| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Two-tailed test (change) |
| Difficulty | Standard +0.8 This S2 question requires understanding of Poisson hypothesis testing with critical regions, actual significance levels, and normal approximation for larger samples. Part (a) demands careful probability calculations for discrete distributions, part (b) requires summing tail probabilities, and part (c) involves recognizing when to apply continuity correction. The multi-part structure and need to work with both exact and approximate methods elevates this above routine hypothesis testing questions. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim Po(8)\) | B1 | For \(Po(8)\) seen or implied |
| e.g. \(P(X \leq 3) = 0.0424\), \(P(X \leq 13) = 0.9658\) so \(P(X \geq 14) = 0.0342\) | M1 | For clear evidence of use of \(Po(8)\); may be implied by correct CR or probability in (b) |
| Critical Region is \(X \leq 3\) or \(X \geq 14\) | A1 A1 | 1st A1: \(X \leq 3\) or \(0 \leq X \leq 3\); 2nd A1: \(X \geq 14\) or \([14,\infty)\). Must be statements with \(X\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.0424 + 0.0342\) | M1 | For adding the two probabilities corresponding to their CR |
| \(= 0.0766\) (or better) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \lambda = 8\) (or \(\mu=80\)), \(H_1: \lambda > 8\) (or \(\mu > 80\)) | B1 | For both hypotheses in terms of \(\lambda\) or \(\mu\); 8 or 80 can be swapped |
| \([R = \text{no. of raisins in 10 muffins}, R \sim Po(80)]\) Use \(Y \sim N(80, 80)\) | M1 A1 | 1st M1: for normal approx; 1st A1: \(E(Y)=80\) and \(\text{Var}(Y)=80\) |
| \(P(R \geq 95) \simeq P(Y \geq 94.5)\) | M1 | For use of continuity correction 94.5 or 95.5 |
| \(= P\!\left(Z > \frac{94.5-80}{\sqrt{80}}\right)\) | M1 | Standardising using their mean and sd; must use 94.5, 95.5 or 95 |
| \(= P(Z > 1.62...) = 1 - 0.9474 =\) awrt \(0.053\) | A1 | For awrt 0.053 or awrt 0.947 |
| Probability is greater than 0.05 so not significant (accept \(H_0\)) | M1 | For correct statement based on their probability and 0.05 |
| Insufficient evidence to support the bakery's claim / insufficient evidence of an increase in the (mean) number of raisins per muffin | A1cso | cso; fully correct contextualised statement. Need either "bakery's claim" or "raisins" and "muffin". NB: if \(P(X=95)\) found: B1 M1 A1 M0 M0 A0 M0 A0 |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim Po(8)$ | B1 | For $Po(8)$ seen or implied |
| e.g. $P(X \leq 3) = 0.0424$, $P(X \leq 13) = 0.9658$ so $P(X \geq 14) = 0.0342$ | M1 | For clear evidence of use of $Po(8)$; may be implied by correct CR or probability in (b) |
| Critical Region is $X \leq 3$ or $X \geq 14$ | A1 A1 | 1st A1: $X \leq 3$ or $0 \leq X \leq 3$; 2nd A1: $X \geq 14$ or $[14,\infty)$. Must be statements with $X$ only |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.0424 + 0.0342$ | M1 | For adding the two probabilities corresponding to their CR |
| $= 0.0766$ (or better) | A1 | |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \lambda = 8$ (or $\mu=80$), $H_1: \lambda > 8$ (or $\mu > 80$) | B1 | For both hypotheses in terms of $\lambda$ or $\mu$; 8 or 80 can be swapped |
| $[R = \text{no. of raisins in 10 muffins}, R \sim Po(80)]$ Use $Y \sim N(80, 80)$ | M1 A1 | 1st M1: for normal approx; 1st A1: $E(Y)=80$ and $\text{Var}(Y)=80$ |
| $P(R \geq 95) \simeq P(Y \geq 94.5)$ | M1 | For use of continuity correction 94.5 or 95.5 |
| $= P\!\left(Z > \frac{94.5-80}{\sqrt{80}}\right)$ | M1 | Standardising using their mean and sd; must use 94.5, 95.5 or 95 |
| $= P(Z > 1.62...) = 1 - 0.9474 =$ awrt $0.053$ | A1 | For awrt 0.053 or awrt 0.947 |
| Probability is greater than 0.05 so not significant (accept $H_0$) | M1 | For correct statement based on their probability and 0.05 |
| Insufficient evidence to support the bakery's claim / insufficient evidence of an increase in the (mean) number of raisins per muffin | A1cso | cso; fully correct contextualised statement. Need either "bakery's claim" or "raisins" and "muffin". **NB**: if $P(X=95)$ found: B1 M1 A1 M0 M0 A0 M0 A0 |
---6. Frugal bakery claims that their packs of 10 muffins contain on average 80 raisins per pack. A Poisson distribution is used to describe the number of raisins per muffin.
A muffin is selected at random to test whether or not the mean number of raisins per muffin has changed.
\begin{enumerate}[label=(\alph*)]
\item Find the critical region for a two-tailed test using a $10 \%$ level of significance. The probability of rejection in each tail should be less than 0.05
\item Find the actual significance level of this test.
The bakery has a special promotion claiming that their muffins now contain even more raisins.
A random sample of 10 muffins is selected and is found to contain a total of 95 raisins.
\item Use a suitable approximation to test the bakery's claim. You should state your hypotheses clearly and use a $5 \%$ level of significance.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2013 Q6 [14]}}