| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Test using proportion |
| Difficulty | Standard +0.3 This is a standard two-proportion hypothesis test with a pooled estimate calculation. Part (i) is straightforward arithmetic (1 mark), and part (ii) follows a routine procedure: state hypotheses, calculate pooled proportion, compute test statistic, compare to critical value. While it requires careful execution across multiple steps, it involves no conceptual difficulty beyond applying the standard S3 formula for comparing two proportions. Slightly above average difficulty due to the multi-step nature and potential for arithmetic errors, but well within typical S3 examination scope. |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Est}(p) = (48+42)/(190+150) (=9/34 \text{ or } 0.265)\) | \(B1\) | Allow consistent use of '-' instead of '+' throughout. |
| \([1]\) | If eg x,y used – must be defined. |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: p_{10} = p_{07}, H_1: p_{10} > p_{07}\) | \(B1\) | |
| where the ps denote relevant population proportions | ||
| Test statistic \(= (48/150 - 42/190)/\) sd | \(M1\) | sd involving 190 and 150 and addition of 'variances' |
| \(sd = \sqrt{(\frac{1}{34} \times \frac{33}{34}(190^{-1} + 150^{-1}))}\) | \(A1\) | Allow A1 ft for sd \(\sqrt{\frac{(48 \times 102/150^2)+(42 \times 148/190^2)}{or TS=2.038\) |
| \(TS = 2.053\) | \(A1\) | |
| Compare with 1.96 and reject \(H_0\); and conclude that there is evidence that the proportion of badgers in the area with TB has increased | \(M1\) | SC Allow for correct comparison with 2.24 for 2-tail test. |
| \(A1ft\) | Conclusion, contextualised, not too assertive. | |
| \([6]\) |
### (i)
$\text{Est}(p) = (48+42)/(190+150) (=9/34 \text{ or } 0.265)$ | $B1$ | Allow consistent use of '-' instead of '+' throughout.
| $[1]$ | If eg x,y used – must be defined.
### (ii)
$H_0: p_{10} = p_{07}, H_1: p_{10} > p_{07}$ | $B1$ |
where the ps denote relevant population proportions | |
Test statistic $= (48/150 - 42/190)/$ sd | $M1$ | sd involving 190 and 150 and addition of 'variances'
$sd = \sqrt{(\frac{1}{34} \times \frac{33}{34}(190^{-1} + 150^{-1}))}$ | $A1$ | Allow A1 ft for sd $\sqrt{\frac{(48 \times 102/150^2)+(42 \times 148/190^2)}{or TS=2.038$
$TS = 2.053$ | $A1$ |
Compare with 1.96 and reject $H_0$; and conclude that there is evidence that the proportion of badgers in the area with TB has increased | $M1$ | SC Allow for correct comparison with 2.24 for 2-tail test.
| $A1ft$ | Conclusion, contextualised, not too assertive.
| $[6]$ |An investigation in 2007 into the incidence of tuberculosis (TB) in badgers in a certain area found that 42 out of a random sample of 190 badgers tested positive for TB.
In 2010, 48 out of a random sample of 150 badgers tested positive for TB.
\begin{enumerate}[label=(\roman*)]
\item Assuming that the population proportions of badgers with TB are the same in 2007 and 2010, obtain the best estimate of this proportion. [1]
\item Carry out a test at the $2\frac{1}{2}\%$ significance level of whether the population proportion of badgers with TB increased from 2007 to 2010. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR S3 2012 Q2 [7]}}