| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2004 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Standard +0.8 This S4 question requires a two-stage hypothesis testing procedure: first an F-test for equality of variances (requiring calculation of sample variances from summary statistics), then a two-sample t-test with pooled variance. The multi-step nature, need to calculate intermediate statistics, interpret results to justify test choice, and work with a directional hypothesis about a difference of 150g makes this moderately challenging, though the procedures themselves are standard for Further Maths Statistics. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Sample size | \(\Sigma x\) | \(\Sigma x ^ { 2 }\) | |
| \(A\) | 11 | 6600 | 3960540 |
| \(B\) | 13 | 9815 | 7410579 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(S_A^2 = \frac{1}{10}\left[3960540 - \frac{6600^2}{11}\right] = 54.0\) | B1 | |
| \(S_B^2 = \frac{1}{12}\left[7410579 - \frac{9815^2}{13}\right] = 21.16\) | B1 | |
| \(H_0: \sigma_A^2 = \sigma_B^2; H_1: \sigma_A^2 \neq \sigma_B^2\) | B1 | |
| CR: \(F_{10,12} > 2.75\) | ||
| \(\frac{S_A^2}{S_B^2} = \frac{54.0}{21.16} = 2.55118\ldots\) | M1 A1 | |
| Since \(2.55118\ldots\) is not in the critical region we can assume that the variances are equal. | B1 | |
| (6) | ||
| (b) \(H_0: \mu_B = \mu_A + 150; H_1: \mu_B > \mu_A + 150\) | B1 (both) | |
| CR: \(t_{22}(0.05) > 1.717\) | 1.717 | B1 |
| \(S_p^2 = \frac{10 \times 54.0 + 12 \times 21.16}{22} = 36.0909\) | M1 A1 | |
| \(t = \frac{1755 - 6001 - 150}{\sqrt{36.0909\ldots\left(\frac{1}{11} + \frac{1}{13}\right)}} = 2.03157\) | M1 A1 | |
| AWRT 2.03 | A1 | |
| Since \(2.03\ldots\) is in the critical region we reject \(H_0\) and conclude that the mean weight of cauliflowers from \(B\) exceeds that from \(A\) by at least 50g. | A1 ft | |
| (8) | ||
| (c) Samples from normal populations | B1 B1 | |
| Equal variances | Any two sensible verifications | |
| Independent samples | (2) | |
| (16 marks) |
| **(a)** $S_A^2 = \frac{1}{10}\left[3960540 - \frac{6600^2}{11}\right] = 54.0$ | B1 |
| $S_B^2 = \frac{1}{12}\left[7410579 - \frac{9815^2}{13}\right] = 21.16$ | B1 |
| $H_0: \sigma_A^2 = \sigma_B^2; H_1: \sigma_A^2 \neq \sigma_B^2$ | B1 |
| CR: $F_{10,12} > 2.75$ | |
| $\frac{S_A^2}{S_B^2} = \frac{54.0}{21.16} = 2.55118\ldots$ | M1 A1 |
| Since $2.55118\ldots$ is not in the critical region we can assume that the variances are equal. | B1 |
| | (6) |
| **(b)** $H_0: \mu_B = \mu_A + 150; H_1: \mu_B > \mu_A + 150$ | B1 (both) |
| CR: $t_{22}(0.05) > 1.717$ | 1.717 | B1 |
| $S_p^2 = \frac{10 \times 54.0 + 12 \times 21.16}{22} = 36.0909$ | M1 A1 |
| $t = \frac{1755 - 6001 - 150}{\sqrt{36.0909\ldots\left(\frac{1}{11} + \frac{1}{13}\right)}} = 2.03157$ | M1 A1 |
| AWRT 2.03 | A1 |
| Since $2.03\ldots$ is in the critical region we reject $H_0$ and conclude that the mean weight of cauliflowers from $B$ exceeds that from $A$ by at least 50g. | A1 ft |
| | (8) |
| **(c)** Samples from normal populations | B1 B1 |
| Equal variances | Any two sensible verifications | |
| Independent samples | | (2) |
| | (16 marks) |7. A grocer receives deliveries of cauliflowers from two different growers, $A$ and $B$. The grocer takes random samples of cauliflowers from those supplied by each grower. He measures the weight $x$, in grams, of each cauliflower. The results are summarised in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
& Sample size & $\Sigma x$ & $\Sigma x ^ { 2 }$ \\
\hline
$A$ & 11 & 6600 & 3960540 \\
\hline
$B$ & 13 & 9815 & 7410579 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show, at the $10 \%$ significance level, that the variances of the populations from which the samples are drawn can be assumed to be equal by testing the hypothesis $\mathrm { H } _ { 0 } : \sigma _ { A } ^ { 2 } = \sigma _ { B } ^ { 2 }$ against hypothesis $\mathrm { H } _ { 1 } : \sigma _ { A } ^ { 2 } \neq \sigma _ { B } ^ { 2 }$.\\
(You may assume that the two samples come from normal populations.)\\
(6)
The grocer believes that the mean weight of cauliflowers provided by $B$ is at least 150 g more than the mean weight of cauliflowers provided by $A$.
\item Use a $5 \%$ significance level to test the grocer's belief.
\item Justify your choice of test.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2004 Q7 [16]}}