AQA S2 2010 January — Question 3 7 marks

Exam BoardAQA
ModuleS2 (Statistics 2)
Year2010
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeSingle sample t-test
DifficultyStandard +0.3 This is a straightforward one-sample t-test with all necessary summary statistics provided. Students must state the hypotheses, calculate the test statistic using given summations, compare to critical value, and state the normality assumption. While it requires multiple steps, each is routine for S2 level with no conceptual challenges or novel problem-solving required, making it slightly easier than average.
Spec5.05c Hypothesis test: normal distribution for population mean

3 Lorraine bought a new golf club. She then practised with this club by using it to hit golf balls on a golf range. After several such practice sessions, she believed that there had been no change from 190 metres in the mean distance that she had achieved when using her old club. To investigate this belief, she measured, at her next practice session, the distance, \(x\) metres, of each of a random sample of 10 shots with her new club. Her results gave $$\sum x = 1840 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 1240$$ Investigate Lorraine's belief at the \(2 \%\) level of significance, stating any assumption that you make.
(7 marks)

Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Assume lengths of shots are normally distributedB1 \(s_n^2 = 124\); \(s_n = 11.1\); iff \(\frac{s_n}{3}\) used
\(\bar{x} = 184\); \(s^2 = \frac{1240}{9} = 137.7\ (s = 11.7)\)B1 CAO; AWFW 137.7 to 138; both \(\bar{x}\) and \(s^2\) (or \(s\))
\(H_0: \mu = 190\), \(H_1: \mu \neq 190\)B1 Both
\(t = \frac{184 - 190}{\sqrt{1240/9 \times 10}}\)M1 \(t = \frac{\bar{x} - 190}{s_{n-1}/\sqrt{10}}\) or \(\frac{\bar{x}-190}{s_n/\sqrt{9}}\)
\(t = -1.62\)A1 AWRT
\(\nu = 9 \Rightarrow t_{\text{crit}} = \pm 2.821\)B1 (accept 2.82)
\(-2.821 < -1.62 < 2.821\), accept \(H_0\); evidence to support Lorraine's belief at 2% level of significanceE1
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Assume lengths of shots are normally distributed | B1 | $s_n^2 = 124$; $s_n = 11.1$; iff $\frac{s_n}{3}$ used |
| $\bar{x} = 184$; $s^2 = \frac{1240}{9} = 137.7\ (s = 11.7)$ | B1 | CAO; AWFW 137.7 to 138; both $\bar{x}$ and $s^2$ (or $s$) |
| $H_0: \mu = 190$, $H_1: \mu \neq 190$ | B1 | Both |
| $t = \frac{184 - 190}{\sqrt{1240/9 \times 10}}$ | M1 | $t = \frac{\bar{x} - 190}{s_{n-1}/\sqrt{10}}$ or $\frac{\bar{x}-190}{s_n/\sqrt{9}}$ |
| $t = -1.62$ | A1 | AWRT |
| $\nu = 9 \Rightarrow t_{\text{crit}} = \pm 2.821$ | B1 | (accept 2.82) |
| $-2.821 < -1.62 < 2.821$, accept $H_0$; evidence to support Lorraine's belief at 2% level of significance | E1 | |

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3 Lorraine bought a new golf club. She then practised with this club by using it to hit golf balls on a golf range.

After several such practice sessions, she believed that there had been no change from 190 metres in the mean distance that she had achieved when using her old club.

To investigate this belief, she measured, at her next practice session, the distance, $x$ metres, of each of a random sample of 10 shots with her new club. Her results gave

$$\sum x = 1840 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 1240$$

Investigate Lorraine's belief at the $2 \%$ level of significance, stating any assumption that you make.\\
(7 marks)

\hfill \mbox{\textit{AQA S2 2010 Q3 [7]}}