| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 19 |
| 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 S1 hypothesis testing question with standard binomial probability calculations. Part (i) requires basic binomial probability formulas, part (ii) is a routine one-tailed test with clear hypotheses, and part (iii) asks for a two-tailed critical region—all standard textbook exercises requiring no novel insight, though the multi-part structure and cumulative probability calculations place it slightly above average difficulty. |
| 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/Working | Marks | Guidance |
| \(X \sim B(10, 0.8)\) | ||
| Either \(P(X=8) = \binom{10}{8} \times 0.8^8 \times 0.2^2 = 0.3020\) (awrt) | M1 | \(0.8^8 \times 0.2^2\) or 0.00671... |
| M1 | \(\binom{10}{8} \times p^8q^2\); \((p+q=1)\) or \(45 \times p^8q^2\) | |
| or \(P(X=8) = P(X \leq 8) - P(X \leq 7) = 0.6242 - 0.3222 = 0.3020\) | A1 CAO | (0.302) not 0.3; OR M2 for \(0.6242 - 0.3222\) |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Either \(P(X \geq 8) = 1 - P(X \leq 7) = 1 - 0.3222 = 0.6778\) | M1 | For \(1 - 0.3222\) (s.o.i.) |
| or \(P(X \geq 8) = P(X=8) + P(X=9) + P(X=10) = 0.3020 + 0.2684 + 0.1074 = 0.6778\) | A1 CAO | awfw 0.677 – 0.678; or M1 for sum of their \(p(X=8)\) plus correct expressions for \(p(X=9)\) and \(p(X=10)\) |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(X \sim B(18, p)\); Let \(p =\) probability of delivery within 24 hours (for population) | B1 | For definition of \(p\) |
| \(H_0: p = 0.8\) | B1 | For \(H_0\) |
| \(H_1: p < 0.8\) | B1 | For \(H_1\) |
| \(P(X \leq 12) = 0.1329 > 5\%\) ref: \([pp = 0.0816]\) | M1 | For probability 0.1329 |
| M1dep | Strictly for comparison of 0.1329 with 5% (seen or clearly implied) | |
| So not enough evidence to reject \(H_0\) | A1dep | On both M's |
| Conclude that there is not enough evidence to indicate that less than 80% of orders will be delivered within 24 hours | E1dep | On M1,M1,A1 for conclusion in context |
| Note: use of critical region method scores M1 for region \(\{0,1,2,...,9,10\}\); M1dep for 12 does not lie in critical region then A1dep E1dep as per scheme | ||
| Total | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \leq 10) = 0.0163 < 2.5\%\) | B1 | for \(0.0163\) or \(0.0513\) seen |
| \(P(X \leq 11) = 0.0513 > 2.5\%\) | seen | |
| M1dep | for either correct comparison with \(2.5\%\) (not 5%) (seen or clearly implied) | |
| A1dep | for correct lower tail CR (must have zero) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \geq 17) = 1 - P(X \leq 16) = 1 - 0.9009 = 0.0991 > 2.5\%\) | B1 | for \(0.0991\) or \(0.0180\) seen |
| \(P(X \geq 18) = 1 - P(X \leq 17) = 1 - 0.9820 = 0.0180 < 2.5\%\) | seen | |
| M1dep | for either correct comparison with \(2.5\%\) (not 5%) (seen or clearly implied) | |
| So critical region is \(\{0,1,2,3,4,5,6,7,8,9,10,18\}\) o.e. | A1dep | for correct upper tail CR |
| Answer | Marks | Guidance |
|---|---|---|
| Correct CR without supportive working scores SC2 max after the 1st B1 (SC1 for each fully correct tail of CR) | Total: 7 | |
| B1 | for \(H_1\) |
## Question 7:
### Part (i)(A):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim B(10, 0.8)$ | | |
| Either $P(X=8) = \binom{10}{8} \times 0.8^8 \times 0.2^2 = 0.3020$ (awrt) | M1 | $0.8^8 \times 0.2^2$ or 0.00671... |
| | M1 | $\binom{10}{8} \times p^8q^2$; $(p+q=1)$ or $45 \times p^8q^2$ |
| or $P(X=8) = P(X \leq 8) - P(X \leq 7) = 0.6242 - 0.3222 = 0.3020$ | A1 CAO | (0.302) not 0.3; OR M2 for $0.6242 - 0.3222$ |
| **Total** | **3** | |
### Part (i)(B):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Either $P(X \geq 8) = 1 - P(X \leq 7) = 1 - 0.3222 = 0.6778$ | M1 | For $1 - 0.3222$ (s.o.i.) |
| or $P(X \geq 8) = P(X=8) + P(X=9) + P(X=10) = 0.3020 + 0.2684 + 0.1074 = 0.6778$ | A1 CAO | awfw 0.677 – 0.678; or M1 for sum of their $p(X=8)$ plus correct expressions for $p(X=9)$ and $p(X=10)$ |
| **Total** | **2** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $X \sim B(18, p)$; Let $p =$ probability of delivery within 24 hours (for population) | B1 | For definition of $p$ |
| $H_0: p = 0.8$ | B1 | For $H_0$ |
| $H_1: p < 0.8$ | B1 | For $H_1$ |
| $P(X \leq 12) = 0.1329 > 5\%$ ref: $[pp = 0.0816]$ | M1 | For probability 0.1329 |
| | M1dep | Strictly for comparison of 0.1329 with 5% (seen or clearly implied) |
| So not enough evidence to reject $H_0$ | A1dep | On both M's |
| Conclude that there is not enough evidence to indicate that less than 80% of orders will be delivered within 24 hours | E1dep | On M1,M1,A1 for conclusion in context |
| Note: use of critical region method scores M1 for region $\{0,1,2,...,9,10\}$; M1dep for 12 does not lie in critical region then A1dep E1dep as per scheme | | |
| **Total** | **7** | |
## Question (iii):
Let $X \sim B(18, 0.8)$
$H_1: p \neq 0.8$
**Lower Tail:**
$P(X \leq 10) = 0.0163 < 2.5\%$ | B1 | for $0.0163$ or $0.0513$ seen
$P(X \leq 11) = 0.0513 > 2.5\%$ | | seen
| M1dep | for either correct comparison with $2.5\%$ **(not 5%)** (seen or clearly implied)
| A1dep | for correct lower tail CR (must have zero)
**Upper Tail:**
$P(X \geq 17) = 1 - P(X \leq 16) = 1 - 0.9009 = 0.0991 > 2.5\%$ | B1 | for $0.0991$ or $0.0180$ seen
$P(X \geq 18) = 1 - P(X \leq 17) = 1 - 0.9820 = 0.0180 < 2.5\%$ | | seen
| M1dep | for either correct comparison with $2.5\%$ **(not 5%)** (seen or clearly implied)
So critical region is $\{0,1,2,3,4,5,6,7,8,9,10,18\}$ o.e. | A1dep | for correct upper tail CR
Condone $X \leq 10$ and $X \geq 18$ or $X = 18$ but **not** $p(X \leq 10)$ and $p(X \geq 18)$
Correct CR without supportive working scores SC2 max after the 1st B1 (SC1 for each fully correct tail of CR) | | **Total: 7**
| B1 | for $H_1$
**TOTAL: 19**7 An online shopping company takes orders through its website. On average $80 \%$ of orders from the website are delivered within 24 hours. The quality controller selects 10 orders at random to check when they are delivered.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that\\
(A) exactly 8 of these orders are delivered within 24 hours,\\
(B) at least 8 of these orders are delivered within 24 hours.
The company changes its delivery method. The quality controller suspects that the changes will mean that fewer than $80 \%$ of orders will be delivered within 24 hours. A random sample of 18 orders is checked and it is found that 12 of them arrive within 24 hours.
\item Write down suitable hypotheses and carry out a test at the $5 \%$ significance level to determine whether there is any evidence to support the quality controller's suspicion.
\item A statistician argues that it is possible that the new method could result in either better or worse delivery times. Therefore it would be better to carry out a 2 -tail test at the $5 \%$ significance level. State the alternative hypothesis for this test. Assuming that the sample size is still 18, find the critical region for this test, showing all of your calculations.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2009 Q7 [19]}}