| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | June |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI from summary stats |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard normal distribution calculations, binomial probability, normal approximation to binomial, and a one-tailed hypothesis test. All parts follow routine procedures with no novel problem-solving required. Part (i) is given as 'show that', parts (ii-iii) are standard applications, and parts (iv-v) follow textbook hypothesis testing methodology. Slightly above average difficulty only due to the multi-part nature and requiring familiarity with multiple topics. |
| Spec | 2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X \sim N(1720, 90^2)\) | ||
| \(P(X < 1700) = P\left(Z < \frac{1700-1720}{90}\right) = P(Z < -0.2222)\) | M1, A1 | M1 for standardising |
| \(= \Phi(-0.2222) = 1 - \Phi(0.2222) = 1 - 0.5879 = 0.4121\) | M1, A1 | M1 use of tables (correct tail), A1 CAO. NB ANSWER GIVEN. Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(2 \text{ of } 4 \text{ below } 1700) = \binom{4}{2} \times 0.4121^2 \times 0.5879^2 = 0.3522\) | M1, M1, A1 | M1 for coefficient; M1 for \(0.4121^2 \times 0.5879^2\); A1 FT (min 2sf). Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Normal approx. with \(\mu = np = 40 \times 0.4121 = 16.48\) | B1 | |
| \(\sigma^2 = npq = 40 \times 0.4121 \times 0.5879 = 9.691\) | B1 | |
| \(P(X \geq 20) = P\left(Z \geq \frac{19.5 - 16.48}{\sqrt{9.691}}\right)\) | B1 | B1 for correct continuity correction |
| \(= P(Z \geq 0.9701) = 1 - \Phi(0.9701)\) | M1 | M1 for correct Normal probability calculation using correct tail |
| \(= 1 - 0.8340 = 0.1660\) | A1 | A1 CAO (but FT wrong or omitted CC). Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu = 1720\) | B1 | |
| \(H_1\) is of this form since the consumer organisation suspects that the mean is below 1720 | E1 | |
| \(\mu\) denotes the mean intensity of 25 Watt low energy bulbs made by this manufacturer | B1 | B1 for definition of \(\mu\). Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Test statistic \(= \frac{1703 - 1720}{90/\sqrt{20}} = \frac{-17}{20.12} = -0.8447\) | M1, A1 | M1 must include \(\sqrt{20}\); A1 FT |
| Lower 5% level 1-tailed critical value \(z = -1.645\) | B1 | B1 for \(-1.645\). Must be \(-1.645\) unless absolute values used |
| \(-0.8447 > -1.645\) so not significant. There is not sufficient evidence to reject \(H_0\) | M1, A1 | M1 for sensible comparison leading to conclusion; FT only candidate's test statistic |
| There is insufficient evidence to conclude that the mean intensity of bulbs made by this manufacturer is less than 1720 | A1 | A1 for conclusion in words in context. Total: 5 |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X \sim N(1720, 90^2)$ | | |
| $P(X < 1700) = P\left(Z < \frac{1700-1720}{90}\right) = P(Z < -0.2222)$ | M1, A1 | M1 for standardising |
| $= \Phi(-0.2222) = 1 - \Phi(0.2222) = 1 - 0.5879 = 0.4121$ | M1, A1 | M1 use of tables (correct tail), A1 CAO. **NB ANSWER GIVEN. Total: 4** |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(2 \text{ of } 4 \text{ below } 1700) = \binom{4}{2} \times 0.4121^2 \times 0.5879^2 = 0.3522$ | M1, M1, A1 | M1 for coefficient; M1 for $0.4121^2 \times 0.5879^2$; A1 FT (min 2sf). **Total: 3** |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Normal approx. with $\mu = np = 40 \times 0.4121 = 16.48$ | B1 | |
| $\sigma^2 = npq = 40 \times 0.4121 \times 0.5879 = 9.691$ | B1 | |
| $P(X \geq 20) = P\left(Z \geq \frac{19.5 - 16.48}{\sqrt{9.691}}\right)$ | B1 | B1 for correct continuity correction |
| $= P(Z \geq 0.9701) = 1 - \Phi(0.9701)$ | M1 | M1 for correct Normal probability calculation using correct tail |
| $= 1 - 0.8340 = 0.1660$ | A1 | A1 CAO (but FT wrong or omitted CC). **Total: 5** |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 1720$ | B1 | |
| $H_1$ is of this form since the consumer organisation suspects that the mean is below 1720 | E1 | |
| $\mu$ denotes the mean intensity of 25 Watt low energy bulbs made by this manufacturer | B1 | B1 for definition of $\mu$. **Total: 3** |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Test statistic $= \frac{1703 - 1720}{90/\sqrt{20}} = \frac{-17}{20.12} = -0.8447$ | M1, A1 | M1 must include $\sqrt{20}$; A1 FT |
| Lower 5% level 1-tailed critical value $z = -1.645$ | B1 | B1 for $-1.645$. Must be $-1.645$ unless absolute values used |
| $-0.8447 > -1.645$ so not significant. There is not sufficient evidence to reject $H_0$ | M1, A1 | M1 for sensible comparison leading to conclusion; FT only candidate's test statistic |
| There is insufficient evidence to conclude that the mean intensity of bulbs made by this manufacturer is less than 1720 | A1 | A1 for conclusion in words in context. **Total: 5** |
---3 Intensity of light is measured in lumens. The random variable $X$ represents the intensity of the light from a standard 100 watt light bulb. $X$ is Normally distributed with mean 1720 and standard deviation 90. You may assume that the intensities for different bulbs are independent.\\
(i) Show that $\mathrm { P } ( X < 1700 ) = 0.4121$.\\
(ii) These bulbs are sold in packs of 4 . Find the probability that the intensities of exactly 2 of the 4 bulbs in a randomly chosen pack are below 1700 lumens.\\
(iii) Use a suitable approximating distribution to find the probability that the intensities of at least 20 out of 40 randomly selected bulbs are below 1700 lumens.
A manufacturer claims that the average intensity of its 25 watt low energy light bulbs is 1720 lumens. A consumer organisation suspects that the true figure may be lower than this. The intensities of a random sample of 20 of these bulbs are measured. A hypothesis test is then carried out to check the claim.\\
(iv) Write down a suitable null hypothesis and explain briefly why the alternative hypothesis should be $\mathrm { H } _ { 1 } : \mu < 1720$. State the meaning of $\mu$.\\
(v) Given that the standard deviation of the intensity of such bulbs is 90 lumens and that the mean intensity of the sample of 20 bulbs is 1703 lumens, carry out the test at the $5 \%$ significance level.
\hfill \mbox{\textit{OCR MEI S2 2009 Q3 [20]}}