| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | F-test for equality of variances |
| Difficulty | Standard +0.3 This is a straightforward application of chi-squared and F-distribution tables/critical values requiring only table lookup skills. Part (a) involves finding a critical value given a probability (routine for S4), while part (b) uses the standard relationship between F-distribution tails. No derivation, proof, or problem-solving insight is required—just mechanical use of statistical tables. |
| Spec | 2.04h Select appropriate distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\chi^2_7 < a) = P(\chi^2_7 < 1.690) + 0.95\) | M1 | Using the chi-squared table, recognising need to find \(P(\chi^2_7 < 1.690)\) |
| \(P(\chi^2_7 < 1.690) = 0.025\) (lower 2.5% point) | ||
| So \(P(\chi^2_7 < a) = 0.975\), giving \(a = 16.013\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Upper 1% critical value for \(F(6,4)\): \(\mathbf{15.21}\) | B1 | From tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Lower 1% critical value \(= \frac{1}{F_{4,6}(0.01)} = \frac{1}{9.15} = 0.1093\) (awrt \(0.109\)) | B1 | Using reciprocal relationship with \(F(4,6)\) |
# Question 1:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\chi^2_7 < a) = P(\chi^2_7 < 1.690) + 0.95$ | M1 | Using the chi-squared table, recognising need to find $P(\chi^2_7 < 1.690)$ |
| $P(\chi^2_7 < 1.690) = 0.025$ (lower 2.5% point) | | |
| So $P(\chi^2_7 < a) = 0.975$, giving $a = 16.013$ | A1 | cao |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Upper 1% critical value for $F(6,4)$: $\mathbf{15.21}$ | B1 | From tables |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Lower 1% critical value $= \frac{1}{F_{4,6}(0.01)} = \frac{1}{9.15} = 0.1093$ (awrt $0.109$) | B1 | Using reciprocal relationship with $F(4,6)$ |
---\begin{enumerate}
\item (a) Find the value of the constant $a$ such that
\end{enumerate}
$$\mathrm { P } \left( 1.690 < \chi _ { 7 } ^ { 2 } < a \right) = 0.95$$
The random variable $Y$ follows an $F$-distribution with 6 and 4 degrees of freedom.\\
(b) (i) Find the upper $1 \%$ critical value for $Y$.\\
(ii) Find the lower $1 \%$ critical value for $Y$.\\
\hfill \mbox{\textit{Edexcel S4 2013 Q1 [4]}}