| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Standard CI with summary statistics |
| Difficulty | Challenging +1.2 This is a multi-part question on hypothesis testing for the difference of means with known population standard deviations, requiring calculation of sample mean, conducting a z-test, finding critical values, and computing power. While it involves several steps and the power calculation in part (c) is less routine, the techniques are all standard S3 material with clear guidance at each stage. The computational demands are moderate and the conceptual framework is well-established, making it above average but not exceptionally challenging. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\bar{y} = 1193\) | B1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(H_0: \mu_Y - \mu_X = 200\) | B1 | 200 is not necessary |
| \(H_1: \mu_Y - \mu_X > 200\) | B1 | 200 is necessary |
| SL \(\alpha = 0.01\) (1%), CV \(z = 2.3263\) | B1 | AWFW 2.32 to 2.33 |
| \(z = \frac{(\bar{y}-\bar{x})-200}{\sqrt{\frac{\sigma_Y^2}{n_y}+\frac{\sigma_X^2}{n_x}}} = \frac{(1193-936)-200}{\sqrt{\frac{65^2}{10}+\frac{45^2}{20}}}\) | M1 M1 A1\(\checkmark\) | Numerator; 200 is not necessary / Denominator / \(\checkmark\) on (a) |
| \(= 2.48\) to \(2.5\) | A1 | AWFW |
| Evidence (at 1% level) to support the claim | A1\(\checkmark\) | \(\checkmark\) on \(z\) and CV |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(CV(\bar{y}-\bar{x}): 200 + z(\text{denominator in (b)})\) | M1 | May be scored in (b) |
| \(200 + 2.3263 \times \sqrt{523.75}\) | ||
| \((= 253.24)\) | A1 | AG; must justify |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Power \(= 1 - P(\text{Type II error})\) | M1 | Use of |
| \(= 1 - P(\text{accept } H_0 \mid H_0 \text{ false})\) | M1 | Use of; or equivalent |
| \(= 1 - P\left(Z < \frac{253.24-275}{\sqrt{523.75}}\right)\) | M1 | Standardising 253.24 using 275 and C's denominator in (b) |
| \(= 1 - \Phi(-0.95) = \Phi(0.95)\) | m1 | Area change |
| \(= 0.83\) | A1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Probability of accepting that difference in mean weights is more than 200 grams | B1 | Not in context \(\Rightarrow\) max of 2 |
| when, in fact, it is 275 grams | B1 | |
| is 0.83 (or 83%) | B1\(\checkmark\) | \(\checkmark\) on (ii) |
# Question 7:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\bar{y} = 1193$ | B1 | CAO |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $H_0: \mu_Y - \mu_X = 200$ | B1 | 200 is not necessary |
| $H_1: \mu_Y - \mu_X > 200$ | B1 | 200 is necessary |
| SL $\alpha = 0.01$ (1%), CV $z = 2.3263$ | B1 | AWFW 2.32 to 2.33 |
| $z = \frac{(\bar{y}-\bar{x})-200}{\sqrt{\frac{\sigma_Y^2}{n_y}+\frac{\sigma_X^2}{n_x}}} = \frac{(1193-936)-200}{\sqrt{\frac{65^2}{10}+\frac{45^2}{20}}}$ | M1 M1 A1$\checkmark$ | Numerator; 200 is not necessary / Denominator / $\checkmark$ on (a) |
| $= 2.48$ to $2.5$ | A1 | AWFW |
| Evidence (at 1% level) to support the claim | A1$\checkmark$ | $\checkmark$ on $z$ and CV |
## Part (c)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $CV(\bar{y}-\bar{x}): 200 + z(\text{denominator in (b)})$ | M1 | May be scored in (b) |
| $200 + 2.3263 \times \sqrt{523.75}$ | | |
| $(= 253.24)$ | A1 | AG; must justify |
## Part (c)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Power $= 1 - P(\text{Type II error})$ | M1 | Use of |
| $= 1 - P(\text{accept } H_0 \mid H_0 \text{ false})$ | M1 | Use of; or equivalent |
| $= 1 - P\left(Z < \frac{253.24-275}{\sqrt{523.75}}\right)$ | M1 | Standardising 253.24 using 275 and C's denominator in (b) |
| $= 1 - \Phi(-0.95) = \Phi(0.95)$ | m1 | Area change |
| $= 0.83$ | A1 | AWRT |
## Part (c)(iii):
| Working | Mark | Guidance |
|---------|------|----------|
| Probability of **accepting** that difference in mean weights **is more than 200** grams | B1 | Not in context $\Rightarrow$ max of 2 |
| when, in fact, **it is 275** grams | B1 | |
| **is 0.83** (or 83%) | B1$\checkmark$ | $\checkmark$ on (ii) |7 A shop sells cooked chickens in two sizes: medium and large.\\
The weights, $X$ grams, of medium chickens may be assumed to be normally distributed with mean $\mu _ { X }$ and standard deviation 45.
The weights, $Y$ grams, of large chickens may be assumed to be normally distributed with mean $\mu _ { Y }$ and standard deviation 65.
A random sample of 20 medium chickens had a mean weight, $\bar { x }$ grams, of 936 .\\
A random sample of 10 large chickens had the following weights in grams:
$$\begin{array} { l l l l l l l l l l }
1165 & 1202 & 1077 & 1144 & 1195 & 1275 & 1136 & 1215 & 1233 & 1288
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Calculate the mean weight, $\bar { y }$ grams, of this sample of large chickens.
\item Hence investigate, at the $1 \%$ level of significance, the claim that the mean weight of large chickens exceeds that of medium chickens by more than 200 grams.
\item \begin{enumerate}[label=(\roman*)]
\item Deduce that, for your test in part (b), the critical value of $( \bar { y } - \bar { x } )$ is 253.24, correct to two decimal places.
\item Hence determine the power of your test in part (b), given that $\mu _ { Y } - \mu _ { X } = 275$.
\item Interpret, in the context of this question, the value that you obtained in part (c)(ii).\\
(3 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2006 Q7 [19]}}