| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample z-test large samples |
| Difficulty | Standard +0.3 This is a standard two-sample t-test with all summary statistics provided. Part (a) is routine variance calculation verification, part (b) follows a textbook hypothesis test procedure, part (c) requires recall of CLT application, and part (d) is basic sampling theory. The question requires no novel insight—just methodical application of learned procedures with large sample sizes that simplify the test. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| \(\Sigma x\) | \(\Sigma x ^ { 2 }\) | \(s ^ { 2 }\) | \(n\) | |
| English films | 10650 | 956909 | 98.5 | 120 |
| French films | 6510 | 615849 | 151 | 70 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s_E^2 = \frac{1}{n-1}\left(\Sigma x^2 - \frac{(\Sigma x)^2}{n}\right) = \frac{1}{119}\left(956909 - \frac{10650^2}{120}\right)\) | M1 | |
| \(= \frac{11721.5}{119} = 98.5\) | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu_E = \mu_F\); \(H_1: \mu_E \neq \mu_F\) | B1 | |
| \(\bar{x}_E = \frac{10650}{120} = 88.75\) and \(\bar{x}_F = \frac{6510}{70} = 93\) | M1 | |
| Test statistic, \(z = \frac{93 - 88.75 - 0}{\sqrt{\frac{151.98}{70}+\frac{98.5}{120}}} = 2.4627...\) | M1A1 | |
| Critical values, \(z = (±)2.5758\) | B1ft | |
| Test stat is not in critical region | M1 | |
| Insufficient evidence to reject \(H_0\) at 1% level | A1ft | (7) |
| No significant evidence of a difference in mean lengths of English and French films | (13 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| By CLT we can assume that the mean of a large sample has a Normal distribution | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| On a list, label English films 1 – 724 and French films 1-473 (oe) | B1 | |
| Use random number table/generator to select | ||
| \(\frac{724}{724+473} \times 190 = 115\) English films and | M1A1 | (3) |
| \(\frac{473}{1197} \times 190 = 75\) French films | (13 marks) |
**5(a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s_E^2 = \frac{1}{n-1}\left(\Sigma x^2 - \frac{(\Sigma x)^2}{n}\right) = \frac{1}{119}\left(956909 - \frac{10650^2}{120}\right)$ | M1 | |
| $= \frac{11721.5}{119} = 98.5$ | A1 | (2) |
**5(b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu_E = \mu_F$; $H_1: \mu_E \neq \mu_F$ | B1 | |
| $\bar{x}_E = \frac{10650}{120} = 88.75$ and $\bar{x}_F = \frac{6510}{70} = 93$ | M1 | |
| Test statistic, $z = \frac{93 - 88.75 - 0}{\sqrt{\frac{151.98}{70}+\frac{98.5}{120}}} = 2.4627...$ | M1A1 | |
| Critical values, $z = (±)2.5758$ | B1ft | |
| Test stat is not in critical region | M1 | |
| Insufficient evidence to reject $H_0$ at 1% level | A1ft | (7) |
| No significant evidence of a difference in mean lengths of English and French films | | (13 marks) |
**5(c)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| By CLT we can assume that the mean of a large sample has a Normal distribution | B1 | (1) |
**5(d)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| On a list, label English films 1 – 724 and French films 1-473 (oe) | B1 | |
| Use random number table/generator to select | | |
| $\frac{724}{724+473} \times 190 = 115$ English films and | M1A1 | (3) |
| $\frac{473}{1197} \times 190 = 75$ French films | | (13 marks) |
**Guidance Notes:**
- Alternative for (a): $s_E^2 = \frac{n}{n-1}\left(\frac{\Sigma x^2}{n} - \bar{x}^2\right) = \frac{120}{119}\left(\frac{956909}{120} - 88.75^2\right) = 98.5$
- (b) 1st B1 needs both $H_0$ and $H_1$, can be in words
- 2nd B1ft on their $H_1$
- 1st M1 for attempt @ both means ($\bar{x}$ may be in (a))
- 2nd M1 for attempt at correct test statistic, ft their values
- 3rd M1 for attempt to compare their test stat and critical values
- A1 ft on their test and critical values but must include comment in context
- (c) Require mention of mean of E or F and normal distribution
- (d) M1 requires use of **random** numbers and attempt to find correct sample sizes; A1 both 115 and 75 found.
---5. A student believes that there is a difference in the mean lengths of English and French films. He goes to the university video library and randomly selects a sample of 120 English films and a sample of 70 French films. He notes the length, $x$ minutes, of each of the films in his samples. His data are summarised in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
& $\Sigma x$ & $\Sigma x ^ { 2 }$ & $s ^ { 2 }$ & $n$ \\
\hline
English films & 10650 & 956909 & 98.5 & 120 \\
\hline
French films & 6510 & 615849 & 151 & 70 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Verify that the unbiased estimate of the variance, $s ^ { 2 }$, of the lengths of English films is 98.5 minutes ${ } ^ { 2 }$
\item Stating your hypotheses clearly, test, at the 1\% level of significance, whether or not the mean lengths of English and French films are different.
\item Explain the significance of the Central Limit Theorem to the test in part (b).
\item The university video library contained 724 English films and 473 French films. Explain how the student could have taken a stratified sample of 190 of these films.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2014 Q5 [13]}}