| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Compare two Poisson means |
| Difficulty | Standard +0.8 This is a Further Maths Statistics question requiring comparison of two Poisson means using a normal approximation. Students must set up hypotheses, calculate sample means, apply the variance formula for difference of Poisson distributions, compute a z-statistic, and reach a conclusion. While methodical, it requires multiple statistical concepts and careful calculation, making it moderately challenging but still a standard S3 exam question. |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \lambda_T = \lambda_S\), \(H_1: \lambda_T > \lambda_S\) | B1 | Both hypotheses correct |
| Under \(H_0\): estimate \(\lambda_S = \frac{940}{40} = 23.5\) per week, \(\lambda_T = \frac{1560}{60} = 26\) per week | M1 | Estimating rates |
| Combined/pooled: under \(H_0\), \(\lambda_T = \lambda_S\), total hits \(= 940 + 1560 = 2500\) over \(100\) weeks, so \(\hat{\lambda} = 25\) | M1 | Correct pooling |
| \(T_{60} \sim \text{Poisson}(60\lambda_S)\), under \(H_0\): \(T_{60} \sim \text{Poisson}(1500)\) | B1 | |
| Use Normal approximation: \(N(1500, 1500)\) | M1 | |
| \(z = \frac{1560 - 1500}{\sqrt{1500}} = \frac{60}{38.73} = 1.549\) | A1 | Correct test statistic (awrt 1.55) |
| Critical value at 2% one-tailed \(= 2.054\) | B1 | |
| \(1.549 < 2.054\), not significant; insufficient evidence to support Tina's claim | A1 | Correct conclusion in context |
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda_T = \lambda_S$, $H_1: \lambda_T > \lambda_S$ | B1 | Both hypotheses correct |
| Under $H_0$: estimate $\lambda_S = \frac{940}{40} = 23.5$ per week, $\lambda_T = \frac{1560}{60} = 26$ per week | M1 | Estimating rates |
| Combined/pooled: under $H_0$, $\lambda_T = \lambda_S$, total hits $= 940 + 1560 = 2500$ over $100$ weeks, so $\hat{\lambda} = 25$ | M1 | Correct pooling |
| $T_{60} \sim \text{Poisson}(60\lambda_S)$, under $H_0$: $T_{60} \sim \text{Poisson}(1500)$ | B1 | |
| Use Normal approximation: $N(1500, 1500)$ | M1 | |
| $z = \frac{1560 - 1500}{\sqrt{1500}} = \frac{60}{38.73} = 1.549$ | A1 | Correct test statistic (awrt 1.55) |
| Critical value at 2% one-tailed $= 2.054$ | B1 | |
| $1.549 < 2.054$, not significant; insufficient evidence to support Tina's claim | A1 | Correct conclusion in context |3 \\
The weekly number of hits, $S$, on Sam's website may be modelled by a Poisson distribution with parameter $\lambda _ { S }$. The weekly number of hits, $T$, on Tina's website may be modelled by a Poisson distribution with parameter $\lambda _ { T }$. \\
During a period of 40 weeks, the number of hits on Sam's website was 940. \\
During a period of 60 weeks, the number of hits on Tina's website was 1560. \\
Assuming that $S$ and $T$ are independent random variables, investigate, at the $2 \%$ level of significance, Tina's claim that the mean weekly number of hits on her website is greater than that on Sam's website. \\
(7 marks) \\
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\hfill \mbox{\textit{AQA S3 2010 Q3 [7]}}