| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two-sample t-test (unknown variances) |
| Difficulty | Standard +0.3 This is a standard two-sample inference question requiring an F-test for equal variances followed by a confidence interval calculation. While it involves multiple parts and careful interpretation, the procedures are routine S4 material with no novel problem-solving required. The conceptual demand is moderate (understanding the link between parts a and c), but execution follows textbook methods directly. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | Sample size | Mean | \(s ^ { 2 }\) |
| Dry feed | 13 | 25.54 | 2.45 |
| Feed with water added | 9 | 27.94 | 1.02 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) \(H_0: \sigma_A^2 = \sigma_B^2\), \(H_1: \sigma_A^2 \neq \sigma_B^2\) | B1 | |
| Critical values \(F_{12,8} = 3.28\) and \(\frac{1}{F_{8,12}} = 0.35\) | B1 | |
| \(\frac{s_B^2}{s_A^2} = 2.40\) \(\left(\frac{s_A^2}{s_B^2} = 0.416\right)\) | M1 A1 | |
| Since 2.40 (0.416) is not in the critical region we accept \(H_0\) and conclude there is no evidence that the two variances are different. | A1ft | |
| (b) \(S_p^2 = \frac{8 \times 1.02 + 12 \times 2.45}{20} = 1.878\) | M1, A1 | |
| \((27.94 - 25.54) \pm 2.086 \times \sqrt{1.878} \times \sqrt{\frac{1}{9} + \frac{1}{13}}\) | B1 M1 A1ft | |
| \((1.16,\ 3.64)\) | A1, A1 | |
| (c) To calculate the confidence interval the variances need to be equal. In part (a) the test showed they are equal. | B1, B1 |
# Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** $H_0: \sigma_A^2 = \sigma_B^2$, $H_1: \sigma_A^2 \neq \sigma_B^2$ | B1 | |
| Critical values $F_{12,8} = 3.28$ and $\frac{1}{F_{8,12}} = 0.35$ | B1 | |
| $\frac{s_B^2}{s_A^2} = 2.40$ $\left(\frac{s_A^2}{s_B^2} = 0.416\right)$ | M1 A1 | |
| Since 2.40 (0.416) is not in the critical region we accept $H_0$ and conclude there is no evidence that the two variances are different. | A1ft | |
| **(b)** $S_p^2 = \frac{8 \times 1.02 + 12 \times 2.45}{20} = 1.878$ | M1, A1 | |
| $(27.94 - 25.54) \pm 2.086 \times \sqrt{1.878} \times \sqrt{\frac{1}{9} + \frac{1}{13}}$ | B1 M1 A1ft | |
| $(1.16,\ 3.64)$ | A1, A1 | |
| **(c)** To calculate the confidence interval the variances need to be equal. In part (a) the test showed they are equal. | B1, B1 | |
---\begin{enumerate}
\item A farmer set up a trial to assess whether adding water to dry feed increases the milk yield of his cows. He randomly selected 22 cows. Thirteen of the cows were given dry feed and the other 9 cows were given the feed with water added. The milk yields, in litres per day, were recorded with the following results.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & Sample size & Mean & $s ^ { 2 }$ \\
\hline
Dry feed & 13 & 25.54 & 2.45 \\
\hline
Feed with water added & 9 & 27.94 & 1.02 \\
\hline
\end{tabular}
\end{center}
You may assume that the milk yield from cows given the dry feed and the milk yield from cows given the feed with water added are from independent normal distributions.\\
(a) Test, at the $10 \%$ level of significance, whether or not the variances of the populations from which the samples are drawn are the same. State your hypotheses clearly.\\
(b) Calculate a $95 \%$ confidence interval for the difference between the two mean milk yields.\\
(c) Explain the importance of the test in part (a) to the calculation in part (b).\\
\hfill \mbox{\textit{Edexcel S4 2009 Q4 [14]}}