OCR S3 2011 January — Question 4 7 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2011
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChi-squared test of independence
TypeStandard 2×2 contingency table
DifficultyStandard +0.8 This is a two-sample proportion test requiring calculation of pooled proportion, test statistic, and finding the exact significance level (p-value) rather than comparing to standard critical values. While the method is standard for S3, the 'smallest significance level' phrasing requires understanding of p-values and careful calculation, making it moderately harder than routine hypothesis testing questions.
Spec5.05c Hypothesis test: normal distribution for population mean

4 A study in 1981 investigated the effect of water fluoridation on children's dental health. In a town with fluoridation, 61 out of a random sample of 107 children showed signs of increased tooth decay after six months. In a town without fluoridation the corresponding number was 106 out of a random sample of 143 children. The population proportions of children with increased tooth decay are denoted by \(p _ { 1 }\) and \(p _ { 2 }\) for the towns with fluoridation and without fluoridation respectively. A test is carried out of the null hypothesis \(p _ { 1 } = p _ { 2 }\) against the alternative hypothesis \(p _ { 1 } < p _ { 2 }\). Find the smallest significance level at which the null hypothesis is rejected.

Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
\(\hat{p}_2 = 106/143,\quad \hat{p}_1 = 61/107\)B1 For both
\(= 0.7413 \qquad = 0.5701\)
Pooled est \(p = 167/250\)B1 Only if used
Variance est \(= \left(\frac{167}{250}\right)\left(\frac{83}{250}\right)(143^{-1}+107^{-1})\)B1
Test statistic \(z = (0.7413 - 0.5701)/\text{SD}\)M1
\(= 2.84(35)\)A1
Smallest significance level \(= 0.23\%\)M1 ART 0.22 or 0.23, Accept 0.0023
SR: No \(p_e\), B1B0B0M1A1(2.84)M1A1 Max 5/7A1\(\sqrt{}\) [7] \(\sqrt{z}\) M1A0 if 0.25%
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\hat{p}_2 = 106/143,\quad \hat{p}_1 = 61/107$ | B1 | For both |
| $= 0.7413 \qquad = 0.5701$ | | |
| Pooled est $p = 167/250$ | B1 | Only if used |
| Variance est $= \left(\frac{167}{250}\right)\left(\frac{83}{250}\right)(143^{-1}+107^{-1})$ | B1 | |
| Test statistic $z = (0.7413 - 0.5701)/\text{SD}$ | M1 | |
| $= 2.84(35)$ | A1 | |
| Smallest significance level $= 0.23\%$ | M1 | ART 0.22 or 0.23, Accept 0.0023 |
| SR: No $p_e$, B1B0B0M1A1(2.84)M1A1 Max 5/7 | A1$\sqrt{}$ **[7]** | $\sqrt{z}$ M1A0 if 0.25% |

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4 A study in 1981 investigated the effect of water fluoridation on children's dental health. In a town with fluoridation, 61 out of a random sample of 107 children showed signs of increased tooth decay after six months. In a town without fluoridation the corresponding number was 106 out of a random sample of 143 children. The population proportions of children with increased tooth decay are denoted by $p _ { 1 }$ and $p _ { 2 }$ for the towns with fluoridation and without fluoridation respectively. A test is carried out of the null hypothesis $p _ { 1 } = p _ { 2 }$ against the alternative hypothesis $p _ { 1 } < p _ { 2 }$. Find the smallest significance level at which the null hypothesis is rejected.

\hfill \mbox{\textit{OCR S3 2011 Q4 [7]}}