are both compatible with these normal methods. Pooling would impose a common-variance assumption that the calculation does not need.

A normal test for a mean difference

The normal test compares the observed difference with its null value in standard-error units, just as a t-test does, but it uses a standard normal reference distribution.

Normal statistic for a mean difference

For independent samples with known variances,

Z=\frac{\bar X-\bar Y-\delta_0} {\sqrt{\sigma_X^2/n_X+\sigma_Y^2/n_Y}}.

Under $H_0:\mu_X-\mu_Y=\delta_0$ , this is standard normal if the populations are normal, and approximately so for sufficiently large samples. For large samples with unknown variances, replace the population variances by their separate unbiased estimates and use an approximate normal test.

This statistic is not printed in the formula booklet. Normal probability tables are provided.

For a 5% upper-tail test the critical value is $1.645$ ; for a 5% two-tail test the critical magnitudes are $1.960$ . These correspond to cumulative normal probabilities $0.95$ and $0.975$ . A lower-tail test uses the negative cutoff.

Worked example 3: higher population mean

Question. Independent normal populations have known variances $9$ and $4$ . Random samples give $n_X=36$ , $\bar x=15.2$ , and $n_Y=25$ , $\bar y=14.0$ . Test at 5% whether $\mu_X>\mu_Y$ , using the normal critical value $1.645$ .

Plan. Use the known variances separately and an upper-tail alternative.

H_0:\mu_X-\mu_Y=0,\qquad H_1:\mu_X-\mu_Y>0.

z=\frac{15.2-14.0}{\sqrt{9/36+4/25}} =\frac{1.2}{\sqrt{0.41}}=1.874085\ldots

Since $1.874085>1.645$ , reject $H_0$ .

Answer. There is sufficient evidence at 5% that the mean of population $X$ exceeds that of population $Y$ .

Check. The rejection boundary for the observed difference is $1.645\sqrt{0.41}=1.053313\ldots$ . The difference $1.2$ exceeds it, giving the same decision without standardising.

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4.2 - Comparing means with a normal distribution

Variance of a difference

A normal test for a mean difference

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