| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Pooled variance estimate calculation |
| Difficulty | Standard +0.8 This is a multi-part S4 question requiring pooled variance calculation, two-sample t-test, F-test for variance equality, and critical evaluation of test validity. While the individual techniques are standard for Further Maths Statistics, the final part requires understanding the logical dependency between tests (that the t-test assumes equal variances, which part (c) contradicts), making this more conceptually demanding than routine hypothesis testing questions. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| \(n\) | \(s ^ { 2 }\) | \(\bar { x }\) | |
| Swimmers | 10 | 48 | 195 |
| Athletes | 13 | 161 | 186 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}\) | M1 | Correct pooled variance formula used |
| \(s_p^2 = \frac{9 \times 48 + 12 \times 161}{10+13-2} = \frac{432+1932}{21} = \frac{2364}{21} = 112.571...= 112.6\) | A1 | Correct calculation shown to 1 d.p. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \mu_S = \mu_A\), \(H_1: \mu_S \neq \mu_A\) | B1 | Both hypotheses correct |
| \(t = \frac{\bar{x}_S - \bar{x}_A}{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} = \frac{195-186}{\sqrt{112.6\left(\frac{1}{10}+\frac{1}{13}\right)}}\) | M1 | Correct structure of test statistic |
| \(= \frac{9}{\sqrt{112.6 \times 0.177...}} = \frac{9}{\sqrt{19.92...}} = \frac{9}{4.463...} = 2.016...\) | A1 | Correct value of test statistic |
| \(\nu = 21\), critical value \(t_{21}(5\%) = 1.721\) (two-tailed 10%) | B1 | Correct critical value |
| Since \(2.016 > 1.721\), reject \(H_0\). There is evidence of a difference in mean arm spans. | A1 | Correct conclusion in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \sigma_S^2 = \sigma_A^2\), \(H_1: \sigma_S^2 \neq \sigma_A^2\) | B1 | Both hypotheses correct |
| \(F = \frac{s_A^2}{s_S^2} = \frac{161}{48} = 3.354...\) | M1 | Correct ratio (larger over smaller) |
| \(F \sim F(12,9)\), critical value \(= 3.07\) (two-tailed 10%) | B1 | Correct distribution and critical value |
| Since \(3.354 > 3.07\), reject \(H_0\) | A1 | Correct comparison |
| There is evidence that the variances are not equal, supporting Bea's belief | A1 | Correct conclusion in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The same data has been used for both tests / the tests are not independent | B1 | Valid reason why Clive is unhappy |
| The test in part (b) is not valid, because it assumes equal variances but part (c) shows the variances are not equal | B1 | Correct identification of invalid test with reason |
## Question 3:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}$ | M1 | Correct pooled variance formula used |
| $s_p^2 = \frac{9 \times 48 + 12 \times 161}{10+13-2} = \frac{432+1932}{21} = \frac{2364}{21} = 112.571...= 112.6$ | A1 | Correct calculation shown to 1 d.p. |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu_S = \mu_A$, $H_1: \mu_S \neq \mu_A$ | B1 | Both hypotheses correct |
| $t = \frac{\bar{x}_S - \bar{x}_A}{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} = \frac{195-186}{\sqrt{112.6\left(\frac{1}{10}+\frac{1}{13}\right)}}$ | M1 | Correct structure of test statistic |
| $= \frac{9}{\sqrt{112.6 \times 0.177...}} = \frac{9}{\sqrt{19.92...}} = \frac{9}{4.463...} = 2.016...$ | A1 | Correct value of test statistic |
| $\nu = 21$, critical value $t_{21}(5\%) = 1.721$ (two-tailed 10%) | B1 | Correct critical value |
| Since $2.016 > 1.721$, reject $H_0$. There is evidence of a difference in mean arm spans. | A1 | Correct conclusion in context |
**Part (c):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \sigma_S^2 = \sigma_A^2$, $H_1: \sigma_S^2 \neq \sigma_A^2$ | B1 | Both hypotheses correct |
| $F = \frac{s_A^2}{s_S^2} = \frac{161}{48} = 3.354...$ | M1 | Correct ratio (larger over smaller) |
| $F \sim F(12,9)$, critical value $= 3.07$ (two-tailed 10%) | B1 | Correct distribution and critical value |
| Since $3.354 > 3.07$, reject $H_0$ | A1 | Correct comparison |
| There is evidence that the variances are not equal, supporting Bea's belief | A1 | Correct conclusion in context |
**Part (d):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| The same data has been used for both tests / the tests are not independent | B1 | Valid reason why Clive is unhappy |
| The test in part (b) is not valid, because it assumes equal variances but part (c) shows the variances are not equal | B1 | Correct identification of invalid test with reason |\begin{enumerate}
\item As part of their research two sports science students, Ali and Bea, select a random sample of 10 adult male swimmers and a random sample of 13 adult male athletes from local sports clubs. They measure the arm span, $x \mathrm {~cm}$, of each person selected.\\
The data are summarised in the table below
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
& $n$ & $s ^ { 2 }$ & $\bar { x }$ \\
\hline
Swimmers & 10 & 48 & 195 \\
\hline
Athletes & 13 & 161 & 186 \\
\hline
\end{tabular}
\end{center}
The students know that the arm spans of adult male swimmers and of adult male athletes may each be assumed to be normally distributed.\\
They decide to share out the data analysis, with Ali investigating the means of the two distributions and Bea investigating the variances of the two distributions.
Ali assumes that the variances of the two distributions are equal. She calculates the pooled estimate of variance, $s _ { p } { } ^ { 2 }$\\
(a) Show that $s _ { p } { } ^ { 2 } = 112.6$ to 1 decimal place.
Ali claims that there is no difference in the mean arm spans of adult male swimmers and of adult male athletes.\\
(b) Stating your hypotheses clearly, test this claim at the $10 \%$ level of significance.\\
(5)
Bea believes that the variances of the arm spans of adult male swimmers and adult male athletes are not equal.\\
(c) Show that, at the $10 \%$ level of significance, the data support Bea's belief. State your hypotheses and show your working clearly.
Ali and Bea combine their work and present their results to their tutor, Clive.\\
(d) Explain why Clive is not happy with their research and state, with a reason, which of the tests in parts (b) and (c) is not valid.\\
\hfill \mbox{\textit{Edexcel S4 2015 Q3 [14]}}