CAIE FP2 2012 June — Question 9 10 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeSingle sample t-test
DifficultyStandard +0.3 This is a straightforward application of the one-sample t-test and confidence interval formula with standard steps: calculate sample mean and standard deviation, find the t-statistic, compare to critical value, then construct the confidence interval. All procedures are routine for Further Maths Statistics with no conceptual challenges or novel problem-solving required, making it slightly easier than average.
Spec5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution

A random sample of 8 observations of a normal random variable \(X\) gave the following summarised data, where \(\overline{x}\) denotes the sample mean. $$\Sigma x = 42.5 \quad \Sigma(x - \overline{x})^2 = 15.519$$ Test, at the 5% significance level, whether the population mean of \(X\) is greater than 4.5. [7] Calculate a 95% confidence interval for the population mean of \(X\). [3]

Question 9:
AnswerMarks
9Calculate sample mean: d = 42⋅5 / 8 = 5⋅3125 M1
Estimate population variance: s 2 = 15⋅519 / 7
(allow biased here: 1⋅940 or 1⋅393 2 ) = 2⋅217 or 1⋅489 2 M1
State hypotheses (A.E.F.): H0: µ = 4⋅5, H1: µ > 4⋅5 B1
Calculate value of t (to 3 sf): t = (d – 4⋅5)/(s/√8) = 1⋅54 M1 *A1
State or use correct tabular t value: t7, 0.95 = 1⋅89[5] *B1
(or can compared with 4⋅5 + 0⋅998 =
5⋅49[8])
Correct conclusion (AEF, dep *A1, *B1): Mean is not greater than 4⋅5 B1
Find confidence interval (allow z in place of
t) e.g.: 5⋅3125 ± t √{15⋅519/(7 × 8)} M1
Use of correct tabular value: t7, 0.975 = 2⋅36[5] A1
AnswerMarks Guidance
Evaluate C.I. correct to 3 s.f.: 5⋅31 ± 1⋅24[5] or [4⋅07, 6⋅56] A17
3[10]
Page 7Mark Scheme: Teachers’ version Syllabus
GCE A LEVEL – May/June 20129231 21
Question 9:
9 | Calculate sample mean: d = 42⋅5 / 8 = 5⋅3125 M1
Estimate population variance: s 2 = 15⋅519 / 7
(allow biased here: 1⋅940 or 1⋅393 2 ) = 2⋅217 or 1⋅489 2 M1
State hypotheses (A.E.F.): H0: µ = 4⋅5, H1: µ > 4⋅5 B1
Calculate value of t (to 3 sf): t = (d – 4⋅5)/(s/√8) = 1⋅54 M1 *A1
State or use correct tabular t value: t7, 0.95 = 1⋅89[5] *B1
(or can compared with 4⋅5 + 0⋅998 =
5⋅49[8])
Correct conclusion (AEF, dep *A1, *B1): Mean is not greater than 4⋅5 B1
Find confidence interval (allow z in place of
t) e.g.: 5⋅3125 ± t √{15⋅519/(7 × 8)} M1
Use of correct tabular value: t7, 0.975 = 2⋅36[5] A1
Evaluate C.I. correct to 3 s.f.: 5⋅31 ± 1⋅24[5] or [4⋅07, 6⋅56] A1 | 7
3 | [10]
Page 7 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – May/June 2012 | 9231 | 21
A random sample of 8 observations of a normal random variable $X$ gave the following summarised data, where $\overline{x}$ denotes the sample mean.
$$\Sigma x = 42.5 \quad \Sigma(x - \overline{x})^2 = 15.519$$

Test, at the 5% significance level, whether the population mean of $X$ is greater than 4.5. [7]

Calculate a 95% confidence interval for the population mean of $X$. [3]

\hfill \mbox{\textit{CAIE FP2 2012 Q9 [10]}}