CAIE FP2 2019 November — Question 8 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2019
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeTwo-sample t-test equal variance
DifficultyChallenging +1.2 This is a standard two-sample t-test with pooled variance requiring calculation of sample statistics from summations, pooling variances, computing the test statistic, and comparing to critical values. While it involves multiple computational steps and careful bookkeeping across two samples, it follows a completely routine procedure taught in Further Maths Statistics with no conceptual challenges or novel problem-solving required. The 9 marks reflect computational length rather than difficulty.
Spec5.05c Hypothesis test: normal distribution for population mean

A random sample of 8 elephants from region \(A\) is taken and their weights, \(x\) tonnes, are recorded. (1 tonne = 1000 kg.) The results are summarised as follows. $$\Sigma x = 32.4 \quad \Sigma x^2 = 131.82$$ A random sample of 10 elephants from region \(B\) is taken. Their weights give a sample mean of 3.78 tonnes and an unbiased variance estimate of 0.1555 tonnes\(^2\). The distributions of the weights of elephants in regions \(A\) and \(B\) are both assumed to be normal with the same population variance. Test at the 10% significance level whether the mean weight of elephants in region \(A\) is the same as the mean weight of elephants in region \(B\). [9]

Question 8:
AnswerMarks Guidance
8x = 32⋅4 / 8 = 4⋅05
AB1 Find sample mean for A
s 2 = (131⋅82 – 32⋅42/8) / 7
A
s 2 = 3/35 (or 0⋅08571 or 0⋅29282 both to 3 s.f.)
AnswerMarks Guidance
AM1 Estimate or imply popln. variance for A
H : µ = µ , H : µ ≠ µ
AnswerMarks Guidance
0 A B 1 A BB1 State hypotheses. AEF
3999
s2 = (7 s 2 + 9 s 2) / 16 = 0⋅12497 or 0⋅35352 or
A B
AnswerMarks Guidance
32000M1 A1 Estimate (pooled) common variance
t = 1⋅746
AnswerMarks Guidance
16, 0.95B1* State or use correct tabular t value
[–] t = (x –x ) / s √(1/8 + 1/10)
AnswerMarks Guidance
A BM1
= 0⋅27 / 0⋅1677 = 1⋅61A1 Find value of t (or can comparex –x = 0⋅27 with 0⋅293)
A B
t < 1⋅75 so [accept H ] mean masses are the same
AnswerMarks Guidance
0DB1 Correct conclusion (FT on t, dep B1*). AEF
9
AnswerMarks Guidance
QuestionAnswer Marks
Question 8:
8 | x = 32⋅4 / 8 = 4⋅05
A | B1 | Find sample mean for A
s 2 = (131⋅82 – 32⋅42/8) / 7
A
s 2 = 3/35 (or 0⋅08571 or 0⋅29282 both to 3 s.f.)
A | M1 | Estimate or imply popln. variance for A
H : µ = µ , H : µ ≠ µ
0 A B 1 A B | B1 | State hypotheses. AEF
3999
s2 = (7 s 2 + 9 s 2) / 16 = 0⋅12497 or 0⋅35352 or
A B
32000 | M1 A1 | Estimate (pooled) common variance
t = 1⋅746
16, 0.95 | B1* | State or use correct tabular t value
[–] t = (x –x ) / s √(1/8 + 1/10)
A B | M1
= 0⋅27 / 0⋅1677 = 1⋅61 | A1 | Find value of t (or can comparex –x = 0⋅27 with 0⋅293)
A B
t < 1⋅75 so [accept H ] mean masses are the same
0 | DB1 | Correct conclusion (FT on t, dep B1*). AEF
9
Question | Answer | Marks | Guidance
A random sample of 8 elephants from region $A$ is taken and their weights, $x$ tonnes, are recorded. (1 tonne = 1000 kg.) The results are summarised as follows.
$$\Sigma x = 32.4 \quad \Sigma x^2 = 131.82$$

A random sample of 10 elephants from region $B$ is taken. Their weights give a sample mean of 3.78 tonnes and an unbiased variance estimate of 0.1555 tonnes$^2$. The distributions of the weights of elephants in regions $A$ and $B$ are both assumed to be normal with the same population variance. Test at the 10% significance level whether the mean weight of elephants in region $A$ is the same as the mean weight of elephants in region $B$.
[9]

\hfill \mbox{\textit{CAIE FP2 2019 Q8 [9]}}