| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2008 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | State meaning of Type I error |
| Difficulty | Standard +0.3 This is a straightforward application of hypothesis testing concepts with normal distributions. Part (i) requires standard definitions, parts (ii)-(iii) involve routine calculations using the normal distribution with known standard deviation (σ/√n = 20/3), and part (iv) requires deriving and evaluating a formula repeatedly. While it's multi-part and requires careful arithmetic, it involves no novel insights—just systematic application of standard S4 techniques. Slightly easier than average due to its procedural nature. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Type I error: rejecting null hypothesis when it is true | B1, B1 | Allow B1 for P(rej \(H_0\) when true) |
| Type II error: accepting null hypothesis when it is false | B1, B1 | Allow B1 for P(acc \(H_0\) when false) |
| OC: P(accepting null hypothesis as a function of the parameter under investigation) | B1, B1 | [P(type II error |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Reject if \(\bar{x} < 735\) or \(\bar{x} > 765\) | ||
| \(\alpha = P(\bar{X} < 735 \text{ or } \bar{X} > 765 \mid \bar{X} \sim N\left(750, \dfrac{20^2}{9}\right))\) | M1 | Might be implicit |
| \(= P\left(Z < \dfrac{(735-750)3}{20} = -2.25\right)\) | A1 | |
| or \(Z > \dfrac{(765-750)3}{20} = 2.25\) | A1 | |
| \(= 2(1-0.9878) = 2 \times 0.0122 = 0.0244\) | A1 | CAO |
| This is the probability of rejecting good output and unnecessarily re-calibrating the machine – seems small [but not very small?] | E1, E1 | Accept any sensible comments |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Accept if \(735 < \bar{x} < 765\), now \(\mu = 725\) | M1 | Might be implicit |
| \(\beta = P(735 < \bar{X} < 765 \mid \bar{X} \sim N\left(725, \dfrac{20^2}{9}\right))\) | ||
| \(= P(1.5 < Z < 6) = 1 - 0.9332 = 0.0668\) | A1, A1, A1 | CAO. If upper limit 765 not considered, max 2 of these 4 marks. If \(\Phi(6)\) not considered, max 3 out of 4 |
| This is the probability of accepting output and carrying on when in fact \(\mu\) has slipped to 725 – small[-ish?] | E1, E1 | Accept sensible comments |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(OC = P\left(735 < \bar{X} < 765 \mid \bar{X} \sim N\left(\mu, \dfrac{20^2}{9}\right)\right)\) | M1 | |
| \(= \Phi\left(\dfrac{(765-\mu)3}{20}\right) - \Phi\left(\dfrac{(735-\mu)3}{20}\right)\) | M1, A1 | Both correct |
| \(\mu=720\): \(\Phi(6.75)-\Phi(2.25)=1-0.9878=0.0122\) | ||
| \(\mu=730\): \(5.25\) and \(0.75 \to 1-0.7734=0.2266\) | ||
| \(\mu=740\): \(3.75\) and \(-0.75 \to 1-(1-0.7734)=0.7734\) | 1 | If any two correct |
| \(\mu=750\): similarly or by write-down from part (ii): \(0.9756\) | 1 | FT |
| \(\mu=760, 770, 780\) by symmetry: \(0.7734,\ 0.2266,\ 0.0122\) | 1 | FT |
# Question 3:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Type I error: rejecting null hypothesis when it is true | B1, B1 | Allow B1 for P(rej $H_0$ when true) |
| Type II error: accepting null hypothesis when it is false | B1, B1 | Allow B1 for P(acc $H_0$ when false) |
| OC: P(accepting null hypothesis as a function of the parameter under investigation) | B1, B1 | [P(type II error | the true value of the parameter) scores B1+B1] |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Reject if $\bar{x} < 735$ or $\bar{x} > 765$ | | |
| $\alpha = P(\bar{X} < 735 \text{ or } \bar{X} > 765 \mid \bar{X} \sim N\left(750, \dfrac{20^2}{9}\right))$ | M1 | Might be implicit |
| $= P\left(Z < \dfrac{(735-750)3}{20} = -2.25\right)$ | A1 | |
| or $Z > \dfrac{(765-750)3}{20} = 2.25$ | A1 | |
| $= 2(1-0.9878) = 2 \times 0.0122 = 0.0244$ | A1 | CAO |
| This is the probability of rejecting good output and unnecessarily re-calibrating the machine – seems small [but not very small?] | E1, E1 | Accept any sensible comments |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Accept if $735 < \bar{x} < 765$, now $\mu = 725$ | M1 | Might be implicit |
| $\beta = P(735 < \bar{X} < 765 \mid \bar{X} \sim N\left(725, \dfrac{20^2}{9}\right))$ | | |
| $= P(1.5 < Z < 6) = 1 - 0.9332 = 0.0668$ | A1, A1, A1 | CAO. If upper limit 765 not considered, max 2 of these 4 marks. If $\Phi(6)$ not considered, max 3 out of 4 |
| This is the probability of accepting output and carrying on when in fact $\mu$ has slipped to 725 – small[-ish?] | E1, E1 | Accept sensible comments |
## Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $OC = P\left(735 < \bar{X} < 765 \mid \bar{X} \sim N\left(\mu, \dfrac{20^2}{9}\right)\right)$ | M1 | |
| $= \Phi\left(\dfrac{(765-\mu)3}{20}\right) - \Phi\left(\dfrac{(735-\mu)3}{20}\right)$ | M1, A1 | Both correct |
| $\mu=720$: $\Phi(6.75)-\Phi(2.25)=1-0.9878=0.0122$ | | |
| $\mu=730$: $5.25$ and $0.75 \to 1-0.7734=0.2266$ | | |
| $\mu=740$: $3.75$ and $-0.75 \to 1-(1-0.7734)=0.7734$ | 1 | If any two correct |
| $\mu=750$: similarly or by write-down from part (ii): $0.9756$ | 1 | FT |
| $\mu=760, 770, 780$ by symmetry: $0.7734,\ 0.2266,\ 0.0122$ | 1 | FT |
---3 (i) Explain the meaning of the following terms in the context of hypothesis testing: Type I error, Type II error, operating characteristic.
A machine fills salt containers that will be sold in shops. The containers are supposed to contain 750 g of salt. The machine operates in such a way that the amount of salt delivered to each container is a Normally distributed random variable with standard deviation 20 g . The machine should be calibrated in such a way that the mean amount delivered, $\mu$, is 750 g .
Each hour, a random sample of 9 containers is taken from the previous hour's output and the sample mean amount of salt is determined. If this is between 735 g and 765 g , the previous hour's output is accepted. If not, the previous hour's output is rejected and the machine is recalibrated.\\
(ii) Find the probability of rejecting the previous hour's output if the machine is properly calibrated. Comment on your result.\\
(iii) Find the probability of accepting the previous hour's output if $\mu = 725 \mathrm {~g}$. Comment on your result.\\
(iv) Obtain an expression for the operating characteristic of this testing procedure in terms of the cumulative distribution function $\Phi ( z )$ of the standard Normal distribution. Evaluate the operating characteristic for the following values (in g) of $\mu$ : 720, 730, 740, 750, 760, 770, 780.
\hfill \mbox{\textit{OCR MEI S4 2008 Q3 [24]}}