| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Standard +0.3 This is a straightforward application of t-distribution confidence intervals with standard formulas. Part (a) involves routine calculation of standard error and constructing a confidence interval from given values. Part (b) requires working backwards from interval width to find the confidence level, which is slightly less routine but still a standard textbook exercise requiring no novel insight. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(d^2 = \frac{93}{12} = 7.75\) | M1, A1 | \(d = \sqrt{\frac{93}{12}} = \sqrt{7.75} \Rightarrow d^2 = 7.75\) |
| Answer | Marks | Guidance |
|---|---|---|
| 80% CI \(= 64.8 \pm 1.363 \times \sqrt{7.75} = 64.8 \pm 3.79 = (61.0, 68.6)\) | B1, M1, A1 | \(t_{11} = 1.363\) or \(1.36\); \(64.8 \pm t_{11}\sqrt{7.75}\) iff \(t_{11} = 1.363\) or \(1.796\); AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| \((64.8 - 5,\ 64.8 + 5) = (59.8, 69.8)\) | B1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| \(w = 2\sqrt{7.75} \times t = 10 \Rightarrow t = 1.796\) | M1, A1 | \(t = 1.79\) to \(1.80\) |
| \(P(X \geq 1.796) = 0.05\), \(P(X \leq -1.796) = 0.05 \Rightarrow P( | X | \leq 1.796) = 0.90\), 90% Confidence Level |
## Question 7(a)(i):
| $d^2 = \frac{93}{12} = 7.75$ | M1, A1 | $d = \sqrt{\frac{93}{12}} = \sqrt{7.75} \Rightarrow d^2 = 7.75$ |
## Question 7(a)(ii):
| 80% CI $= 64.8 \pm 1.363 \times \sqrt{7.75} = 64.8 \pm 3.79 = (61.0, 68.6)$ | B1, M1, A1 | $t_{11} = 1.363$ or $1.36$; $64.8 \pm t_{11}\sqrt{7.75}$ iff $t_{11} = 1.363$ or $1.796$; AWRT |
## Question 7(b)(i):
| $(64.8 - 5,\ 64.8 + 5) = (59.8, 69.8)$ | B1 | AWRT |
## Question 7(b)(ii):
| $w = 2\sqrt{7.75} \times t = 10 \Rightarrow t = 1.796$ | M1, A1 | $t = 1.79$ to $1.80$ |
| $P(X \geq 1.796) = 0.05$, $P(X \leq -1.796) = 0.05 \Rightarrow P(|X| \leq 1.796) = 0.90$, 90% Confidence Level | M1, A1 | iff $t = 1.796$ correct |
---7 Jim , a mathematics teacher, knows that the marks, $X$, achieved by his students can be modelled by a normal distribution with unknown mean $\mu$ and unknown variance $\sigma ^ { 2 }$.
Jim selects 12 students at random and from their marks he calculates that $\bar { x } = 64.8$ and $s ^ { 2 } = 93.0$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item An estimate for the standard error of the sample mean is $d$. Show that $d ^ { 2 } = 7.75$.
\item Construct an $80 \%$ confidence interval for $\mu$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Write down a confidence interval for $\mu$, based on Jim's sample of 12 students, which has a width of 10 marks.
\item Determine the percentage confidence level for the interval found in part (b)(i).
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2010 Q7 [10]}}