| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2011 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample confidence interval difference of means |
| Difficulty | Standard +0.3 This is a standard two-sample hypothesis testing question requiring an F-test for equal variances followed by a pooled t-interval. While it involves multiple parts and calculations from raw data, the procedures are routine S4 material with no novel problem-solving required. The question clearly signposts each step, making it slightly easier than average for A-level statistics. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s_x^2 = \frac{1559691 - 6 \times \left(\frac{3059}{6}\right)^2}{5} = 22.1666...\) | M1 | Use of correct formula for \(s_x^2\) with reasonable attempt at \(\sum x^2\) and \(\sum x\) |
| \(H_0: \sigma_x^2 = \sigma_y^2 \quad H_1: \sigma_x^2 \neq \sigma_y^2\) | B1 | |
| \(\frac{s_x^2}{s_y^2} = 1.895...\) | M1 | Use of correct test statistic; allow use of \(3.42\) instead of \(3.42^2\); top must be their variance |
| \(F_{5,4} = 6.26\) | B1 | |
| \(\frac{s_x^2}{s_y^2} = 1.895...\) awrt 1.90 and comment: not significant — variances of weights of the two boxes can be assumed equal | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = 509.833... \Rightarrow \bar{x} - \bar{y} = 5.03333\) | M1 | Attempting \(\bar{x} - \bar{y}\); can follow through their \(\bar{x}\) |
| \(s_p^2 = \frac{5s_x^2 + 4s_y^2}{9} = 17.513...\) awrt 17.5 | M1A1 | For attempt to find pooled estimate of variance |
| 5% two tail \(t\) value is \(t_9 = 1.833\) | B1 | |
| 90% confidence interval is \(5.03... \pm 1.833 \times \sqrt{17.513...} \times \sqrt{\frac{1}{6} + \frac{1}{5}}\) | M1 | For use of correct formula for CI; allow any \(t\) value and ft their \(\bar{x}\) and \(s_p\) |
| \((0.388...,\ 9.6782...)\) awrt (0.388, 9.68) | A1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Zero is not in CI, there is evidence to reject the manufacturer's claim | B1ft | |
| Or the weight of the contents of the boxes has changed | B1ft |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s_x^2 = \frac{1559691 - 6 \times \left(\frac{3059}{6}\right)^2}{5} = 22.1666...$ | M1 | Use of correct formula for $s_x^2$ with reasonable attempt at $\sum x^2$ and $\sum x$ |
| $H_0: \sigma_x^2 = \sigma_y^2 \quad H_1: \sigma_x^2 \neq \sigma_y^2$ | B1 | |
| $\frac{s_x^2}{s_y^2} = 1.895...$ | M1 | Use of correct test statistic; allow use of $3.42$ instead of $3.42^2$; top must be their variance |
| $F_{5,4} = 6.26$ | B1 | |
| $\frac{s_x^2}{s_y^2} = 1.895...$ awrt **1.90** and comment: not significant — variances of **weights** of the two **boxes** can be assumed equal | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = 509.833... \Rightarrow \bar{x} - \bar{y} = 5.03333$ | M1 | Attempting $\bar{x} - \bar{y}$; can follow through their $\bar{x}$ |
| $s_p^2 = \frac{5s_x^2 + 4s_y^2}{9} = 17.513...$ awrt **17.5** | M1A1 | For attempt to find pooled estimate of variance |
| 5% two tail $t$ value is $t_9 = 1.833$ | B1 | |
| 90% confidence interval is $5.03... \pm 1.833 \times \sqrt{17.513...} \times \sqrt{\frac{1}{6} + \frac{1}{5}}$ | M1 | For use of correct formula for CI; allow any $t$ value and ft their $\bar{x}$ and $s_p$ |
| $(0.388...,\ 9.6782...)$ awrt **(0.388, 9.68)** | A1, A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Zero is not in CI, there **is** evidence to **reject** the manufacturer's claim | B1ft | |
| Or the weight of the contents of the boxes has changed | B1ft | |
---\begin{enumerate}
\item The weights of the contents of breakfast cereal boxes are normally distributed.
\end{enumerate}
A manufacturer changes the style of the boxes but claims that the weight of the contents remains the same.\\
A random sample of 6 old style boxes had contents with the following weights (in grams).
$$\begin{array} { l l l l l l }
512 & 503 & 514 & 506 & 509 & 515
\end{array}$$
The weights, $y$ grams, of the contents of an independent random sample of 5 new style boxes gave
$$\bar { y } = 504.8 \text { and } s _ { y } = 3.420$$
(a) Use a two-tail test to show, at the $10 \%$ level of significance, that the variances of the weights of the contents of the old and new style boxes can be assumed to be equal. State your hypotheses clearly.\\
(b) Showing your working clearly, find a $90 \%$ confidence interval for $\mu _ { x } - \mu _ { y }$, where $\mu _ { x }$ and $\mu _ { y }$ are the mean weights of the contents of old and new style boxes respectively.\\
(c) With reference to your confidence interval comment on the manufacturer's claim.\\
\hfill \mbox{\textit{Edexcel S4 2011 Q5 [14]}}