| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Hypothesis test for mean |
| Difficulty | Moderate -0.8 This is a straightforward confidence interval interpretation question requiring only reading values from a spreadsheet and comparing them to a given value (2.6%). No calculations needed, just basic statistical reasoning about whether 2.6 falls within the given interval. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| A | B | C | D | E | F | G | H | 1 | J | K | |
| 1 | Original brightness | 1075 | 1121 | 1106 | 1095 | 1101 | 1109 | 1114 | 1123 | 1108 | 1115 |
| 2 | After 10000 hours | 1042 | 1084 | 1076 | 1065 | 1070 | 1079 | 1081 | 1091 | 1080 | 1082 |
| 3 | Percentage reduction | 3.07 | 3.30 | 2.71 | 2.74 | 2.82 | 2.71 | 2.96 | 2.85 | 2.53 | 2.96 |
| 4 | |||||||||||
| 5 | |||||||||||
| 6 | Sample mean (\%) | 2.8650 | |||||||||
| 7 | Sample sd (\%) | 0.2179 | |||||||||
| 8 | SE | 0.0689 | |||||||||
| 9 | DF | 9 | |||||||||
| 10 | tvalue | 2.262 | |||||||||
| 11 | Lower limit | 2.709 | |||||||||
| 12 | Upper limit | 3.021 | |||||||||
| 1.3 | |||||||||||
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | 2.709 < µ < 3.021 |
| [1] | 1.1 | |
| 1 | (b) | Confidence interval does suggest that there is a |
| Answer | Marks |
|---|---|
| …since the interval does not contain 2.6 | B1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (c) | SE (standard error) is the standard deviation of the |
| Answer | Marks |
|---|---|
| 0.2179 | E1 |
| Answer | Marks |
|---|---|
| [2] | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (d) | =The popul a(t i=on0 o.0f 6d8if9fe)rences must be Normally |
| Answer | Marks |
|---|---|
| distributed. | E1 |
| Answer | Marks |
|---|---|
| [2] | 2.3 |
| 1.2 | For population of differences |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (e) | The pairing will eliminate any differences in |
| Answer | Marks |
|---|---|
| compare the brightness before and after | E1 |
| Answer | Marks |
|---|---|
| [2] | 2.2b |
| 2.2b | Give 1 mark for any valid comment. |
Question 1:
1 | (a) | 2.709 < µ < 3.021 | B1
[1] | 1.1
1 | (b) | Confidence interval does suggest that there is a
difference in the average percentage reduction in
brightness …
…since the interval does not contain 2.6 | B1
B1
[2] | 3.4
2.2b
1 | (c) | SE (standard error) is the standard deviation of the
sample mean = sample sd /√n
0.2179 | E1
E1
[2] | 2.4
1.1
1 | (d) | =The popul a(t i=on0 o.0f 6d8if9fe)rences must be Normally
√10
distributed. | E1
E1
[2] | 2.3
1.2 | For population of differences
For Normally distributed
1 | (e) | The pairing will eliminate any differences in
brightness of different bulbs and so will only
compare the brightness before and after | E1
E1
[2] | 2.2b
2.2b | Give 1 mark for any valid comment.
For 2 marks must include pairing1 Over time LED light bulbs gradually lose brightness. For a particular type of LED bulb, it is known that the mean reduction in brightness after 10000 hours is $2.6 \%$. A manufacturer produces a new version of this bulb, which costs less to make, but is claimed to have the same reduction in brightness after 10000 hours as the previous version.
In order to check this claim, a random sample of 10 bulbs is selected. For each bulb, the original brightness and the brightness after 10000 hours are measured, in suitable units. A spreadsheet is used to produce a $95 \%$ confidence interval for the mean percentage reduction in brightness. A screenshot of the spreadsheet is shown in Fig. 1.
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{A} & B & C & D & E & F & G & H & 1 & J & K \\
\hline
1 & Original brightness & 1075 & 1121 & 1106 & 1095 & 1101 & 1109 & 1114 & 1123 & 1108 & 1115 \\
\hline
2 & After 10000 hours & 1042 & 1084 & 1076 & 1065 & 1070 & 1079 & 1081 & 1091 & 1080 & 1082 \\
\hline
3 & Percentage reduction & 3.07 & 3.30 & 2.71 & 2.74 & 2.82 & 2.71 & 2.96 & 2.85 & 2.53 & 2.96 \\
\hline
4 & & & & & & & & & & & \\
\hline
5 & & & & & & & & & & & \\
\hline
6 & Sample mean (\%) & 2.8650 & & & & & & & & & \\
\hline
7 & Sample sd (\%) & 0.2179 & & & & & & & & & \\
\hline
8 & SE & 0.0689 & & & & & & & & & \\
\hline
9 & DF & 9 & & & & & & & & & \\
\hline
10 & tvalue & 2.262 & & & & & & & & & \\
\hline
11 & Lower limit & 2.709 & & & & & & & & & \\
\hline
12 & Upper limit & 3.021 & & & & & & & & & \\
\hline
1.3 & & & & & & & & & & & \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item State the confidence interval in the form $a < \mu < b$.
\item Explain whether the confidence interval suggests that the mean percentage reduction in brightness after 10000 hours is different from 2.6\%.
\item Explain how the value in cell B8 was calculated.
\item State an assumption necessary for this confidence interval to be calculated.
\item Explain the advantage of using the same bulbs for both measurements.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2021 Q1 [9]}}