| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample z-test large samples |
| Difficulty | Moderate -0.3 This is a standard two-sample t-test with all summary statistics provided. Students need to state hypotheses, calculate the pooled variance and test statistic using given formulas, compare to critical values, and conclude. Part (b) is trivial. While it requires careful calculation, it's a routine textbook exercise with no conceptual challenges beyond applying the standard procedure. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| \multirow{2}{*}{} | \multirow[b]{2}{*}{Sample size} | Length (cm) | ||
| Sample mean | Sample standard deviation | |||
| \multirow{2}{*}{Cucumber variety} | Fanfare | 50 | 22.0 | 1.31 |
| Marketmore | 75 | 21.6 | 0.702 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu_F = \mu_M\), \(H_1: \mu_F \neq \mu_M\) | B1 | Both hypotheses correct |
| \(s_p^2 = \frac{49(1.31)^2 + 74(0.702)^2}{123} = \frac{83.9249 + 36.4659}{123} = \frac{120.3908}{123} \approx 0.979\) | M1 A1 | Pooled variance; correct calculation |
| \(z = \frac{22.0 - 21.6}{\sqrt{0.979\left(\frac{1}{50}+\frac{1}{75}\right)}} = \frac{0.4}{\sqrt{0.979 \times 0.03\overline{3}}} = \frac{0.4}{0.1808} \approx 2.21\) | M1 A1 | Correct test statistic formula and value |
| Critical value \(z = 2.576\) (1% two-tailed) | B1 | Correct critical value |
| \(2.21 < 2.576\), do not reject \(H_0\); no significant difference in mean lengths | A1ft | Correct conclusion in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Any valid characteristic e.g. diameter/width, weight/mass, colour | B1 | Must be measurable characteristic |
# Question 2:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_F = \mu_M$, $H_1: \mu_F \neq \mu_M$ | B1 | Both hypotheses correct |
| $s_p^2 = \frac{49(1.31)^2 + 74(0.702)^2}{123} = \frac{83.9249 + 36.4659}{123} = \frac{120.3908}{123} \approx 0.979$ | M1 A1 | Pooled variance; correct calculation |
| $z = \frac{22.0 - 21.6}{\sqrt{0.979\left(\frac{1}{50}+\frac{1}{75}\right)}} = \frac{0.4}{\sqrt{0.979 \times 0.03\overline{3}}} = \frac{0.4}{0.1808} \approx 2.21$ | M1 A1 | Correct test statistic formula and value |
| Critical value $z = 2.576$ (1% two-tailed) | B1 | Correct critical value |
| $2.21 < 2.576$, do not reject $H_0$; no significant difference in mean lengths | A1ft | Correct conclusion in context |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Any valid characteristic e.g. diameter/width, weight/mass, colour | B1 | Must be measurable characteristic |
---2 As part of a comparison of two varieties of cucumber, Fanfare and Marketmore, random samples of harvested cucumbers of each variety were selected and their lengths measured, in centimetres. The results are summarised in the table.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multirow[b]{2}{*}{Sample size} & \multicolumn{2}{|c|}{Length (cm)} \\
\hline
& & & Sample mean & Sample standard deviation \\
\hline
\multirow{2}{*}{Cucumber variety} & Fanfare & 50 & 22.0 & 1.31 \\
\hline
& Marketmore & 75 & 21.6 & 0.702 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Test, at the $1 \%$ level of significance, the hypothesis that there is no difference between the mean length of harvested Fanfare cucumbers and that of harvested Marketmore cucumbers.
\item In addition to length, name one other characteristic of cucumbers that could be used for comparative purposes.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2012 Q2 [7]}}