| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2012 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Challenging +1.2 This is a standard S4 two-sample t-test question with a type II error calculation. Part (a) requires routine application of pooled variance t-test procedures. Part (b)(i) involves working backwards from known population variance to find acceptance region boundaries using normal distribution (straightforward algebra). Part (b)(ii) requires calculating P(Type II error) by finding probability that test statistic falls in acceptance region under alternative hypothesis—a textbook S4 technique. While multi-step with 16 marks total, all components follow standard S4 procedures without requiring novel insight or particularly complex reasoning. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Number in sample | Sample mean \(\bar{x}\) | \(\sum x^2\) | |
| Female | 6 | 19.6 | 2308.01 |
| Male | 12 | 13.7 | 2262.57 |
| Answer | Marks |
|---|---|
| \(S_F^2 = \frac{1}{5}(2308.01 - 6 \times 19.6^2) = 0.61\) | B1 |
| \(S_M^2 = \frac{1}{11}(2262.57 - 12 \times 13.7^2) = 0.93545..\) | B1 |
| \(H_0: \mu_F = \mu_M + 5\); \(H_1: \mu_F \neq \mu_M + 5\) both | B1 |
| CR: \(t_{16}(0.025) > 2.120\) 2.12 | B1 |
| \(S_p^2 = \frac{5 \times 0.61 + 11 \times 0.93545...}{16} = 0.83375\) | M1 A1 |
| \(t = \frac{19.6 - 13.7 - 5}{\sqrt{0.83375(\frac{1}{6} + \frac{1}{12})}} = 1.971\) | M1 A1ftA1 |
| Since 1.971 is not in the critical region we accept \(H_0\) and conclude that the mean shell length of female turtles does exceed the shell length of male turtles by 5cm. (or Biologists claim is correct) | A1 ft |
| (10 marks total) |
| Answer | Marks |
|---|---|
| \(-1.96 < \frac{\bar{X}_f - \bar{X}_m - 5}{\sqrt{(\frac{0.9}{6} + \frac{0.9}{12})}} < 1.96\) | B1 M1 |
| \(4.07 < \bar{X}_f - \bar{X}_m < 5.93\) | A1cso |
| (6 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| P(Type II error) \(= P(4.07 < \bar{X}_f - \bar{X}_m < 5.93 | N(6, 0.225))\) | M1 |
| \(= P\left(\frac{4.07 - 6}{\sqrt{0.225}} < z < \frac{5.93 - 6}{\sqrt{0.225}}\right)\) | M1 | |
| \(= 0.44\) awrt 0.44 | A1 | |
| (6 marks total) |
**Part (a)**
| $S_F^2 = \frac{1}{5}(2308.01 - 6 \times 19.6^2) = 0.61$ | B1 |
| $S_M^2 = \frac{1}{11}(2262.57 - 12 \times 13.7^2) = 0.93545..$ | B1 |
| $H_0: \mu_F = \mu_M + 5$; $H_1: \mu_F \neq \mu_M + 5$ both | B1 |
| CR: $t_{16}(0.025) > 2.120$ 2.12 | B1 |
| $S_p^2 = \frac{5 \times 0.61 + 11 \times 0.93545...}{16} = 0.83375$ | M1 A1 |
| $t = \frac{19.6 - 13.7 - 5}{\sqrt{0.83375(\frac{1}{6} + \frac{1}{12})}} = 1.971$ | M1 A1ftA1 |
| Since 1.971 is not in the critical region we accept $H_0$ and conclude that the mean shell length of female turtles does exceed the shell length of male turtles by 5cm. (or Biologists claim is correct) | A1 ft |
| | (10 marks total) |
**Part (b)(i)**
| $-1.96 < \frac{\bar{X}_f - \bar{X}_m - 5}{\sqrt{(\frac{0.9}{6} + \frac{0.9}{12})}} < 1.96$ | B1 M1 |
| $4.07 < \bar{X}_f - \bar{X}_m < 5.93$ | A1cso |
| | (6 marks total) |
**Part (b)(ii)**
| P(Type II error) $= P(4.07 < \bar{X}_f - \bar{X}_m < 5.93 | N(6, 0.225))$ | M1 |
| $= P\left(\frac{4.07 - 6}{\sqrt{0.225}} < z < \frac{5.93 - 6}{\sqrt{0.225}}\right)$ | M1 |
| $= 0.44$ awrt 0.44 | A1 |
| | (6 marks total) |
**Guidance notes for 2(a):**
- B1 awrt 0.61
- B1 awrt 0.935
- Both may be implied by correct $t$ value or $S_p$
- B1 allow rearrangements e.g. $\mu_F - \mu_M = 5$. If M and F not used then they must make clear what each letter is.
- B1 CV (if using one tail test allow 1.746)
- M1 $\frac{5\times\text{their }0.61+11\times\text{their }0.93545...}{16}$
- A1 awrt 0.834
- M1 $\pm\frac{19.6-13.7-5}{\sqrt{p(\frac{1}{6}+\frac{1}{12})}}$ where $p$ is either their 0.61 or 0.94 or their $S_p^2$ (awrt 0.834) (Allow 13.7 - 19.6 - 5)
- A1 ft their $S_p^2$
- A1 awrt 1.97
- B1 1.96
- M1 must use $z$ value
**Guidance notes for 2(b):**
- B1 awrt 0.61
- B1 awrt 0.935
- Both may be implied by correct $t$ value or $S_p$
- B1 allow rearrangements e.g. $\mu_F - \mu_M = 5$. If M and F not used then they must make clear what each letter is.
- B1 CV (if using one tail test allow 1.746)
- M1 $\frac{5\times\text{their }0.61+11\times\text{their }0.93545...}{16}$
- A1 awrt 0.834
- M1 writing or using $N(6, 0.225)$
- M1 finding correct area and standardising (must use 6 but allow use of 0.9 and (0.9/18) for var)
**Total: 16 marks**
---A biologist investigating the shell size of turtles takes random samples of adult female and adult male turtles and records the length, $x$ cm, of the shell. The results are summarised below.
\begin{tabular}{|c|c|c|c|}
\hline
& Number in sample & Sample mean $\bar{x}$ & $\sum x^2$ \\
\hline
Female & 6 & 19.6 & 2308.01 \\
\hline
Male & 12 & 13.7 & 2262.57 \\
\hline
\end{tabular}
You may assume that the samples come from independent normal distributions with the same variance.
The biologist claims that the mean shell length of adult female turtles is 5 cm longer than the mean shell length of adult male turtles.
\begin{enumerate}[label=(\alph*)]
\item Test the biologist's claim at the 5\% level of significance. [10]
\item Given that the true values for the variance of the population of adult male turtles and adult female turtles are both 0.9 cm$^2$,
\begin{enumerate}[label=(\roman*)]
\item show that when samples of size 6 and 12 are used with a 5\% level of significance, the biologist's claim will be accepted if $4.07 < \bar{X}_F - \bar{X}_M < 5.93$ where $\bar{X}_F$ and $\bar{X}_M$ are the mean shell lengths of females and males respectively.
\item Hence find the probability of a type II error for this test if in fact the true mean shell length of adult female turtles is 6 cm more than the mean shell length of adult male turtles. [6]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2012 Q2 [16]}}