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 one-sample t-test with unknown variance requiring calculation of sample statistics, correct hypothesis formulation, test statistic computation, and critical value comparison. While methodologically straightforward for Further Maths students, it requires careful execution of multiple steps (calculating mean and variance from summations, applying the t-test formula, using t-tables) with potential for arithmetic errors, placing it moderately above average difficulty.
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. $$\Sigma x = 72.0 \quad \Sigma x^2 = 542.0$$ 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.

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

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]}}