AQA S2 2007 January — Question 3 8 marks

Exam BoardAQA
ModuleS2 (Statistics 2)
Year2007
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeTwo-tail z-test
DifficultyStandard +0.3 This is a straightforward two-tailed hypothesis test with clearly stated hypotheses, given summary statistics, and standard procedure. Students must calculate the sample mean (83.5), use the normal distribution with known variance, compute a z-statistic, and compare to critical values. While it requires multiple steps, each is routine for S2 level with no conceptual challenges or novel problem-solving required—slightly easier than average due to its mechanical nature.
Spec5.05c Hypothesis test: normal distribution for population mean

3 The handicap committee of a golf club has indicated that the mean score achieved by the club's members in the past was 85.9 . A group of members believes that recent changes to the golf course have led to a change in the mean score achieved by the club's members and decides to investigate this belief. A random sample of the scores, \(x\), of 100 club members was taken and is summarised by $$\sum x = 8350 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 15321$$ where \(\bar { x }\) denotes the sample mean.
Test, at the \(5 \%\) level of significance, the group's belief that the mean score of 85.9 has changed.

Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\bar{x} = 83.5\)B1
\(s^2 = \dfrac{1}{99}(15321) = 154.76\), \(s = 12.44\)B1 \(154 < s^2 \leq 155\), \((12.4 \leq s \leq 12.45)\)
\(H_0: \mu = 85.9\), \(H_1: \mu \neq 85.9\)B1
Under \(H_0\), \(\bar{X} \sim N\!\left(85.9, \dfrac{12.44^2}{100}\right)\)
\(z_{\text{crit}} = \pm 1.96\)B1 \(z = 1.96 + 2\) tail test used
\(z = \dfrac{83.5 - 85.9}{12.44/10} = -1.929\)M1 \(\dfrac{(\text{their } \bar{x}) - 85.9}{(\text{their } s)/10}\)
A1AWFW \(-1.94\) to \(1.92\)
Accept \(H_0\), reject the claimA1\(\checkmark\) On their \(z\)
Insufficient evidence to suggest that the mean has changed from 85.9 at the 5% level of significance.E1\(\checkmark\) Total: 8
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = 83.5$ | B1 | |
| $s^2 = \dfrac{1}{99}(15321) = 154.76$, $s = 12.44$ | B1 | $154 < s^2 \leq 155$, $(12.4 \leq s \leq 12.45)$ |
| $H_0: \mu = 85.9$, $H_1: \mu \neq 85.9$ | B1 | |
| Under $H_0$, $\bar{X} \sim N\!\left(85.9, \dfrac{12.44^2}{100}\right)$ | | |
| $z_{\text{crit}} = \pm 1.96$ | B1 | $z = 1.96 + 2$ tail test used |
| $z = \dfrac{83.5 - 85.9}{12.44/10} = -1.929$ | M1 | $\dfrac{(\text{their } \bar{x}) - 85.9}{(\text{their } s)/10}$ |
| | A1 | AWFW $-1.94$ to $1.92$ |
| Accept $H_0$, reject the claim | A1$\checkmark$ | On their $z$ |
| Insufficient evidence to suggest that the mean has changed from 85.9 at the 5% level of significance. | E1$\checkmark$ | **Total: 8** |

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3 The handicap committee of a golf club has indicated that the mean score achieved by the club's members in the past was 85.9 .

A group of members believes that recent changes to the golf course have led to a change in the mean score achieved by the club's members and decides to investigate this belief.

A random sample of the scores, $x$, of 100 club members was taken and is summarised by

$$\sum x = 8350 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 15321$$

where $\bar { x }$ denotes the sample mean.\\
Test, at the $5 \%$ level of significance, the group's belief that the mean score of 85.9 has changed.

\hfill \mbox{\textit{AQA S2 2007 Q3 [8]}}