| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Pooled variance estimation |
| Difficulty | Standard +0.3 This is a straightforward application of confidence intervals for the difference of two normal means with known standard deviations. Students need to calculate sample means, apply the standard formula for the difference of independent normals, and use z-critical values. Part (b) requires basic understanding of sampling bias. The calculations are routine and the concepts are standard S3 material, making it slightly easier than average. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x} = \frac{1055+1045+1000+985+1040+1025+1005+1030+1015+1060}{10} = \frac{10260}{10} = 1026\) | B1 | Correct value of \(\bar{x}\) |
| \(\bar{y} = \frac{1085+1055+1055+1000+1035+1050+1005+1075}{8} = \frac{8360}{8} = 1045\) | B1 | Correct value of \(\bar{y}\) |
| \(\bar{y} - \bar{x} = 1045 - 1026 = 19\) | B1 | Correct difference |
| \(SE = \sqrt{\frac{30^2}{8} + \frac{25^2}{10}} = \sqrt{\frac{900}{8} + \frac{625}{10}} = \sqrt{112.5 + 62.5} = \sqrt{175}\) | M1 A1 | Method for standard error using given \(\sigma_X = 25\), \(\sigma_Y = 30\) |
| \(19 \pm 1.96\sqrt{175}\) | M1 | Using \(z = 1.96\) for 95% CI |
| \(= 19 \pm 25.92...\) | ||
| \((-7, 45)\) or \((-5, 45)\) to nearest 5 kg | A1 | Correct interval rounded to nearest 5 kg |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Fred may have chosen bags that looked heavier/fuller (or similar non-random selection comment) | B1 | Valid justification of possible bias |
| Use a random number generator / systematic sampling to select the bags to be weighed | B1 | Valid method to eliminate bias |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = \frac{1055+1045+1000+985+1040+1025+1005+1030+1015+1060}{10} = \frac{10260}{10} = 1026$ | B1 | Correct value of $\bar{x}$ |
| $\bar{y} = \frac{1085+1055+1055+1000+1035+1050+1005+1075}{8} = \frac{8360}{8} = 1045$ | B1 | Correct value of $\bar{y}$ |
| $\bar{y} - \bar{x} = 1045 - 1026 = 19$ | B1 | Correct difference |
| $SE = \sqrt{\frac{30^2}{8} + \frac{25^2}{10}} = \sqrt{\frac{900}{8} + \frac{625}{10}} = \sqrt{112.5 + 62.5} = \sqrt{175}$ | M1 A1 | Method for standard error using given $\sigma_X = 25$, $\sigma_Y = 30$ |
| $19 \pm 1.96\sqrt{175}$ | M1 | Using $z = 1.96$ for 95% CI |
| $= 19 \pm 25.92...$ | | |
| $(-7, 45)$ or $(-5, 45)$ to nearest 5 kg | A1 | Correct interval rounded to nearest 5 kg |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Fred may have chosen bags that looked heavier/fuller (or similar non-random selection comment) | B1 | Valid justification of possible bias |
| Use a random number generator / systematic sampling to select the bags to be weighed | B1 | Valid method to eliminate bias |
---3 A builders' merchant's depot has two machines, X and Y , each of which can be used for filling bags with sand or gravel.
The weight, in kilograms, delivered by machine X may be modelled by a normal distribution with mean $\mu _ { \mathrm { X } }$ and standard deviation 25 .
The weight, in kilograms, delivered by machine Y may be modelled by a normal distribution with mean $\mu _ { \mathrm { Y } }$ and standard deviation 30 .
Fred, the depot's yardman, records the weights, in kilograms, of a random sample of 10 bags of sand delivered by machine X as\\
$\begin{array} { l l l l l l l l l l } 1055 & 1045 & 1000 & 985 & 1040 & 1025 & 1005 & 1030 & 1015 & 1060 \end{array}$\\
He also records the weights, in kilograms, of a random sample of 8 bags of gravel delivered by machine Y as
$$\begin{array} { l l l l l l l l }
1085 & 1055 & 1055 & 1000 & 1035 & 1050 & 1005 & 1075
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Construct a $95 \%$ confidence interval for $\mu _ { \mathrm { Y } } - \mu _ { \mathrm { X } }$, giving the limits to the nearest 5 kg .
\item Dot, the depot's manager, commented that Fred's data collection may have been biased.
Justify her comment and explain how the possible bias could have been eliminated.\\
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2013 Q3 [9]}}