| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 16 |
| 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 hypothesis testing question covering standard S1 content. Part (i) involves routine binomial probability calculations using formulas or tables. Parts (ii) and (iii) are textbook two-tailed hypothesis tests with the critical probability value provided in (iii), requiring only mechanical application of the test procedure. No novel insight or complex reasoning is needed—just systematic application of standard techniques taught in S1. |
| Spec | 2.04b 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 |
| Let \(X \sim B(20, 0.35)\); Let \(p\) = probability of a customer using the internet (for population) | B1 | For definition of \(p\) in context. Minimum needed: \(p\) = probability of using internet. Must include word probability/chance/proportion/percentage/likelihood but NOT possibility. Do NOT allow '\(p\) = the probability of using internet is different' |
| \(H_0: p = 0.35\) | B1 | Allow \(p=35\%\), allow only \(p\) or \(\theta\) or \(\pi\) or \(\rho\) if defined. Allow \(H_0 = p=0.35\), Allow \(H_0: p = \frac{7}{20}\) or \(p = \frac{35}{100}\). Do not allow \(H_0: P(X=x) = 0.35\), \(P(0.35)\), \(p(x)=0.35\), \(x=0.35\) (unless \(x\) correctly defined as probability). For hypotheses in words allow Maximum B0B1B1 |
| \(H_1: p \neq 0.35\) | B1 | Allow '\(p < 0.35\) or \(p > 0.35\)' in place of \(p \neq 0.35\). Do not allow if \(H_1\) wrong |
| \(H_1\) has this form because the test investigates whether the proportion is different (rather than lower or higher) | E1 | Do not allow if \(H_1\) wrong |
| \(P(X \geq 10)\) | B1 | For notation \(P(X \geq 10)\) or \(P(X > 9)\) or \(1 - P(X \leq 9)\) (as long as no incorrect notation). No further marks if point probs used. DO NOT FT wrong \(H_1\) |
| \(= 1 - 0.8782 = 0.1218\) | B1* | For 0.1218. Allow 0.12. Or for \(1 - 0.8782\). Independent of previous mark |
| \(> 2.5\%\) | M1* dep | For comparison with 2.5% |
| So not significant. Conclude that there is not enough evidence to indicate that the probability is different. (Must state 'probability', not just 'p') | A1*, E1* dep on A1 | Allow 'accept \(H_0\)' or 'reject \(H_1\)'. Must include 'sufficient evidence' or something similar |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| LOWER TAIL: \(P(X \leq 2) = 0.0121 < 2.5\%\); \(P(X \leq 3) = 0.0444 > 2.5\%\) | B1 | For either probability. Do not insist on correct notation |
| UPPER TAIL: \(P(X \geq 11) = 1 - P(X \leq 10) = 1 - 0.9468 = 0.0532 > 2.5\%\); \(P(X \geq 12) = 1 - P(X \leq 11) = 1 - 0.9804 = 0.0196 < 2.5\%\) | B1 | For either probability |
| So critical region is \(\{0,1,2,12,13,14,15,16,17,18,19,20\}\) | M1* dep | cao dep on at least one correct comparison with 2.5%. Condone \(\{0,1,2,12,\ldots 20\}\), \(X \leq 2\), \(X \geq 12\) oe |
| So not significant. Conclude that there is not enough evidence to indicate that the probability is different. | A1*, E1* dep on A1 | NB If CR found correctly then \(P(X=10)\) subsequently found but candidate says '10 not in CR' then allow up to all last five marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.0022 < 2.5\%\), so reject \(H_0\). Significant. | B1 | For either reject \(H_0\) or significant, dep on correct comparison |
| Conclude that there is enough evidence to indicate that the probability is different. | E1* dep | Dep on good attempt at correct hypotheses in part (ii). If they have \(H_1: p > 0.35\), allow SC1 if all correct including comparison with 5% |
# Question 1(ii):
**Defining the distribution and hypotheses**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $X \sim B(20, 0.35)$; Let $p$ = probability of a customer using the internet (for population) | B1 | For definition of $p$ in context. Minimum needed: $p$ = probability of using internet. Must include word probability/chance/proportion/percentage/likelihood but NOT possibility. Do NOT allow '$p$ = the probability of using internet is different' |
| $H_0: p = 0.35$ | B1 | Allow $p=35\%$, allow only $p$ or $\theta$ or $\pi$ or $\rho$ if defined. Allow $H_0 = p=0.35$, Allow $H_0: p = \frac{7}{20}$ or $p = \frac{35}{100}$. Do not allow $H_0: P(X=x) = 0.35$, $P(0.35)$, $p(x)=0.35$, $x=0.35$ (unless $x$ correctly defined as probability). For hypotheses in words allow Maximum B0B1B1 |
| $H_1: p \neq 0.35$ | B1 | Allow '$p < 0.35$ or $p > 0.35$' in place of $p \neq 0.35$. Do not allow if $H_1$ wrong |
| $H_1$ has this form because the test investigates whether the proportion is different (rather than lower or higher) | E1 | Do not allow if $H_1$ wrong |
| $P(X \geq 10)$ | B1 | For notation $P(X \geq 10)$ or $P(X > 9)$ or $1 - P(X \leq 9)$ (as long as no incorrect notation). No further marks if point probs used. DO NOT FT wrong $H_1$ |
| $= 1 - 0.8782 = 0.1218$ | B1* | For 0.1218. Allow 0.12. Or for $1 - 0.8782$. Independent of previous mark |
| $> 2.5\%$ | M1* dep | For comparison with 2.5% |
| So not significant. Conclude that there is not enough evidence to indicate that the probability is different. (Must state 'probability', not just 'p') | A1*, E1* dep on A1 | Allow 'accept $H_0$' or 'reject $H_1$'. Must include 'sufficient evidence' or something similar |
**Alternative Method (Critical Region):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| LOWER TAIL: $P(X \leq 2) = 0.0121 < 2.5\%$; $P(X \leq 3) = 0.0444 > 2.5\%$ | B1 | For either probability. Do not insist on correct notation |
| UPPER TAIL: $P(X \geq 11) = 1 - P(X \leq 10) = 1 - 0.9468 = 0.0532 > 2.5\%$; $P(X \geq 12) = 1 - P(X \leq 11) = 1 - 0.9804 = 0.0196 < 2.5\%$ | B1 | For either probability |
| So critical region is $\{0,1,2,12,13,14,15,16,17,18,19,20\}$ | M1* dep | cao dep on at least one correct comparison with 2.5%. Condone $\{0,1,2,12,\ldots 20\}$, $X \leq 2$, $X \geq 12$ oe |
| So not significant. Conclude that there is not enough evidence to indicate that the probability is different. | A1*, E1* dep on A1 | NB If CR found correctly then $P(X=10)$ subsequently found but candidate says '10 not in CR' then allow up to all last five marks |
**[Total: 9 marks]**
---
# Question 1(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.0022 < 2.5\%$, so reject $H_0$. Significant. | B1 | For either reject $H_0$ or significant, dep on correct comparison |
| Conclude that there is enough evidence to indicate that the probability is different. | E1* dep | Dep on good attempt at correct hypotheses in part (ii). If they have $H_1: p > 0.35$, allow SC1 if all correct including comparison with 5% |
**[Total: 2 marks]**
---1 A coffee shop provides free internet access for its customers. It is known that the probability that a randomly selected customer is accessing the internet is 0.35 , independently of all other customers.
\begin{enumerate}[label=(\roman*)]
\item 10 customers are selected at random.\\
(A) Find the probability that exactly 5 of them are accessing the internet.\\
(B) Find the probability that at least 5 of them are accessing the internet.\\
(C) Find the expected number of these customers who are accessing the internet.
Another coffee shop also provides free internet access. It is suspected that the probability that a randomly selected customer at this coffee shop is accessing the internet may be different from 0.35 . A random sample of 20 customers at this coffee shop is selected. Of these, 10 are accessing the internet.
\item Carry out a hypothesis test at the $5 \%$ significance level to investigate whether the probability for this coffee shop is different from 0.35 . Give a reason for your choice of alternative hypothesis.
\item To get a more reliable result, a much larger random sample of 200 customers is selected over a period of time, and another hypothesis test is carried out. You are given that 90 of the 200 customers were accessing the internet. You are also given that, if $X$ has the binomial distribution with parameters $n = 200$ and $p = 0.35$, then $\mathrm { P } ( X \geqslant 90 ) = 0.0022$. Using the same hypotheses and significance level which you used in part (ii), complete this test.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q1 [16]}}