| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2011 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared distribution theory and properties |
| Difficulty | Standard +0.8 This is a Further Maths S4 question requiring MGF verification through integration, differentiation of MGFs for moments, proof using MGF properties for sums of independent variables, and CLT application. While systematic, it demands fluency with advanced statistical theory (MGFs, chi-squared properties, CLT) and multi-step technical execution across four parts, placing it moderately above average difficulty. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(n=2\): \(f(x) = \frac{1}{2}e^{-x/2}\) | ||
| \(M(\theta) = E(e^{\theta X}) = \int_0^{\infty} \frac{1}{2} e^{-x(\frac{1}{2}-\theta)} dx\) | A1 | Any equivalent form |
| \(= \frac{1}{2}\left[\frac{e^{-x(\frac{1}{2}-\theta)}}{-(\frac{1}{2}-\theta)}\right]_0^{\infty}\) | A1 | |
| \(= \frac{\frac{1}{2}}{\frac{1}{2}-\theta}\) | A1 | |
| \(= (1-2\theta)^{-1}\) | A1 | beware printed answer |
| \(n=4\): \(f(x) = \frac{1}{4}xe^{-x/2}\) | ||
| \(M(\theta) = \int_0^{\infty} \frac{1}{4} x e^{-x(\frac{1}{2}-\theta)} dx\) | M1 | attempt to integrate by parts |
| \(= \frac{1}{4}\left\{\left[\frac{xe^{-x(\frac{1}{2}-\theta)}}{-(\frac{1}{2}-\theta)}\right]_0^{\infty} - \int_0^{\infty} \frac{e^{-x(\frac{1}{2}-\theta)}}{-(\frac{1}{2}-\theta)} dx\right\}\) | A1, A1 | for each component as shown |
| \(= \frac{1}{4}\left\{[0-0] + \frac{1}{\frac{1}{2}-\theta} \cdot 2(1-2\theta)^{-1}\right\}\) | A1, A1 | for each component as shown |
| \(= \frac{1}{2} \cdot \frac{1}{(1-2\theta)} (1-2\theta)^{-1} = (1-2\theta)^{-2}\) | A1 | beware printed answer |
| [10] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mean \(= M'(0)\): \(M'(\theta) = -2(-\frac{n}{2})(1-2\theta)^{-\frac{n}{2}-1} = n(1-2\theta)^{-\frac{n}{2}-1}\) | M1 A1 | |
| \(\therefore\) mean \(= n\) | A1 | |
| Variance \(= M''(0) - \{M'(0)\}^2\) | ||
| \(M''(\theta) = n(-\frac{n}{2}-1)(-2)(1-2\theta)^{-\frac{n}{2}-2} = n(n+2)(1-2\theta)^{-\frac{n}{2}-2}\) | M1 A1 | |
| \(\therefore M''(0) = n(n+2)\) | A1 | |
| \(\therefore\) variance \(= n(n+2) - n^2 = 2n\) | A1 | |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| By convolution theorem | M1 | |
| \(M_W(\theta) = \{(1-2\theta)^{-\frac{1}{2}}\}^k = (1-2\theta)^{-k/2}\) | B1 | |
| This is the mgf of \(\chi^2_k\), so by uniqueness of mgfs | M1 | |
| \(W \sim \chi^2_k\) | B1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(W \sim \chi^2_{100}\) has mean 100, variance 200. Can regard \(W\) as sum of large "random sample" of \(\chi^2_1\) variates. | ||
| \(P(\chi^2_{100} < 118.5) \approx P\left(N(0,1) < \frac{118.5-100}{\sqrt{200}} = 1.308\right)\) | M1, A1 c.a.o. | for use of \(N(0,1)\); for 1.308 |
| \(= 0.9045\) | A1 c.a.o. | |
| [3] |
# Question 2 (4769 June 2011):
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $n=2$: $f(x) = \frac{1}{2}e^{-x/2}$ | | |
| $M(\theta) = E(e^{\theta X}) = \int_0^{\infty} \frac{1}{2} e^{-x(\frac{1}{2}-\theta)} dx$ | A1 | Any equivalent form |
| $= \frac{1}{2}\left[\frac{e^{-x(\frac{1}{2}-\theta)}}{-(\frac{1}{2}-\theta)}\right]_0^{\infty}$ | A1 | |
| $= \frac{\frac{1}{2}}{\frac{1}{2}-\theta}$ | A1 | |
| $= (1-2\theta)^{-1}$ | A1 | beware printed answer |
| $n=4$: $f(x) = \frac{1}{4}xe^{-x/2}$ | | |
| $M(\theta) = \int_0^{\infty} \frac{1}{4} x e^{-x(\frac{1}{2}-\theta)} dx$ | M1 | attempt to integrate by parts |
| $= \frac{1}{4}\left\{\left[\frac{xe^{-x(\frac{1}{2}-\theta)}}{-(\frac{1}{2}-\theta)}\right]_0^{\infty} - \int_0^{\infty} \frac{e^{-x(\frac{1}{2}-\theta)}}{-(\frac{1}{2}-\theta)} dx\right\}$ | A1, A1 | for each component as shown |
| $= \frac{1}{4}\left\{[0-0] + \frac{1}{\frac{1}{2}-\theta} \cdot 2(1-2\theta)^{-1}\right\}$ | A1, A1 | for each component as shown |
| $= \frac{1}{2} \cdot \frac{1}{(1-2\theta)} (1-2\theta)^{-1} = (1-2\theta)^{-2}$ | A1 | beware printed answer |
| **[10]** | | |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mean $= M'(0)$: $M'(\theta) = -2(-\frac{n}{2})(1-2\theta)^{-\frac{n}{2}-1} = n(1-2\theta)^{-\frac{n}{2}-1}$ | M1 A1 | |
| $\therefore$ mean $= n$ | A1 | |
| Variance $= M''(0) - \{M'(0)\}^2$ | | |
| $M''(\theta) = n(-\frac{n}{2}-1)(-2)(1-2\theta)^{-\frac{n}{2}-2} = n(n+2)(1-2\theta)^{-\frac{n}{2}-2}$ | M1 A1 | |
| $\therefore M''(0) = n(n+2)$ | A1 | |
| $\therefore$ variance $= n(n+2) - n^2 = 2n$ | A1 | |
| **[7]** | | |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| By convolution theorem | M1 | |
| $M_W(\theta) = \{(1-2\theta)^{-\frac{1}{2}}\}^k = (1-2\theta)^{-k/2}$ | B1 | |
| This is the mgf of $\chi^2_k$, so by uniqueness of mgfs | M1 | |
| $W \sim \chi^2_k$ | B1 | |
| **[4]** | | |
## Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $W \sim \chi^2_{100}$ has mean 100, variance 200. Can regard $W$ as sum of large "random sample" of $\chi^2_1$ variates. | | |
| $P(\chi^2_{100} < 118.5) \approx P\left(N(0,1) < \frac{118.5-100}{\sqrt{200}} = 1.308\right)$ | M1, A1 c.a.o. | for use of $N(0,1)$; for 1.308 |
| $= 0.9045$ | A1 c.a.o. | |
| **[3]** | | |
---2 The random variable $X$ has the $\chi _ { n } ^ { 2 }$ distribution. This distribution has moment generating function $\mathrm { M } ( \theta ) = ( 1 - 2 \theta ) ^ { - \frac { 1 } { 2 } n }$, where $\theta < \frac { 1 } { 2 }$.\\
(i) Verify the expression for $\mathrm { M } ( \theta )$ quoted above for the cases $n = 2$ and $n = 4$, given that the probability density functions of $X$ in these cases are as follows.
$$\begin{array} { l l }
n = 2 : & \mathrm { f } ( x ) = \frac { 1 } { 2 } \mathrm { e } ^ { - \frac { 1 } { 2 } x } \quad ( x > 0 ) \\
n = 4 : & \mathrm { f } ( x ) = \frac { 1 } { 4 } x \mathrm { e } ^ { - \frac { 1 } { 2 } x } \quad ( x > 0 )
\end{array}$$
(ii) For the general case, use $\mathrm { M } ( \theta )$ to find the mean and variance of $X$ in terms of $n$.\\
(iii) $Y _ { 1 } , Y _ { 2 } , \ldots , Y _ { k }$ are independent random variables, each with the $\chi _ { 1 } ^ { 2 }$ distribution. Show that $W = \sum _ { i = 1 } ^ { k } Y _ { i }$ has the $\chi _ { k } ^ { 2 }$ distribution.\\
(iv) Use the Central Limit Theorem to find an approximation for $\mathrm { P } ( W < 118.5 )$ for the case $k = 100$.
\hfill \mbox{\textit{OCR MEI S4 2011 Q2 [24]}}