CAIE FP2 2017 June — Question 7 7 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2017
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeSingle sample t-test
DifficultyStandard +0.8 This is a standard one-sample t-test with unknown variance requiring calculation of sample statistics from summations, correct hypothesis formulation, test statistic computation, and critical value comparison. While methodical, it's a direct application of the t-test procedure with no conceptual surprises, making it moderately above average difficulty for A-level but routine for Further Maths students who have learned this topic.
Spec5.05c Hypothesis test: normal distribution for population mean

A farmer grows a particular type of fruit tree. On average, the mass of fruit produced per tree has been 6.2 kg. He has developed a new kind of soil and claims that the mean mass of fruit produced per tree when growing in this new soil has increased. A random sample of 10 trees grown in the new soil is chosen. The masses, \(x\) kg, of fruit produced are summarised as follows. $$\Sigma x = 72.0 \qquad \Sigma x^2 = 542.0$$ Test at the 5% significance level whether the farmer's claim is justified, assuming a normal distribution. [7]

Question 7:
AnswerMarks Guidance
7x = 7⋅2 B1
s2 = (542 – 722 /10) / 9
AnswerMarks Guidance
[ = 118/45 or 2⋅622 or 1⋅6192 ]M1 Estimate population variance
(allow biased here: 2⋅36 or 1⋅5362)
H : µ = 6⋅2, H : µ > 6⋅2 (AEF)
AnswerMarks Guidance
0 1B1 State hypotheses (B0 forx …)
t = 1⋅83[3]
AnswerMarks Guidance
9, 0.95B1 State or use correct tabular t-value
t = (x – 6⋅2) / (s/√10) = 1⋅95
[Accept H :]
AnswerMarks Guidance
1M1 A1 Find value of t (or can comparex with 6⋅2 + 0⋅939 =
7⋅14 )
Consistent conclusion
AnswerMarks Guidance
Claim (of mean mass increased) is justified (AEF)B1 FT (FT on both t-values)
Total:7
QuestionAnswer Marks
Question 7:
7 | x = 7⋅2 | B1 | Find sample mean
s2 = (542 – 722 /10) / 9
[ = 118/45 or 2⋅622 or 1⋅6192 ] | M1 | Estimate population variance
(allow biased here: 2⋅36 or 1⋅5362)
H : µ = 6⋅2, H : µ > 6⋅2 (AEF)
0 1 | B1 | State hypotheses (B0 forx …)
t = 1⋅83[3]
9, 0.95 | B1 | State or use correct tabular t-value
t = (x – 6⋅2) / (s/√10) = 1⋅95
[Accept H :]
1 | M1 A1 | Find value of t (or can comparex with 6⋅2 + 0⋅939 =
7⋅14 )
Consistent conclusion
Claim (of mean mass increased) is justified (AEF) | B1 FT | (FT on both t-values)
Total: | 7
Question | Answer | Marks | Guidance
A farmer grows a particular type of fruit tree. On average, the mass of fruit produced per tree has been 6.2 kg. He has developed a new kind of soil and claims that the mean mass of fruit produced per tree when growing in this new soil has increased. A random sample of 10 trees grown in the new soil is chosen. The masses, $x$ kg, of fruit produced are summarised as follows.

$$\Sigma x = 72.0 \qquad \Sigma x^2 = 542.0$$

Test at the 5% significance level whether the farmer's claim is justified, assuming a normal distribution.
[7]

\hfill \mbox{\textit{CAIE FP2 2017 Q7 [7]}}