Pre-U Pre-U 9795/2 2010 June — Question 8 8 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2010
SessionJune
Marks8
TopicConfidence intervals
TypeComment on claim using CI
DifficultyStandard +0.3 This is a standard two-sample t-test confidence interval question with pooled variance. Part (i) requires routine application of a formula (pooled variance, standard error, t-critical value) with straightforward arithmetic. Part (ii) is a simple interpretation of whether zero lies in the interval. While it's a Further Maths topic, the execution is mechanical with no novel insight required, making it slightly easier than average A-level difficulty.
Spec5.05d Confidence intervals: using normal distribution

8 Two groups of Year 12 pupils, one at each of schools \(A\) and \(B\), are given the same mathematics test. The scores, \(x\) and \(y\), of pupils at schools \(A\) and \(B\) respectively are summarised as follows.
School \(A\)\(n _ { A } = 15\)\(\bar { x } = 53\)\(\Sigma ( x - \bar { x } ) ^ { 2 } = 925\)
School \(B\)\(n _ { B } = 12\)\(\bar { y } = 47\)\(\Sigma ( y - \bar { y } ) ^ { 2 } = 850\)
  1. Assuming that the two groups are random samples from independent normal populations with means \(\mu _ { A }\) and \(\mu _ { B }\) respectively and a common, but unknown, variance, construct a \(98 \%\) confidence interval for \(\mu _ { A } - \mu _ { B }\).
  2. Comment, with a reason, on any difference in ability between the two schools.

(i) Assuming common variance is \(\sigma^2\)
\(s^2 = \frac{925 + 850}{15 + 12 - 2} = \frac{1775}{25} = 71\) — M1A1
\((\nu = 25)\ t = 2.485\) — B1
98% confidence limits are \(6 \pm 2.485\sqrt{71}\sqrt{\frac{1}{15} + \frac{1}{12}}\) — M1A1\(\checkmark\)
98% confidence interval is \((-2.11,\ 14.1)\) AWRT — A1 [6]
(ii) \(0 \in \text{CI} \Rightarrow\) there is insufficient evidence to demonstrate a difference in ability on this test. — M1A1\(\checkmark\) [2]
[Total: 8]
**(i)** Assuming common variance is $\sigma^2$

$s^2 = \frac{925 + 850}{15 + 12 - 2} = \frac{1775}{25} = 71$ — M1A1

$(\nu = 25)\ t = 2.485$ — B1

98% confidence limits are $6 \pm 2.485\sqrt{71}\sqrt{\frac{1}{15} + \frac{1}{12}}$ — M1A1$\checkmark$

98% confidence interval is $(-2.11,\ 14.1)$ AWRT — A1 [6]

**(ii)** $0 \in \text{CI} \Rightarrow$ there is insufficient evidence to demonstrate a difference in ability on this test. — M1A1$\checkmark$ [2]

**[Total: 8]**
8 Two groups of Year 12 pupils, one at each of schools $A$ and $B$, are given the same mathematics test. The scores, $x$ and $y$, of pupils at schools $A$ and $B$ respectively are summarised as follows.

\begin{center}
\begin{tabular}{ l l l l }
School $A$ & $n _ { A } = 15$ & $\bar { x } = 53$ & $\Sigma ( x - \bar { x } ) ^ { 2 } = 925$ \\
School $B$ & $n _ { B } = 12$ & $\bar { y } = 47$ & $\Sigma ( y - \bar { y } ) ^ { 2 } = 850$ \\
\end{tabular}
\end{center}

(i) Assuming that the two groups are random samples from independent normal populations with means $\mu _ { A }$ and $\mu _ { B }$ respectively and a common, but unknown, variance, construct a $98 \%$ confidence interval for $\mu _ { A } - \mu _ { B }$.\\
(ii) Comment, with a reason, on any difference in ability between the two schools.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2010 Q8 [8]}}