Pre-U Pre-U 9795/2 2013 June — Question 4 10 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2013
SessionJune
Marks10
TopicConfidence intervals
TypeCI from raw data list
DifficultyStandard +0.8 This requires calculating a two-sample t-confidence interval from raw data, involving computing means, pooled variance, and appropriate degrees of freedom. While the calculations are systematic, the small sample sizes, need for pooled variance assumption, and correct application of t-distribution make this more demanding than standard single-sample CI questions.
Spec5.05d Confidence intervals: using normal distribution

4 The broadband speed in village \(P\) was measured on 8 randomly selected occasions and the broadband speed in village \(Q\) was measured on 6 randomly selected occasions. The results, measured in megabits per second, are shown below.
Village \(P :\)4.83.52.93.74.24.65.13.3
Village \(Q :\)2.41.92.33.12.72.9
  1. Calculate a \(90 \%\) confidence interval for the difference in mean broadband speed in these two villages.
  2. State two assumptions that you have made in carrying out the calculation.

(i) \(P\): \(\bar{x}_1 = 4.0125\), S.D.\(_P = 0.73218\ldots\) or \(0.78273\ldots\) (or variance) B1
\(Q\): \(\bar{x}_2 = 2.55\), S.D.\(_Q = 0.39895\ldots\) or \(0.43703\ldots\) (or variance) B1
\(\hat{\sigma}^2 = \frac{4.28875 + 0.955}{8 + 6 - 2} = 0.43697\) M1A1
\(t_{12}(0.95) = 1.782\) B1
90% confidence limits are:
\(1.4625 \pm (1.782 \times 0.66104\ldots \times \sqrt{8^{-1} + 6^{-1}})\) (ft on \(t\) value.) M1A1
90% confidence interval is \((0.826, 2.10)\) (Accept 0.827 from \(t = 1.78\).) A1 [8]
(ii) Distributions of broadband speeds are normal. B1
The populations have a common variance. B1 [2]
(i) $P$: $\bar{x}_1 = 4.0125$, S.D.$_P = 0.73218\ldots$ or $0.78273\ldots$ (or variance) **B1**

$Q$: $\bar{x}_2 = 2.55$, S.D.$_Q = 0.39895\ldots$ or $0.43703\ldots$ (or variance) **B1**

$\hat{\sigma}^2 = \frac{4.28875 + 0.955}{8 + 6 - 2} = 0.43697$ **M1A1**

$t_{12}(0.95) = 1.782$ **B1**

90% confidence limits are:
$1.4625 \pm (1.782 \times 0.66104\ldots \times \sqrt{8^{-1} + 6^{-1}})$ (ft on $t$ value.) **M1A1**✓

90% confidence interval is $(0.826, 2.10)$ (Accept 0.827 from $t = 1.78$.) **A1** [8]

(ii) Distributions of broadband speeds are normal. **B1**
The populations have a common variance. **B1** [2]
4 The broadband speed in village $P$ was measured on 8 randomly selected occasions and the broadband speed in village $Q$ was measured on 6 randomly selected occasions. The results, measured in megabits per second, are shown below.

\begin{center}
\begin{tabular}{ l l l l l l l l l }
Village $P :$ & 4.8 & 3.5 & 2.9 & 3.7 & 4.2 & 4.6 & 5.1 & 3.3 \\
Village $Q :$ & 2.4 & 1.9 & 2.3 & 3.1 & 2.7 & 2.9 &  &  \\
\end{tabular}
\end{center}

(i) Calculate a $90 \%$ confidence interval for the difference in mean broadband speed in these two villages.\\
(ii) State two assumptions that you have made in carrying out the calculation.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2013 Q4 [10]}}