| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| 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 two hypothesis tests: an F-test for variance equality followed by a two-sample t-test. While the calculations are standard for Further Maths Statistics, students must correctly identify which test to use, interpret the F-test result to determine pooled vs unpooled variance for part (b), and explain the connection between tests. The multi-stage reasoning and need to link parts (a) and (c) to inform part (b) elevates this above routine application. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Site | Sample size \(( n )\) | Sample mean \(( \bar { x } )\) | Standard deviation \(( s )\) |
| \(A\) | 7 | 8.43 | 4.24 |
| \(B\) | 13 | 14.31 | 4.37 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \sigma_A^2 = \sigma_B^2\), \(H_1: \sigma_A^2 \neq \sigma_B^2\) | B1 | Both hypotheses correct |
| \(F = \frac{s_B^2}{s_A^2} = \frac{4.37^2}{4.24^2} = \frac{19.0969}{17.9776} = 1.0623\) | M1 A1 | Correct ratio (larger over smaller) |
| Critical value: \(F_{12,6}\) at 1% (two-tailed): \(F = 7.72\) | B1 | Correct critical value |
| \(1.0623 < 7.72\), do not reject \(H_0\). No evidence of difference in variances. | A1 | Correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \mu_B = \mu_A\), \(H_1: \mu_B > \mu_A\) | B1 | Both hypotheses (one-tailed) |
| Pooled variance: \(s_p^2 = \frac{6 \times 4.24^2 + 12 \times 4.37^2}{18} = \frac{107.866 + 228.817}{18} = \frac{336.683}{18} = 18.705\) | M1 A1 | Correct pooled estimate |
| \(t = \frac{14.31 - 8.43}{\sqrt{18.705\left(\frac{1}{13}+\frac{1}{7}\right)}} = \frac{5.88}{\sqrt{18.705 \times 0.2198}} = \frac{5.88}{\sqrt{4.111}} = \frac{5.88}{2.0276}\) | M1 | Correct test statistic formula |
| \(t = 2.900\) | A1 | Correct value |
| Critical value: \(t_{18}\) at 1% one-tail \(= 2.552\) | B1 | Correct critical value |
| \(2.900 > 2.552\), reject \(H_0\). Evidence that mean compression strength is greater at site \(B\) (younger site). | A1 | Correct conclusion in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The test in (a) justifies the use of a pooled \(t\)-test in (b), since equal variances are assumed for the pooled test. | B1 | Accept equivalent statements about justifying equal variance assumption |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \sigma_A^2 = \sigma_B^2$, $H_1: \sigma_A^2 \neq \sigma_B^2$ | B1 | Both hypotheses correct |
| $F = \frac{s_B^2}{s_A^2} = \frac{4.37^2}{4.24^2} = \frac{19.0969}{17.9776} = 1.0623$ | M1 A1 | Correct ratio (larger over smaller) |
| Critical value: $F_{12,6}$ at 1% (two-tailed): $F = 7.72$ | B1 | Correct critical value |
| $1.0623 < 7.72$, do not reject $H_0$. No evidence of difference in variances. | A1 | Correct conclusion |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu_B = \mu_A$, $H_1: \mu_B > \mu_A$ | B1 | Both hypotheses (one-tailed) |
| Pooled variance: $s_p^2 = \frac{6 \times 4.24^2 + 12 \times 4.37^2}{18} = \frac{107.866 + 228.817}{18} = \frac{336.683}{18} = 18.705$ | M1 A1 | Correct pooled estimate |
| $t = \frac{14.31 - 8.43}{\sqrt{18.705\left(\frac{1}{13}+\frac{1}{7}\right)}} = \frac{5.88}{\sqrt{18.705 \times 0.2198}} = \frac{5.88}{\sqrt{4.111}} = \frac{5.88}{2.0276}$ | M1 | Correct test statistic formula |
| $t = 2.900$ | A1 | Correct value |
| Critical value: $t_{18}$ at 1% one-tail $= 2.552$ | B1 | Correct critical value |
| $2.900 > 2.552$, reject $H_0$. Evidence that mean compression strength is greater at site $B$ (younger site). | A1 | Correct conclusion in context |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The test in (a) justifies the use of a pooled $t$-test in (b), since equal variances are assumed for the pooled test. | B1 | Accept equivalent statements about justifying equal variance assumption |
---3. An archaeologist is studying the compression strength of bricks at some ancient European sites. He took random samples from two sites $A$ and $B$ and recorded the compression strength of these bricks in appropriate units. The results are summarised below.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
Site & Sample size $( n )$ & Sample mean $( \bar { x } )$ & Standard deviation $( s )$ \\
\hline
$A$ & 7 & 8.43 & 4.24 \\
\hline
$B$ & 13 & 14.31 & 4.37 \\
\hline
\end{tabular}
\end{center}
It can be assumed that the compression strength of bricks is normally distributed.
\begin{enumerate}[label=(\alph*)]
\item Test, at the $2 \%$ level of significance, whether or not there is evidence of a difference in the variances of compression strength of the bricks between these two sites. State your hypotheses clearly.\\
(5)
Site $A$ is older than site $B$ and the archaeologist claims that the mean compression strength of the bricks was greater at the younger site.
\item Stating your hypotheses clearly and using a $1 \%$ level of significance, test the archaeologist's claim.
\item Explain briefly the importance of the test in part (a) to the test in part (b).
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2013 Q3 [12]}}