A survey of 400 people finds 220 in favour. Test H_0: p = 0.5 against H_1: p ≠ 0.5; find the test statistic. [2 marks]

SQA AH style: two proportions-style practice: In sample 1, 40 of 200 succeed; in sample 2, 30 of 200 succeed. Explain how to form the pooled proportion and the test statistic for H_0: p_1 = p_2.

Under H_0 the two proportions are equal, so pool them: p = 40 + 30200 + 200 = 70400 = 0.175 (1 mark). The standard error of the difference uses the pooled proportion: p(1 - p)(1n_1 + 1n_2) (1 mark). The test statistic is z = p_1 - p_2p(1 - p)((1)/(n_1) + (1)/(n_2)), compared with the standard normal; here p_1 = 0.20, p_2 = 0.15 (1 mark). Markers reward the pooled proportion, the standard error using it, and the correct test statistic.

ScotlandStatisticsSyllabus dot point

How do you test a claim about a population proportion, or compare two proportions?

Carry out hypothesis tests for a single population proportion and for the difference between two proportions, using the normal approximation, stating the hypotheses, computing the test statistic and interpreting the result.

A focused answer to the SQA Advanced Higher Statistics proportion test content: testing a single population proportion and the difference between two proportions using the normal approximation, with the test statistics, the pooled estimate for two samples, and how to interpret the outcome.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

For a single proportion, test $H_0: p = p_0$ with $z = \dfrac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}}$ , where $\hat{p}$ is the sample proportion and the standard error uses the hypothesised $p_0$ . For two proportions, test $H_0: p_1 = p_2$ using the pooled proportion $\hat{p} = \dfrac{x_1 + x_2}{n_1 + n_2}$ in the standard error: $z = \dfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$ . Both rely on the normal approximation, valid when the counts of successes and failures are large enough (each expected count above about $5$ ).

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What this dot point is asking
Testing a single proportion
Comparing two proportions
When the normal approximation applies
Try this

What this dot point is asking

Many questions are about proportions rather than means: the fraction of voters favouring a policy, the success rate of a treatment, the bias of a coin. The SQA wants you to test a single proportion against a claimed value and to compare two proportions, in both cases using the normal approximation to the binomial, stating hypotheses, computing a z-statistic and concluding in context.

Testing a single proportion

The one-sample proportion test asks whether an observed success rate is consistent with a claimed value.

The crucial detail is that the standard error uses $p_0$ , the hypothesised proportion, not the sample $\hat{p}$ , because the whole calculation is carried out under the assumption that $H_0$ is true.

Test a single proportion

A factory claims at most 10% of its items are defective. We have sample data that suggest a higher rate, and we want to know whether the evidence is strong enough to reject the claim. The key detail is that the standard error is computed using the hypothesised proportion $p_0$ , not the sample proportion, because all calculations are done under the assumption that $H_0$ is true.

Step 1: State the hypotheses and sample proportion

The factory's claim becomes $H_0: p = 0.10$ (at most 10% defective); we suspect the true rate is higher, so $H_1: p > 0.10$ (one-tailed test at $\alpha = 0.05$ ). From the sample, $\hat{p} = \dfrac{28}{200} = 0.14$ .

Step 2: Compute the standard error under $H_0$ and form the test statistic

Use the hypothesised $p_0 = 0.10$ in the standard error formula, then subtract the hypothesised value and divide.

\text{SE} = \sqrt{\dfrac{p_0(1 - p_0)}{n}} = \sqrt{\dfrac{0.10 \times 0.90}{200}} = \sqrt{0.00045} \approx 0.0212

z = \dfrac{\hat{p} - p_0}{\text{SE}} = \dfrac{0.14 - 0.10}{0.0212} \approx 1.886

Step 3: Compare with the critical value and conclude

For a one-tailed test at $\alpha = 0.05$ , the critical value is $z_{0.05} = 1.645$ .

1.886 > 1.645 \implies \text{reject } H_0

Final answer: there is significant evidence at the 5% level that more than 10% of items are defective.

Comparing two proportions

The two-sample proportion test asks whether two groups share the same underlying proportion.

Compare two proportions

We want to know whether a treatment group has a different recovery rate from a control group. Under $H_0: p_1 = p_2$ , the best estimate of the common underlying proportion combines both samples into a single pooled estimate, which is then used in the standard error of the difference.

Step 1: Compute each sample proportion and the pooled estimate

Treatment: $45$ of $150$ recover, so $\hat{p}_1 = \dfrac{45}{150} = 0.30$ . Control: $30$ of $150$ recover, so $\hat{p}_2 = \dfrac{30}{150} = 0.20$ . Under $H_0$ both proportions are equal, so we pool the successes and the totals.

\hat{p} = \dfrac{45 + 30}{150 + 150} = \dfrac{75}{300} = 0.25

Step 2: Compute the standard error of the difference

Use the pooled proportion $\hat{p}$ in the formula. The two equal sample sizes make the bracket $\dfrac{1}{n_1}+\dfrac{1}{n_2}=\dfrac{2}{150}=0.01\overline{3}$ .

\text{SE} = \sqrt{\hat{p}(1-\hat{p})\!\left(\dfrac{1}{n_1}+\dfrac{1}{n_2}\right)} = \sqrt{0.25\times 0.75\times 0.01\overline{3}} = \sqrt{0.0025} = 0.05

Step 3: Form the test statistic and state the conclusion

Divide the observed difference in proportions by the standard error, then compare with the critical value for a two-tailed test at $5\%$ .

z = \dfrac{\hat{p}_1 - \hat{p}_2}{\text{SE}} = \dfrac{0.30 - 0.20}{0.05} = 2.0

Since $2.0 > 1.96$ , reject $H_0$ .

Final answer: there is significant evidence at the 5% level that the recovery proportions differ between the treatment and control groups.

When the normal approximation applies

Both tests approximate the binomial by a normal, so they need the samples to be large enough.

Try this

Q1. A survey of $400$ people finds $220$ in favour. Test $H_0: p = 0.5$ against $H_1: p \neq 0.5$ ; find the test statistic. [2 marks]

Cue. $\hat{p} = 0.55$ , standard error $= \sqrt{\dfrac{0.5 \times 0.5}{400}} = 0.025$ , so $z = \dfrac{0.55 - 0.5}{0.025} = 2.0$ .

Q2. State the pooled proportion when sample 1 has $18$ successes in $60$ and sample 2 has $12$ successes in $40$ . [1 mark]

Cue. $\hat{p} = \dfrac{18 + 12}{60 + 40} = \dfrac{30}{100} = 0.30$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: one proportion4 marksA coin is tossed

100

times and lands heads

59

times. Test at the

5\%

level whether the coin is biased toward heads. (Use

z_{0.05} = 1.645

Show worked answer →

Hypotheses: $H_0: p = 0.5$ against $H_1: p > 0.5$ (one-tailed), where $p$ is the true probability of heads (1 mark).

Sample proportion $\hat{p} = \dfrac{59}{100} = 0.59$ ; standard error under $H_0$ is $\sqrt{\dfrac{p_0(1 - p_0)}{n}} = \sqrt{\dfrac{0.5 \times 0.5}{100}} = 0.05$ (1 mark).

Test statistic: $z = \dfrac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}} = \dfrac{0.59 - 0.5}{0.05} = 1.8$ (1 mark).

Since $1.8 > 1.645$ , reject $H_0$ : there is significant evidence at the $5\%$ level that the coin is biased toward heads (1 mark). Markers reward the hypotheses, the standard error under $H_0$ , the test statistic and the conclusion.

AH style: two proportions3 marksIn sample 1,

40

200

succeed; in sample 2,

30

200

succeed. Explain how to form the pooled proportion and the test statistic for

H_0: p_1 = p_2

Show worked answer →

Under $H_0$ the two proportions are equal, so pool them: $\hat{p} = \dfrac{40 + 30}{200 + 200} = \dfrac{70}{400} = 0.175$ (1 mark).

The standard error of the difference uses the pooled proportion: $\sqrt{\hat{p}(1 - \hat{p})\left(\dfrac{1}{n_1} + \dfrac{1}{n_2}\right)}$ (1 mark).

The test statistic is $z = \dfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$ , compared with the standard normal; here $\hat{p}_1 = 0.20$ , $\hat{p}_2 = 0.15$ (1 mark). Markers reward the pooled proportion, the standard error using it, and the correct test statistic.

Related dot points

Sources & how we know this

SQA Advanced Higher Statistics Course Specification (C803 77) — SQA (2023)
SQA Advanced Higher Statistics Data Booklet — SQA (2019)