AQA S2 2007 June — Question 3 8 marks

Exam BoardAQA
ModuleS2 (Statistics 2)
Year2007
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeOne-tail z-test (lower tail)
DifficultyStandard +0.3 This is a straightforward one-sample z-test with all necessary information provided. Students must calculate the sample mean (34.6), recognize this as a one-tailed test (H₁: μ < 36), compute the test statistic using the given variance, and compare to critical value. While it requires multiple steps and proper hypothesis test structure, it's a standard S2 textbook exercise with no conceptual surprises or novel problem-solving required.
Spec5.05c Hypothesis test: normal distribution for population mean

3 David is the professional coach at the golf club where Becki is a member. He claims that, after having a series of lessons with him, the mean number of putts that Becki takes per round of golf will reduce from her present mean of 36 . After having the series of lessons with David, Becki decides to investigate his claim.
She therefore records, for each of a random sample of 50 rounds of golf, the number of putts, \(x\), that she takes to complete the round. Her results are summarised below, where \(\bar { x }\) denotes the sample mean. $$\sum x = 1730 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 784$$ Using a \(z\)-test and the \(1 \%\) level of significance, investigate David's claim.

Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(H_0: \mu = 36\); \(H_1: \mu < 36\)B1
\(\bar{x} = \frac{1730}{50} = 34.6\)B1
\(s^2 = \frac{784}{49} = 16\)B1
Test statistic: \(z = \frac{34.6 - 36}{\frac{4}{\sqrt{50}}} = -2.47\)M1, A1 \(-2.48\) to \(-2.47\)
\(z_{\text{crit}} = -2.3263\)B1
Reject \(H_0\)\(A1\checkmark\)
Sufficient evidence at 1% level of significance to support David's claim\(E1\checkmark\)
# Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu = 36$; $H_1: \mu < 36$ | B1 | |
| $\bar{x} = \frac{1730}{50} = 34.6$ | B1 | |
| $s^2 = \frac{784}{49} = 16$ | B1 | |
| Test statistic: $z = \frac{34.6 - 36}{\frac{4}{\sqrt{50}}} = -2.47$ | M1, A1 | $-2.48$ to $-2.47$ |
| $z_{\text{crit}} = -2.3263$ | B1 | |
| Reject $H_0$ | $A1\checkmark$ | |
| Sufficient evidence at 1% level of significance to support David's claim | $E1\checkmark$ | |

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3 David is the professional coach at the golf club where Becki is a member. He claims that, after having a series of lessons with him, the mean number of putts that Becki takes per round of golf will reduce from her present mean of 36 .

After having the series of lessons with David, Becki decides to investigate his claim.\\
She therefore records, for each of a random sample of 50 rounds of golf, the number of putts, $x$, that she takes to complete the round. Her results are summarised below, where $\bar { x }$ denotes the sample mean.

$$\sum x = 1730 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 784$$

Using a $z$-test and the $1 \%$ level of significance, investigate David's claim.

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