CAIE S2 2022 November — Question 2 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2022
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeOne-tail z-test (lower tail)
DifficultyModerate -0.3 This is a straightforward one-tailed z-test with known variance following a standard template: state hypotheses, calculate test statistic using given values (z = (9.8-10.3)/(2.6/√100) = -1.923), compare with critical value (-2.054 at 2% level), and conclude. All values are provided, requiring only routine application of the z-test procedure with no conceptual challenges or problem-solving insight.
Spec5.05c Hypothesis test: normal distribution for population mean

In the past, the mean length of a particular variety of worm has been 10.3 cm, with standard deviation 2.6 cm. Following a change in the climate, it is thought that the mean length of this variety of worm has decreased. The lengths of a random sample of 100 worms of this variety are found and the mean of this sample is found to be 9.8 cm. Assuming that the standard deviation remains at 2.6 cm, carry out a test at the 2% significance level of whether the mean length has decreased. [5]

Question 2:
AnswerMarks
2H : Population mean length = 10.3 cm
o
H : Population mean length < 10.3 cm
AnswerMarks Guidance
1B1 or μ = 10.3 (not just ‘mean’).
μ < 10.3
9.8−10.3
AnswerMarks Guidance
2.6/ 100M1 If ± 1.923 (or 0.0272) seen allow M1 implied.
= –1.923A1 Accept ± .
Accept 3sf.
AnswerMarks Guidance
–1.923 > –2.054 or –2.055M1 OE For a valid comparison.
Or compare 1 – ϕ(‘1.923’’) with 0.02 e.g. 0.0272 > 0.02
Use of CV 9.8 > 9.766 scores M1 A1 for 9.766 and M1
for comparison.
[Not reject H ] No evidence that [mean] length has decreased
AnswerMarks Guidance
0A1 FT FT their z. No contradictions, not definite, in context.
5
AnswerMarks Guidance
QuestionAnswer Marks
Question 2:
2 | H : Population mean length = 10.3 cm
o
H : Population mean length < 10.3 cm
1 | B1 | or μ = 10.3 (not just ‘mean’).
μ < 10.3
9.8−10.3

2.6/ 100 | M1 | If ± 1.923 (or 0.0272) seen allow M1 implied.
= –1.923 | A1 | Accept ± .
Accept 3sf.
–1.923 > –2.054 or –2.055 | M1 | OE For a valid comparison.
Or compare 1 – ϕ(‘1.923’’) with 0.02 e.g. 0.0272 > 0.02
Use of CV 9.8 > 9.766 scores M1 A1 for 9.766 and M1
for comparison.
[Not reject H ] No evidence that [mean] length has decreased
0 | A1 FT | FT their z. No contradictions, not definite, in context.
5
Question | Answer | Marks | Guidance
In the past, the mean length of a particular variety of worm has been 10.3 cm, with standard deviation 2.6 cm. Following a change in the climate, it is thought that the mean length of this variety of worm has decreased. The lengths of a random sample of 100 worms of this variety are found and the mean of this sample is found to be 9.8 cm.

Assuming that the standard deviation remains at 2.6 cm, carry out a test at the 2% significance level of whether the mean length has decreased. [5]

\hfill \mbox{\textit{CAIE S2 2022 Q2 [5]}}