CAIE S2 2008 November — Question 2 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2008
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeTwo-tail z-test
DifficultyModerate -0.3 This is a straightforward two-tail z-test with known population standard deviation. Students must state hypotheses, calculate the sample mean (35 minutes), compute the test statistic using z = (x̄ - μ)/(σ/√n), and compare with critical values at 10% significance. While it requires correct application of the normal distribution test procedure, it's a standard textbook exercise with small sample size (n=3) making calculations simple. Slightly easier than average due to minimal computation and being a direct application of a learned technique.
Spec2.05e Hypothesis test for normal mean: known variance

2 The times taken for the pupils in Ming's year group to do their English homework have a normal distribution with standard deviation 15.7 minutes. A teacher estimates that the mean time is 42 minutes. The times taken by a random sample of 3 students from the year group were 27, 35 and 43 minutes. Carry out a hypothesis test at the \(10 \%\) significance level to determine whether the teacher's estimate for the mean should be accepted, stating the null and alternative hypotheses.

\(H_0: \mu = 42\)
\(H_1: \mu \neq 42\)
AnswerMarks Guidance
Test statistic \(z = \frac{35 - 42}{(15.7/\sqrt{3})} = -0.772\)B1 Correct \(H_0\) and \(H_1\)
M1Standardising attempt, must have \(\sqrt{3}\) used correctly
A1Correct test statistic (±)
M1Correct comparison. ±1.645 seen or ±1.64 or ±1.65 must compare + with + or – with – (or 1.282 if one-tail test being followed)
A1ft [5]Correct conclusion. (ft) No contradictions.
OR: \(42 \pm1.645(15.7/\sqrt{3})\) (27.1, 56.9) 27.1 < 35 < 56.9 Teacher's estimate can be accepted.M1
A1Correct comparison
M1
A1ftCorrect conclusion
$H_0: \mu = 42$
$H_1: \mu \neq 42$

Test statistic $z = \frac{35 - 42}{(15.7/\sqrt{3})} = -0.772$ | B1 | Correct $H_0$ and $H_1$
| M1 | Standardising attempt, must have $\sqrt{3}$ used correctly
| A1 | Correct test statistic (±)
| M1 | Correct comparison. ±1.645 seen or ±1.64 or ±1.65 must compare + with + or – with – (or 1.282 if one-tail test being followed)
| A1ft [5] | Correct conclusion. (ft) No contradictions.

OR: $42 \pm1.645(15.7/\sqrt{3})$ (27.1, 56.9) 27.1 < 35 < 56.9 Teacher's estimate can be accepted. | M1 | 
| A1 | Correct comparison
| M1 | 
| A1ft | Correct conclusion

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2 The times taken for the pupils in Ming's year group to do their English homework have a normal distribution with standard deviation 15.7 minutes. A teacher estimates that the mean time is 42 minutes. The times taken by a random sample of 3 students from the year group were 27, 35 and 43 minutes. Carry out a hypothesis test at the $10 \%$ significance level to determine whether the teacher's estimate for the mean should be accepted, stating the null and alternative hypotheses.

\hfill \mbox{\textit{CAIE S2 2008 Q2 [5]}}