AQA S2 2006 January — Question 6 8 marks

Exam BoardAQA
ModuleS2 (Statistics 2)
Year2006
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeOne-tail z-test (lower tail)
DifficultyModerate -0.3 This is a straightforward one-sample z-test with known variance following a standard template: state hypotheses, calculate test statistic using given values (σ=9, n=35, x̄=61.5), compare to critical value, and conclude. Part (b) requires recalling the definition of Type I error in context. The question is slightly easier than average because it's a direct application of a standard procedure with no complications, though it does require proper hypothesis test structure.
Spec5.05c Hypothesis test: normal distribution for population mean

6 In previous years, the marks obtained in a French test by students attending Topnotch College have been modelled satisfactorily by a normal distribution with a mean of 65 and a standard deviation of 9 . Teachers in the French department at Topnotch College suspect that this year their students are, on average, underachieving. In order to investigate this suspicion, the teachers selected a random sample of 35 students to take the French test and found that their mean score was 61.5.
  1. Investigate, at the \(5 \%\) level of significance, the teachers' suspicion.
  2. Explain, in the context of this question, the meaning of a Type I error.

6(a)
AnswerMarks Guidance
\(H_0: \mu = 65\)B1
\(H_1: \mu < 65\)
\[\bar{X} \sim N\left(65, \frac{81}{35}\right)\]
\(z_{\text{crit}} = -1.6449\)B1
\[z = \frac{61.5 - 65}{9/\sqrt{35}} = -2.30\]M1A1
Reject \(H_0\) at 5% level of significanceA1√
Evidence to suggest students may be under-achievingE1 6 marks
6(b)
AnswerMarks Guidance
Reject \(H_0\) when \(H_0\) true ⇒ Conclude that students are under-achieving when in fact they are notE1, E1 2 marks
Question 6 Total: 8 marks
**6(a)**
$H_0: \mu = 65$ | B1 | | 1-tailed test

$H_1: \mu < 65$ | |

$$\bar{X} \sim N\left(65, \frac{81}{35}\right)$$ | |

$z_{\text{crit}} = -1.6449$ | B1 | |

$$z = \frac{61.5 - 65}{9/\sqrt{35}} = -2.30$$ | M1A1 | | for σ√n used

Reject $H_0$ at 5% level of significance | A1√ | | (on their z-values)

Evidence to suggest students may be under-achieving | E1 | 6 marks

**6(b)**
Reject $H_0$ when $H_0$ true ⇒ Conclude that students are under-achieving when in fact they are not | E1, E1 | 2 marks

**Question 6 Total: 8 marks**

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6 In previous years, the marks obtained in a French test by students attending Topnotch College have been modelled satisfactorily by a normal distribution with a mean of 65 and a standard deviation of 9 .

Teachers in the French department at Topnotch College suspect that this year their students are, on average, underachieving.

In order to investigate this suspicion, the teachers selected a random sample of 35 students to take the French test and found that their mean score was 61.5.
\begin{enumerate}[label=(\alph*)]
\item Investigate, at the $5 \%$ level of significance, the teachers' suspicion.
\item Explain, in the context of this question, the meaning of a Type I error.
\end{enumerate}

\hfill \mbox{\textit{AQA S2 2006 Q6 [8]}}