9709 · 6.5

Hypothesis tests — practice questions

Practice and worked examples for 9709 Hypothesis tests. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

The mass of a certain type of chocolate bar is claimed to be 50g. The masses are known to be normally distributed with a standard deviation of 1.5g. A quality control manager suspects that the machine is producing underweight bars. She takes a random sample of 10 bars and finds their mean mass is 49.2g. Test her suspicion at the 5% significance level.

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Hypotheses: Let $\mu$ be the population mean mass of the chocolate bars. $H_0: \mu = 50$ (The mean mass is 50g). $H_1: \mu < 50$ (The mean mass is less than 50g, i.e., underweight). This is a one-tailed test.
Significance Level: $\alpha = 0.05$ (5%).
Critical Value: For a one-tailed test at the 5% level, we need the Z-value that gives a lower tail probability of 0.05. From tables, this corresponds to $P(Z < -1.645) = 0.05$ . So, the critical value is -1.645. The critical region is $Z \le -1.645$ .
Test Statistic: The population is $X \sim N(\mu, 1.5^2)$ . We have $n=10$ , $\bar{x}=49.2$ , $\sigma=1.5$ . $Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{49.2 - 50}{1.5 / \sqrt{10}} = \frac{-0.8}{0.4743...} = -1.686...$
Comparison and Conclusion: The test statistic $Z = -1.687$ is less than the critical value of $-1.645$ . Therefore, the result falls in the critical region. We reject $H_0$ . There is sufficient evidence at the 5% significance level to suggest that the machine is producing underweight chocolate bars.

Worked example 2

In the past, 30% of customers at a coffee shop ordered oat milk. After a marketing campaign, the manager wants to know if this proportion has changed. In a random sample of 150 customers, 57 ordered oat milk. Test at the 10% significance level whether the proportion has changed. Also, find the p-value of this test.

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Hypotheses: Let $p$ be the population proportion of customers ordering oat milk. $H_0: p = 0.3$ $H_1: p \neq 0.3$ (The proportion has 'changed'). This is a two-tailed test.
Significance Level: $\alpha = 0.10$ (10%). For a two-tailed test, this means 5% in each tail.
Test Statistic: Sample proportion $\hat{p} = \frac{57}{150} = 0.38$ . We use a Normal approximation to the Binomial distribution. Under $H_0$ , the distribution of the sample proportion is approximately $N(p, \frac{p(1-p)}{n})$ . Mean $\mu = p = 0.3$ . Variance $\sigma^2 = \frac{p(1-p)}{n} = \frac{0.3 \times 0.7}{150} = 0.0014$ . Standard deviation $\sigma = \sqrt{0.0014} \approx 0.03742$ . $Z = \frac{\hat{p} - p}{\sqrt{p(1-p)/n}} = \frac{0.38 - 0.3}{0.03742} = 2.138...$
Comparison (p-value method): The p-value is the probability of getting a result at least as extreme as the one observed. Since this is a two-tailed test, we find the probability of $Z > 2.138$ and multiply by 2. $P(Z > 2.138) = 1 - \Phi(2.138) = 1 - 0.9837 = 0.0163$ . p-value $= 2 \times 0.0163 = 0.0326$ .
Conclusion: The p-value (0.0326) is less than the significance level $\alpha$ (0.10). Therefore, we reject $H_0$ . There is sufficient evidence at the 10% significance level to conclude that the proportion of customers ordering oat milk has changed.

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