OCR S3 2015 June — Question 2 7 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2015
SessionJune
Marks7
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TopicModelling and Hypothesis Testing
TypeConfidence intervals for proportions
DifficultyStandard +0.3 This is a standard two-proportion hypothesis test with clearly stated hypotheses, straightforward calculations of pooled proportion and test statistic, and routine comparison to critical values. While it requires multiple steps, it follows a well-practiced procedure with no conceptual surprises, making it slightly easier than average.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean
2 In a poll of people aged 18-21, 46 out of 200 randomly chosen university students agreed with a proposition. 51 out of 300 randomly chosen others who were not university students agreed with it. Test, at the \(5 \%\) significance level, whether the proportion of university students who agree with the proposition differs from the proportion of those who are not university students.
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: p_a = p_n \quad H_1: p_a \neq p_n\)B1 If words, must have 'population'. SC1 1-tail
\(\hat{p} = \frac{97}{500} = 0.194\)B1 B0B1M1A1A1M1(\(>\)1.645 rej)A0
\(TS = \frac{0.23 - 0.17}{\sqrt{0.194 \times 0.806 \times (\frac{1}{200} + \frac{1}{300})}}\)M1 Allow 1 error in var formula for M1. SC2 not pooled: \(sd = \sqrt{(\frac{46\times154}{200^3} + \frac{51\times249}{300^3})}\)
Correct var \((= 0.0361)\)A1 \(TS = 1.63\)
1.66A1 B1B0M1A1ftA0M1A1ft0.0516/7\(>\)2.5%
"\(1.66\)" \(< 1.96\), accept \(H_0\)M1 ft TS; 0.0965\(>\)5% or 0.0483\(>\)2.5%. SC3 (both errors): B0B0M1A1ftA0M1(\(<\)1.645 acc)A0
Insufficient evidence to suggest that there is a difference between the proportions.A1 Not over-assertive. CWO (accept from 1.63) 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: p_a = p_n \quad H_1: p_a \neq p_n$ | B1 | If words, must have 'population'. SC1 1-tail |
| $\hat{p} = \frac{97}{500} = 0.194$ | B1 | B0B1M1A1A1M1($>$1.645 rej)A0 |
| $TS = \frac{0.23 - 0.17}{\sqrt{0.194 \times 0.806 \times (\frac{1}{200} + \frac{1}{300})}}$ | M1 | Allow 1 error in var formula for M1. SC2 not pooled: $sd = \sqrt{(\frac{46\times154}{200^3} + \frac{51\times249}{300^3})}$ |
| Correct var $(= 0.0361)$ | A1 | $TS = 1.63$ |
| 1.66 | A1 | B1B0M1A1ftA0M1A1ft0.0516/7$>$2.5% |
| "$1.66$" $< 1.96$, accept $H_0$ | M1 | ft TS; 0.0965$>$5% or 0.0483$>$2.5%. SC3 (both errors): B0B0M1A1ftA0M1($<$1.645 acc)A0 |
| Insufficient evidence to suggest that there is a difference between the proportions. | A1 | Not over-assertive. CWO (accept from 1.63) |

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2 In a poll of people aged 18-21, 46 out of 200 randomly chosen university students agreed with a proposition. 51 out of 300 randomly chosen others who were not university students agreed with it. Test, at the $5 \%$ significance level, whether the proportion of university students who agree with the proposition differs from the proportion of those who are not university students.

\hfill \mbox{\textit{OCR S3 2015 Q2 [7]}}
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