OCR S3 2012 June — Question 3 6 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2012
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModelling and Hypothesis Testing
TypeConfidence intervals for proportions
DifficultyChallenging +1.2 This is a two-sample proportion hypothesis test with a non-zero null hypothesis value, requiring calculation of sample proportions, pooled standard error under H₀, a z-test statistic, and interpretation. While it involves multiple steps and careful handling of the specified difference (0.15), it's a standard S3 procedure with straightforward calculations and no conceptual surprises—moderately above average difficulty due to the non-standard null hypothesis and multi-step nature.
Spec5.05c Hypothesis test: normal distribution for population mean

3 A charity raises money by sending letters asking for donations. Because of recent poor responses, the charity's fund-raiser, Anna, decides to alter the letter's appearance and designs two possible alternatives, one colourful and the other plain. She believes that the colourful letter will be more successful. Anna sends 60 colourful letters and 40 plain letters to 100 people randomly chosen from the charity's database. There were 39 positive responses to the colourful letter and 12 positive responses to the plain letter. The population proportions of positive responses to the colourful and plain letters are denoted by \(p _ { C }\) and \(p _ { P }\) respectively. Test the null hypothesis \(p _ { C } - p _ { P } = 0.15\) against the alternative hypothesis \(p _ { C } - p _ { P } > 0.15\) at the \(2 \frac { 1 } { 2 } \%\) significance level and state what Anna could report to her manager.

Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(TS = (39/60 - 12/40 - 0.15)/s\)M1 No marks for hyps in this Q.
\(s^2 = 0.65 \times 0.35/60 + 0.3 \times 0.7/40\ (=0.0090417\ \text{(5sf)})\)M1A1 M1 for "variance" involving 40 and 60. eg "pooled" \(s^2=0.0104125\), \(TS=1.95998\) A0A0
\(TS = 2.10(3)\)A1
Compare TS with \(1.96\) and reject \(H_0\)M1 Or "\(1.78\%\)"\(<2.5\%\)
Reports that there is evidence (at 5% SL) that popn resp to coloured more than 15% greater than popn resp to plainA1ft Ft TS. Not over-assertive
[6]
# Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $TS = (39/60 - 12/40 - 0.15)/s$ | M1 | No marks for hyps in this Q. |
| $s^2 = 0.65 \times 0.35/60 + 0.3 \times 0.7/40\ (=0.0090417\ \text{(5sf)})$ | M1A1 | M1 for "variance" involving 40 and 60. eg "pooled" $s^2=0.0104125$, $TS=1.95998$ A0A0 |
| $TS = 2.10(3)$ | A1 | |
| Compare TS with $1.96$ and reject $H_0$ | M1 | Or "$1.78\%$"$<2.5\%$ |
| Reports that there is evidence (at 5% SL) that popn resp to coloured more than 15% greater than popn resp to plain | A1ft | Ft TS. Not over-assertive |
| | **[6]** | |

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3 A charity raises money by sending letters asking for donations. Because of recent poor responses, the charity's fund-raiser, Anna, decides to alter the letter's appearance and designs two possible alternatives, one colourful and the other plain. She believes that the colourful letter will be more successful. Anna sends 60 colourful letters and 40 plain letters to 100 people randomly chosen from the charity's database. There were 39 positive responses to the colourful letter and 12 positive responses to the plain letter. The population proportions of positive responses to the colourful and plain letters are denoted by $p _ { C }$ and $p _ { P }$ respectively. Test the null hypothesis $p _ { C } - p _ { P } = 0.15$ against the alternative hypothesis $p _ { C } - p _ { P } > 0.15$ at the $2 \frac { 1 } { 2 } \%$ significance level and state what Anna could report to her manager.

\hfill \mbox{\textit{OCR S3 2012 Q3 [6]}}