| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two-sample z-test (known variances) |
| Difficulty | Challenging +1.2 This is a standard two-sample z-test with known variance, requiring routine application of hypothesis testing procedures including finding critical values and calculating Type II error probability. While it involves multiple parts and the Type II error calculation requires careful work with the normal distribution, all techniques are standard S3 material with no novel insight required. The computational steps are straightforward given the formulas, placing it moderately above average difficulty for A-level. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu_A = \mu_B\) (or \(\mu_A - \mu_B = 0\)) | B1 | Both hypotheses required |
| \(H_1: \mu_A \neq \mu_B\) (or \(\mu_A - \mu_B \neq 0\)) | Two-tailed test | |
| Under \(H_0\), \(\bar{x} - \bar{y} \sim N\left(0, \frac{285^2}{80} + \frac{285^2}{120}\right)\) | M1 | Correct form of variance |
| \(\text{Var}(\bar{x} - \bar{y}) = \frac{285^2}{80} + \frac{285^2}{120} = 285^2\left(\frac{1}{80}+\frac{1}{120}\right)\) | A1 | Correct variance expression |
| \(= 285^2 \times \frac{5}{240} = 1692.1875\) | ||
| \(z = \frac{3770 - 3695}{\sqrt{1692.1875}} = \frac{75}{41.136...} = 1.823\) | M1 A1 | Correct standardisation |
| Critical value \(z = \pm 1.96\) | B1 | Stated or implied |
| Since \(1.823 < 1.96\), do not reject \(H_0\); insufficient evidence of a difference in mean tensile strength | A1 | Correct conclusion in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Because the sample sizes are large (CLT applies), so the distribution of \(\bar{x} - \bar{y}\) is approximately Normal regardless of the underlying distributions | B1 | Must reference large samples / CLT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Critical values: \(\pm 1.96 \times \sqrt{1692.1875}\) | M1 | Using \(\pm 1.96 \times \sqrt{\text{their variance}}\) |
| \(= \pm 1.96 \times 41.136... = \pm 80.63\) | A1 | Correct answer to 2 d.p. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Type II error: fail to reject \(H_0\) when true difference is 125 | M1 | Correct definition applied |
| \(P(\bar{x} - \bar{y} < 80.63 \mid \mu_A - \mu_B = 125)\) | M1 | Correct probability statement (two critical regions considered) |
| \(= P\left(Z < \frac{80.63 - 125}{41.136}\right) = P(Z < -1.079)\) | A1 | Correct standardisation |
| \(= 1 - \Phi(1.079) \approx 1 - 0.8598 = 0.1402\) | A1 | Accept answers in range \([0.139, 0.141]\) |
# Question 8:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_A = \mu_B$ (or $\mu_A - \mu_B = 0$) | B1 | Both hypotheses required |
| $H_1: \mu_A \neq \mu_B$ (or $\mu_A - \mu_B \neq 0$) | | Two-tailed test |
| Under $H_0$, $\bar{x} - \bar{y} \sim N\left(0, \frac{285^2}{80} + \frac{285^2}{120}\right)$ | M1 | Correct form of variance |
| $\text{Var}(\bar{x} - \bar{y}) = \frac{285^2}{80} + \frac{285^2}{120} = 285^2\left(\frac{1}{80}+\frac{1}{120}\right)$ | A1 | Correct variance expression |
| $= 285^2 \times \frac{5}{240} = 1692.1875$ | | |
| $z = \frac{3770 - 3695}{\sqrt{1692.1875}} = \frac{75}{41.136...} = 1.823$ | M1 A1 | Correct standardisation |
| Critical value $z = \pm 1.96$ | B1 | Stated or implied |
| Since $1.823 < 1.96$, do not reject $H_0$; insufficient evidence of a difference in mean tensile strength | A1 | Correct conclusion in context |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Because the sample sizes are large (CLT applies), so the distribution of $\bar{x} - \bar{y}$ is approximately Normal regardless of the underlying distributions | B1 | Must reference large samples / CLT |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Critical values: $\pm 1.96 \times \sqrt{1692.1875}$ | M1 | Using $\pm 1.96 \times \sqrt{\text{their variance}}$ |
| $= \pm 1.96 \times 41.136... = \pm 80.63$ | A1 | Correct answer to 2 d.p. |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Type II error: fail to reject $H_0$ when true difference is 125 | M1 | Correct definition applied |
| $P(\bar{x} - \bar{y} < 80.63 \mid \mu_A - \mu_B = 125)$ | M1 | Correct probability statement (two critical regions considered) |
| $= P\left(Z < \frac{80.63 - 125}{41.136}\right) = P(Z < -1.079)$ | A1 | Correct standardisation |
| $= 1 - \Phi(1.079) \approx 1 - 0.8598 = 0.1402$ | A1 | Accept answers in range $[0.139, 0.141]$ |8 The tensile strength of rope is measured in kilograms.
The standard deviation of the tensile strength of a particular design of 10 mm diameter rope is known to be 285 kilograms. A retail organisation, which buys such rope from two manufacturers, A and B , wishes to compare their ropes for mean tensile strength.
The mean tensile strength, $\bar { x }$, of a random sample of 80 lengths from manufacturer A was 3770 kilograms.
The mean tensile strength, $\bar { y }$, of a random sample of 120 lengths from manufacturer B was 3695 kilograms.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Test, at the $5 \%$ level of significance, the hypothesis that there is no difference between the mean tensile strength of rope from manufacturer A and that of rope from manufacturer B.
\item Why was it not necessary to know the distributions of tensile strength in order for your test in part (a)(i) to be valid?
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Deduce that, for your test in part (a)(i), the critical values of $( \bar { x } - \bar { y } )$ are $\pm 80.63$, correct to two decimal places.
\item In fact, the mean tensile strength of rope from manufacturer A exceeds that of rope from manufacturer B by 125 kilograms.
Determine the probability of a Type II error for a test of the hypothesis in part (a)(i) at the $5 \%$ level of significance, based upon a random sample of 80 lengths from manufacturer A and a random sample of 120 lengths from manufacturer B. (4 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2011 Q8 [13]}}