| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Multiple binomial probability calculations |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard binomial probability calculations and a routine hypothesis test with normal approximation. Parts (a)-(b) require direct use of binomial tables or calculator functions. Part (c) needs recognizing that total points = 90 - 2F, then finding P(F > 10). Part (d) is a textbook one-tailed test with continuity correction. All techniques are standard S2 content with no novel insight required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(F \geq 12) = 1 - P(F_n\ 11)\) | M1 | Writing or using \(1 - P(F_n\ 11)\) |
| \(= 0.34517\ldots\) awrt \(0.345\) | A1 | awrt 0.345 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(8_n\ F < 15) = P(F_n\ 14) - P(F_n\ 7)\) | M1 | \(P(F_n\ 14) - P(F_n\ 7)\) |
| \(= 0.81104\ldots\) awrt \(0.811\) | A1 | awrt 0.811 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3(30-F) + F < 70\) or \(F > 10\) OR \(3(R) + 30 - R < 70\) or \(R < 20\) | M1 | Allow equation instead of inequality (may be implied by 2nd M1) |
| \(P(F > 10) = 1 - P(F_n\ 10)\) OR \(P(R < 20) = P(R_n\ 19)\) | M1 | Writing or using \(1 - P(F_n\ 10)\) ft their 10 but must be finding the correct tail |
| \(= 0.4922\ldots\) awrt \(0.492\) | A1 | awrt 0.492 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p = 0.35 \quad H_1: p > 0.35\) | B1 | Both hypotheses in terms of \(p\) or \(\pi\) |
| Let \(Y\) be the number of customers who do not buy free range eggs. \(Y \sim N(70,\ 45.5)\) | M1 | Writing or using a normal distribution with a mean of 70 |
| Standardising using \(85.5/86/86.5\), their mean and their sd | M1 | Standardising using \(85.5/86/86.5\), their mean and their sd |
| \(P(Y \geq 86) \approx P\!\left(Z > \dfrac{85.5 - 70}{\sqrt{45.5}}\right)\) or \(\pm\dfrac{x - 0.5 - 70}{\sqrt{45.5}} = 1.6449\) | M1 | Using a continuity correction \(86 \pm 0.5\) |
| \(\approx 0.01078\ldots\) or \(81.595\ldots\) | A1 | Correct probability awrt \(0.0108\) or awrt \(0.0107\) or \(x\) value awrt \(82\); or allow awrt \(2.29\ldots\) and \(1.6449\) seen. NB exact Binomial \(0.01156\), \(P_0\) approx. \(0.0352\) |
| There is evidence to reject \(H_0\) (in the critical region) | dM1 | (dep on 1st M1) A correct statement based on comparing 86 with their CR, or their prob with \(0.05\) [condone \(0.989 > 0.95\)] – contradicting non-contextual comments M0 |
| There is evidence to support the manager's belief / the proportion of customers who do not buy free range eggs is more than \(35\%\) | A1 | A correct statement in context. NB award M1A1 for a correct contextual statement on its own. |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(F \geq 12) = 1 - P(F_n\ 11)$ | M1 | Writing or using $1 - P(F_n\ 11)$ |
| $= 0.34517\ldots$ awrt $0.345$ | A1 | awrt 0.345 |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(8_n\ F < 15) = P(F_n\ 14) - P(F_n\ 7)$ | M1 | $P(F_n\ 14) - P(F_n\ 7)$ |
| $= 0.81104\ldots$ awrt $0.811$ | A1 | awrt 0.811 |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3(30-F) + F < 70$ or $F > 10$ OR $3(R) + 30 - R < 70$ or $R < 20$ | M1 | Allow equation instead of inequality (may be implied by 2nd M1) |
| $P(F > 10) = 1 - P(F_n\ 10)$ OR $P(R < 20) = P(R_n\ 19)$ | M1 | Writing or using $1 - P(F_n\ 10)$ ft their 10 but must be finding the correct tail |
| $= 0.4922\ldots$ awrt $0.492$ | A1 | awrt 0.492 |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.35 \quad H_1: p > 0.35$ | B1 | Both hypotheses in terms of $p$ or $\pi$ |
| Let $Y$ be the number of customers who do **not** buy free range eggs. $Y \sim N(70,\ 45.5)$ | M1 | Writing or using a normal distribution with a mean of 70 |
| Standardising using $85.5/86/86.5$, their mean and their sd | M1 | Standardising using $85.5/86/86.5$, their mean and their sd |
| $P(Y \geq 86) \approx P\!\left(Z > \dfrac{85.5 - 70}{\sqrt{45.5}}\right)$ or $\pm\dfrac{x - 0.5 - 70}{\sqrt{45.5}} = 1.6449$ | M1 | Using a continuity correction $86 \pm 0.5$ |
| $\approx 0.01078\ldots$ or $81.595\ldots$ | A1 | Correct probability awrt $0.0108$ or awrt $0.0107$ or $x$ value awrt $82$; or allow awrt $2.29\ldots$ and $1.6449$ seen. **NB** exact Binomial $0.01156$, $P_0$ approx. $0.0352$ |
| There is evidence to reject $H_0$ (in the critical region) | dM1 | (dep on 1st M1) A correct statement based on comparing 86 with their CR, or their prob with $0.05$ [condone $0.989 > 0.95$] – contradicting non-contextual comments M0 |
| There is evidence to support the **manager's** belief / the proportion of customers who do **not** buy free range eggs is more than $35\%$ | A1 | A correct statement in context. **NB** award M1A1 for a correct contextual statement on its own. |\begin{enumerate}
\item A research project into food purchases found that $35 \%$ of people who buy eggs do not buy free range eggs.
\end{enumerate}
A random sample of 30 people who bought eggs is taken. The random variable $F$ denotes the number of people who do not buy free range eggs.\\
(a) Find $\mathrm { P } ( F \geqslant 12 )$\\
(b) Find $\mathrm { P } ( 8 \leqslant F < 15 )$
A farm shop gives 3 loyalty points with every purchase of free range eggs. With every purchase of eggs that are not free range the farm shop gives 1 loyalty point. A random sample of 30 customers who buy eggs from the farm shop is taken.\\
(c) Find the probability that the total number of points given to these customers is less than 70
The manager of the farm shop believes that the proportion of customers who buy eggs but do not buy free range eggs is more than $35 \%$
In a survey of 200 customers who buy eggs, 86 do not buy free range eggs.
Using a suitable test and a normal approximation,\\
(d) determine, at the $5 \%$ level of significance, whether there is evidence to support the manager's belief. State your hypotheses clearly.
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\hfill \mbox{\textit{Edexcel S2 2021 Q1 [14]}}