CAIE Further Paper 4 2021 June — Question 1 6 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2021
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeSingle sample t-test
DifficultyStandard +0.3 This is a straightforward one-sample t-test with clearly stated hypotheses (μ > 14.8), given summary statistics requiring calculation of sample mean and standard deviation, then comparison with critical value from tables. While it's a Further Maths topic, the execution is mechanical with no conceptual challenges beyond standard procedure.
Spec5.05c Hypothesis test: normal distribution for population mean

Farmer A grows apples of a certain variety. Each tree produces 14.8 kg of apples, on average, per year. Farmer B grows apples of the same variety and claims that his apple trees produce a higher mass of apples per year than Farmer A's trees. The masses of apples from Farmer B's trees may be assumed to be normally distributed. A random sample of 10 trees from Farmer B is chosen. The masses, \(x\) kg, of apples produced in a year are summarised as follows. $$\sum x = 152.0 \qquad \sum x^2 = 2313.0$$ Test, at the 5% significance level, whether Farmer B's claim is justified. [6]

Question 1:
AnswerMarks
11 1522 
s2 = 2313− =0.28889
9 10
AnswerMarks Guidance
 M1 13
, accept unsimplified.
45
H : μ = 14.8 H : μ > 14.8
AnswerMarks Guidance
0 1B1 If μ not used, ‘population mean’ required.
Must see 14.8, must be = and >.
152
−14.8
[ t= ] 10
AnswerMarks Guidance
s/ 10M1 Using unbiased estimate.
p-value is 0.0215.
[ ]
AnswerMarks Guidance
t= 2.35A1 Rounds to 2.35.
Tabular value = 1.833
‘2.35’ > 1.833
Reject H
AnswerMarks Guidance
0M1 Comparison with 1.833 and correct FT conclusion.
Sufficient evidence to accept Farmer B’s claim oeA1 Correct conclusion, in context, following correct work. Level
of uncertainty in language is used.
Allow ±1 difference in third significant figure of their t for this
mark.
6
AnswerMarks Guidance
QuestionAnswer Marks
Question 1:
1 | 1 1522 
s2 = 2313− =0.28889

9 10
  | M1 | 13
, accept unsimplified.
45
H : μ = 14.8 H : μ > 14.8
0 1 | B1 | If μ not used, ‘population mean’ required.
Must see 14.8, must be = and >.
152
−14.8
[ t= ] 10
s/ 10 | M1 | Using unbiased estimate.
p-value is 0.0215.
[ ]
t= 2.35 | A1 | Rounds to 2.35.
Tabular value = 1.833
‘2.35’ > 1.833
Reject H
0 | M1 | Comparison with 1.833 and correct FT conclusion.
Sufficient evidence to accept Farmer B’s claim oe | A1 | Correct conclusion, in context, following correct work. Level
of uncertainty in language is used.
Allow ±1 difference in third significant figure of their t for this
mark.
6
Question | Answer | Marks | Guidance
Farmer A grows apples of a certain variety. Each tree produces 14.8 kg of apples, on average, per year. Farmer B grows apples of the same variety and claims that his apple trees produce a higher mass of apples per year than Farmer A's trees. The masses of apples from Farmer B's trees may be assumed to be normally distributed.

A random sample of 10 trees from Farmer B is chosen. The masses, $x$ kg, of apples produced in a year are summarised as follows.

$$\sum x = 152.0 \qquad \sum x^2 = 2313.0$$

Test, at the 5% significance level, whether Farmer B's claim is justified. [6]

\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q1 [6]}}