| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | State hypotheses with additional parts |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring standard binomial test procedures. Part (a) involves stating a binomial distribution and writing its expansion (basic recall), part (b) is a routine one-tailed test at 5% level, and part (c) asks for a standard critique of sampling method. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04b Binomial probabilities: link to binomial expansion2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(B(150, 0.08)\) or \(B\left(150, \frac{2}{25}\right)\) oe | B1 [1] | or \(n=150\), \(p=0.08\); Ignore all else |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((0.92 + 0.08)^{150}\) oe; Allow \(a=0.92\), \(b=0.08\), \(n=150\) | B1 [1] | or \((0.08+0.92)^{150}\); Not eg \(\binom{150}{x}0.92^{(150-x)}0.08^x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: p = 0.08\) where \(p =\) proportion of adults (in the town) who use coffee shop. Allow "probability". Allow 8%. | B1 | 1.1: Allow "where \(p\) is the population proportion"; Subtract B1 for each error; Allow other letters including \(x\) if defined as proportion |
| \(H_1: p > 0.08\) | B1 | 2.5: 2-tail B1B0; Not include 0.08 B0B0; undefined \(p\) B1B0 eg \(H_0=0.08\) etc: B0B0; not in terms of parameter B1B0; \(p=\) sample proportion B1B0 |
| \(B(150, 0.08)\) & \(X=18\) stated or used; Allow \(X=17\) or 19 | M1 | 3.3: Correct distribution and value of \(X\); stated or implied eg by 0.055 or 0.031 (\(X \geq 19\)) or 0.0923 (\(X \geq 17\)) or 0.0239 (\(X=18\)) or 0.945 or 0.969 or 0.908 or 0.976; even if within incorrect statement eg \(P(X=18)=0.0552\) |
| \(P(X \geq 18) = (1 - 0.945 =) 0.055(1)\) (2 sf) | A1 | 3.4: cao BC |
| comp 0.05 | A1 | 1.1: Explicit comparison, dep 0.055(1) or 0.031 or 0.0923 |
| Not reject \(H_0\); Allow Accept \(H_0\) or Reject \(H_1\) | M1 | 1.1: dep \(P(X \geq 18\) or \(19\) or \(17)\) seen or 0.055 or 0.031 or 0.0923; Might be implied by conclusion; NB Allow opposite conclusion on ft from 0.031 |
| Insufficient evidence that proportion who use coffee shop is more than 0.08 (or has increased) | A1f [7] | 2.2b: Any equivalent statement in context; allow "likelihood", "percentage"; Ignore all else; Dep \(P(X \geq 18)\) oe or 0.055 seen; ft their \(P(X \geq 18)\) only (not 19 or 17); Not definite; Not "Insufficient evidence that proportion has changed" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Hypotheses as above | B1B0 | |
| Calculation as above | M1A1 | |
| Comparison: compare 0.025 oe | A1 | No more marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply \(p\) not constant or sample not representative, or not random, with reason. EG: Fri population (or probability) may be different from Sat; Fri/Sat may be different from other days; Those shopping are more likely to use coffee shop; People who work full-time more likely to shop on Sat; People who work can't buy coffee on Fri; People may buy coffee on the way to work, not on Sat; People in town may not be representative of everyone. OR State or imply that one person using the shop may not be independent (of other people using the shop) with reason. EG: Friends (groups, families etc) may visit the shop together; A person may have been in the shop on both Fri & Sat | B1 [1] | 3.5b: NOT just "Not random"; NOT There may be more people on Fri & Sat; NOT People in town may not live in the town. NOT just "Shoppers may not be independent"; Ignore all else |
## Question 9(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $B(150, 0.08)$ or $B\left(150, \frac{2}{25}\right)$ oe | B1 [1] | or $n=150$, $p=0.08$; Ignore all else |
---
## Question 9(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0.92 + 0.08)^{150}$ oe; Allow $a=0.92$, $b=0.08$, $n=150$ | B1 [1] | or $(0.08+0.92)^{150}$; Not eg $\binom{150}{x}0.92^{(150-x)}0.08^x$ |
---
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: p = 0.08$ where $p =$ proportion of adults (in the town) who use coffee shop. Allow "probability". Allow 8%. | B1 | 1.1: Allow "where $p$ is the population proportion"; Subtract B1 for each error; Allow other letters including $x$ if defined as proportion |
| $H_1: p > 0.08$ | B1 | 2.5: 2-tail B1B0; Not include 0.08 B0B0; undefined $p$ B1B0 eg $H_0=0.08$ etc: B0B0; not in terms of parameter B1B0; $p=$ sample proportion B1B0 |
| $B(150, 0.08)$ & $X=18$ stated or used; Allow $X=17$ or 19 | M1 | 3.3: Correct distribution and value of $X$; stated or implied eg by 0.055 or 0.031 ($X \geq 19$) or 0.0923 ($X \geq 17$) or 0.0239 ($X=18$) or 0.945 or 0.969 or 0.908 or 0.976; even if within incorrect statement eg $P(X=18)=0.0552$ |
| $P(X \geq 18) = (1 - 0.945 =) 0.055(1)$ (2 sf) | A1 | 3.4: cao BC |
| comp 0.05 | A1 | 1.1: Explicit comparison, dep 0.055(1) or 0.031 or 0.0923 |
| Not reject $H_0$; Allow Accept $H_0$ or Reject $H_1$ | M1 | 1.1: dep $P(X \geq 18$ or $19$ or $17)$ seen or 0.055 or 0.031 or 0.0923; Might be implied by conclusion; NB Allow opposite conclusion on ft from 0.031 |
| Insufficient evidence that proportion who use coffee shop is more than 0.08 (or has increased) | A1f [7] | 2.2b: Any equivalent statement in context; allow "likelihood", "percentage"; Ignore all else; Dep $P(X \geq 18)$ oe or 0.055 seen; ft their $P(X \geq 18)$ only (not 19 or 17); Not definite; Not "Insufficient evidence that proportion has changed" |
**Alternative (incorrect) method using 2 tails:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Hypotheses as above | B1B0 | |
| Calculation as above | M1A1 | |
| Comparison: compare 0.025 oe | A1 | No more marks |
---
## Question 9(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $p$ not constant or sample not representative, or not random, with reason. EG: Fri population (or probability) may be different from Sat; Fri/Sat may be different from other days; Those shopping are more likely to use coffee shop; People who work full-time more likely to shop on Sat; People who work can't buy coffee on Fri; People may buy coffee on the way to work, not on Sat; People in town may not be representative of everyone. OR State or imply that one person using the shop may not be independent (of other people using the shop) with reason. EG: Friends (groups, families etc) may visit the shop together; A person may have been in the shop on both Fri & Sat | B1 [1] | 3.5b: NOT just "Not random"; NOT There may be more people on Fri & Sat; NOT People in town may not live in the town. NOT just "Shoppers may not be independent"; Ignore all else |
---9 Last year, market research showed that $8 \%$ of adults living in a certain town used a particular local coffee shop. Following an advertising campaign, it was expected that this proportion would increase. In order to test whether this had happened, a random sample of 150 adults in the town was chosen.
The random variable $X$ denotes the number of these 150 adults who said that they used the local coffee shop.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Assuming that the proportion of adults using the local coffee shop is unchanged from the previous year, state a suitable binomial distribution with which to model the variable $X$.
\item The probabilities given by this model are the terms of the binomial expansion of an expression of the form $( a + b ) ^ { n }$.
Write down this expression, using appropriate values of $a , b$ and $n$.
It was found that 18 of these 150 adults said that they use the local coffee shop.
\end{enumerate}\item Test, at the 5\% significance level, whether the proportion of adults in the town who use the local coffee shop has increased.
It was later discovered by a statistician that the random sample of 150 adults had been chosen from shoppers in the town on a Friday and a Saturday.
\item Explain why this suggests that the assumptions made when using a binomial model for $X$ may not be valid in this context.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q9 [10]}}