S4 June 2013 (R) Q8

EdexcelOld spec12 marksConfidence Intervals

8. A random sample \(W_1, W_2, \ldots, W_n\) is taken from a distribution with mean \(\mu\) and variance \(\sigma^2\)

(a) Write down \(\mathrm{E}\left(\displaystyle\sum_{i=1}^{n} W_i\right)\) and show that \(\mathrm{E}\left(\displaystyle\sum_{i=1}^{n} {W_i}^2\right) = n(\sigma^2 + \mu^2)\) (4)

An estimator for \(\mu\) is

\[\bar{X} = \frac{1}{n}\sum_{i=1}^{n} W_i\]

(b) Show that \(\bar{X}\) is a consistent estimator for \(\mu\). (3)

An estimator of \(\sigma^2\) is

\[U = \frac{1}{n}\sum_{i=1}^{n} {W_i}^2 - \left(\frac{1}{n}\sum_{i=1}^{n} W_i\right)^2\]

(d) Write down an unbiased estimator of \(\sigma^2\) in the form \(kU\), where \(k\) is in terms of \(n\). (1)

Mark scheme (a)

Scheme	Marks
\(\mathrm{E}\left(\displaystyle\sum_{i=1}^{n} W_i\right) = n\mu\)	B1
\(\mathrm{E}\left({W_i}^2\right) = \mathrm{Var}(W_i) + \left(\mathrm{E}(W_i)\right)^2\)	M1
\(= \sigma^2 + \mu^2\)	A1
\(\mathrm{E}\left(\displaystyle\sum_{i=1}^{n} {W_i}^2\right) = \mathrm{E}\left({W_1}^2 + {W_2}^2 + \ldots\ {W_n}^2\right)\)
\(= n\left(\sigma^2 + \mu^2\right)\)	A1 cso
	(4)

Notes

1st M1 using \(\mathrm{E}\left({W_i}^2\right) = \mathrm{Var}(W_i) + \left(\mathrm{E}(W_i)\right)^2\)

Mark scheme (b)

Scheme	Marks
\(\mathrm{E}\left(\dfrac{1}{n}\displaystyle\sum_{i=1}^{n} W_i\right) = \dfrac{1}{n}\mathrm{E}\left(\displaystyle\sum_{i=1}^{n} W_i\right)\)
\(= \mu\)	B1
\(\mathrm{Var}\left(\dfrac{1}{n}\displaystyle\sum_{i=1}^{n} W_i\right) = \dfrac{1}{n^2}\mathrm{Var}\left(W_1 + W_2 + \ldots + W_n\right)\)
\(= \dfrac{1}{n^2}n\sigma^2\)
\(= \dfrac{\sigma^2}{n},\ \to 0 \text{ as } n \to \infty\)	B1,B1d
	(3)

Notes

2nd B1 stating \(\mathrm{Var}\left(\dfrac{1}{n}\displaystyle\sum_{i=1}^{n} W_i\right) = \dfrac{\sigma^2}{n}\)

3rd B1 dependent on 2nd B1, stating \(\dfrac{\sigma^2}{n} \to 0\) as \(n \to \infty\)

Mark scheme (c)

Scheme	Marks
\(\mathrm{E}\left[\dfrac{1}{n}\left(\sum {w_i}^2\right) - \left(\bar{w}\right)^2\right] = \dfrac{1}{n} \times n\left(\sigma^2 + \mu^2\right) - \mathrm{E}(\bar{w}^2)\)	M1
\(\mathrm{Var}(\bar{w}) = \mathrm{E}(\bar{w}^2) - [\mathrm{E}(\bar{w})]^2 \Rightarrow \mathrm{E}(\bar{w}^2) - \mu^2 = \dfrac{\sigma^2}{n}\)	M1
Hence expected value is \(\left(\sigma^2 + \mu^2\right) - \dfrac{\sigma^2}{n} - \mu^2 = \dfrac{(n-1)\sigma^2}{n}\)	A1
Bias \(= (-)\dfrac{\sigma^2}{n}\)	A1
	(4)

Notes

1st M1 attempting correct method with their answer to part (a) – award for \(\left(\sigma^2 + \mu^2\right) - E\left(\dfrac{1}{n}\displaystyle\sum_{i=1}^{n} w_i\right)^2\)

2nd M1 using \(\mathrm{Var}(\bar{w}) = \mathrm{E}(\bar{w}^2) - [\mathrm{E}(\bar{w})]^2\)

Scheme	Marks
\(\dfrac{n}{(n-1)}U\)	B1
	(1)
	(12 marks)