OCR Further Statistics 2021 November — Question 4 9 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2021
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentral limit theorem
TypeJustifying CLT for confidence intervals
DifficultyStandard +0.3 This is a straightforward Further Statistics question requiring standard calculations (sample mean, unbiased variance estimate, confidence interval) and conceptual understanding of when CLT applies. The calculations are routine, and the conceptual part (c) tests basic understanding rather than deep insight. Slightly above average difficulty due to being Further Maths content, but mechanically simple.
Spec5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution

4 A random sample of 160 observations of a random variable \(X\) is selected. The sample can be summarised as follows. \(n = 160 \quad \sum x = 2688 \quad \sum x ^ { 2 } = 48398\)
  1. Calculate unbiased estimates of the following.
    1. \(\mathrm { E } ( X )\)
    2. \(\operatorname { Var } ( X )\)
  2. Find a 99\% confidence interval for \(\mathrm { E } ( X )\), giving the end-points of the interval correct to 4 significant figures.
  3. Explain whether it was necessary to use the Central Limit Theorem in answering
    1. part (a),
    2. part (b).

Question 4:
AnswerMarks Guidance
4(a) (i)
[1]1.1 Or exact equivalent
(ii)48398
−16.82 [= 20.2475]
160
× 160
159
AnswerMarks
= 20.3748…M1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1If single formula used, full marks if correct; M0M1 if
wrong but divisor 159 seen anywhere
Awrt 20.4, www
AnswerMarks Guidance
4(b) x ±z σ2 /160
z = 2.576
AnswerMarks
(15.88, 17.72)M1
A1
A1
AnswerMarks
[3]3.3
1.1
AnswerMarks
3.4Any z from Φ–1, 160 needed, allow √ errors
Or better, e.g. 2.575829
Both, 4 sf required by question, www (NB: σ 2 = 20.2475
gives same end-points to 4 SF but this gets M1A1A0)
AnswerMarks Guidance
4(c) (i)
distributionB1
[1]2.4 Mention at least one of E(X) and Var(X) explicitly, or
“not relevant to X ”
AnswerMarks Guidance
(ii)Needed in (b) as parent distribution not stated to be normal B1
[1]2.4 Must make it clear that two distributions are involved.
“n is large” etc: B0
Question 4:
4 | (a) | (i) | µˆ =x =16.8 | B1
[1] | 1.1 | Or exact equivalent
(ii) | 48398
−16.82 [= 20.2475]
160
× 160
159
= 20.3748… | M1
M1
A1
[3] | 1.1
1.1
1.1 | If single formula used, full marks if correct; M0M1 if
wrong but divisor 159 seen anywhere
Awrt 20.4, www
4 | (b) | x ±z σ2 /160
z = 2.576
(15.88, 17.72) | M1
A1
A1
[3] | 3.3
1.1
3.4 | Any z from Φ–1, 160 needed, allow √ errors
Or better, e.g. 2.575829
Both, 4 sf required by question, www (NB: σ 2 = 20.2475
gives same end-points to 4 SF but this gets M1A1A0)
4 | (c) | (i) | Not needed in (a) as E(X) and Var(X) are independent of the
distribution | B1
[1] | 2.4 | Mention at least one of E(X) and Var(X) explicitly, or
“not relevant to X ”
(ii) | Needed in (b) as parent distribution not stated to be normal | B1
[1] | 2.4 | Must make it clear that two distributions are involved.
“n is large” etc: B0
4 A random sample of 160 observations of a random variable $X$ is selected. The sample can be summarised as follows.\\
$n = 160 \quad \sum x = 2688 \quad \sum x ^ { 2 } = 48398$
\begin{enumerate}[label=(\alph*)]
\item Calculate unbiased estimates of the following.
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
\end{enumerate}\item Find a 99\% confidence interval for $\mathrm { E } ( X )$, giving the end-points of the interval correct to 4 significant figures.
\item Explain whether it was necessary to use the Central Limit Theorem in answering
\begin{enumerate}[label=(\roman*)]
\item part (a),
\item part (b).
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR Further Statistics 2021 Q4 [9]}}