| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Binomial parameters from given information |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard binomial distribution properties and hypothesis testing. Part (a) uses mean=np and variance=np(1-p) to find parameters through simple algebra. Part (b) is a routine binomial probability calculation. Part (c) is a textbook one-tailed hypothesis test with clearly stated hypotheses. All parts require only direct application of standard formulas with no novel insight or complex problem-solving. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(np = 4.2\), \(np(1-p) = 3.57\) | M1 | 1st M1 correct expressions for mean and variance |
| leading to \((1-p) = 0.85\) | M1 | 2nd M1 for attempting to solve simultaneously |
| \(p = 0.15\), \(n = 28\) | A1 A1 | 1st A1 \(p=0.15\); 2nd A1 \(n=28\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X\sim\text{B}(25,\ 0.35)\), \(E(X)=8.75\) | B1 | B1 for \(E(X)=8.75\) (may be implied by M1) |
| \(P(X>8.75) = P(X\geq 9) = 1-P(X\leq 8) = 1-0.4668 = 0.5332\) awrt 0.533 | M1 A1 | M1 for \(1-P(X\leq 8)\) with B(25,0.35); A1 awrt 0.533 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: p=0.1\), \(H_1: p<0.1\) | B1 | B1 both hypotheses correct (must use \(p\) or \(\pi\)) |
| \(Y\sim\text{B}(40,\ 0.1)\), \(P(Y\leq 1) = 0.080473...\) | M1 | 1st M1 for awrt 0.0805 or for stating critical region is \(Y=0\) from B(40,0.1) |
| Do not reject \(H_0\) / Not significant | dM1 | 2nd dM1 dependent on previous M; correct statement (no contradicting statements) |
| The proportion of customers buying more than 2 bags of sweets is not less than 10%/not less than the shop's claim; or: The shop's claim is not rejected | A1cso | A1cso: must include proportion/number/percentage/probability and 10%/shop's claim |
## Question 4:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $np = 4.2$, $np(1-p) = 3.57$ | M1 | 1st M1 correct expressions for mean and variance |
| leading to $(1-p) = 0.85$ | M1 | 2nd M1 for attempting to solve simultaneously |
| $p = 0.15$, $n = 28$ | A1 A1 | 1st A1 $p=0.15$; 2nd A1 $n=28$ |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X\sim\text{B}(25,\ 0.35)$, $E(X)=8.75$ | B1 | B1 for $E(X)=8.75$ (may be implied by M1) |
| $P(X>8.75) = P(X\geq 9) = 1-P(X\leq 8) = 1-0.4668 = 0.5332$ **awrt 0.533** | M1 A1 | M1 for $1-P(X\leq 8)$ with B(25,0.35); A1 awrt 0.533 |
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: p=0.1$, $H_1: p<0.1$ | B1 | B1 both hypotheses correct (must use $p$ or $\pi$) |
| $Y\sim\text{B}(40,\ 0.1)$, $P(Y\leq 1) = 0.080473...$ | M1 | 1st M1 for awrt 0.0805 or for stating critical region is $Y=0$ from B(40,0.1) |
| Do not reject $H_0$ / Not significant | dM1 | 2nd dM1 dependent on previous M; correct statement (no contradicting statements) |
| The proportion of customers buying more than 2 bags of sweets is not less than 10%/not less than the shop's claim; or: The shop's claim is not rejected | A1cso | A1cso: must include proportion/number/percentage/probability and 10%/shop's claim |
---4. A sweet shop produces different coloured sweets and sells them in bags.
The proportion of green sweets produced is $p$
Each bag is filled with a random sample of $n$ sweets.
The mean number of green sweets in a bag is 4.2 and the variance is 3.57
\begin{enumerate}[label=(\alph*)]
\item Find the value of $n$ and the value of $p$
The proportion of red sweets produced by the shop is 0.35
\item Find the probability that, in a random sample of 25 sweets, the number of red sweets exceeds the expected number of red sweets.
The shop claims that $10 \%$ of its customers buy more than two bags of sweets. A random sample of 40 customers is taken and 1 customer buys more than two bags of sweets.
\item Test, at the $5 \%$ level of significance, whether or not there is evidence that the proportion of customers who buy more than two bags of sweets is less than the shop's claim. State your hypotheses clearly.\\
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\hfill \mbox{\textit{Edexcel S2 2018 Q4 [11]}}