Question

Q5  (please also show the steps):

CLT = Central Limit Theorem

Q5 Consider a problem of estimating the difference of proportions for two populations. In sample 1, out of n subjects, Si of them are "successes" and the rest are "failures". In sample 2, out of n2 subjects, S2 of them are "successes" and the rest are "failures". It is known that Si~ B(ni,P) and S2 ~ B(n2, p). We are interested in estimating P1 - P2. 1. Denote fi = 5 and 62 = 42. Show that 81 - Êz is an unbiased estimator for P1 – P2. And what is the mean squared error of it? 2. Explain how CLT would be applied to approximate the distribution of P1 - 2 so that P1 - D2N (P1-P2, ~ N(>1 Pi(1.- Pa) + 12 47 m)) approximately MacBoo

Answer 1

Solution,

(1)

Si ~ Bn1, p1)
and $S_{}\sim B(n_{},p_{})$

$\hat{p_{1}}=\frac{S_{1}}{n_{1}}$ and   $\hat{p_{2}}=\frac{S_{2}}{n_{2}}$

S1 S2 Epi - p2) = E - E n1 n2

$E(\hat{p_{1}}-\hat{p_{2}})=\frac{n_{1}p_{1}}{n_{1}}-\frac{n_{2}p_{2}}{n_{2}}=p_{1}-p_{2}$

Therefore, $\hat{p_{1}}-\hat{p_{2}}$ is an unbiased estimator of p₁ - p_2.

mean square error is given by-

$E[(\hat{p_{1}}-\hat{p_{2}})-(p_{1}-p_{2})]^{2}$

$MSE=E[(\hat{p_{1}}-p_{1})-(\hat{p_{2}}-p_{2})]^{2}$

$=E[(\hat{p_{1}}-p_{1})^{2}+(\hat{p_{2}}-p_{2})^{2}-2(\hat{p_{1}}-p_{1})(\hat{p_{2}}-p_{2})]$$=E[(\hat{p_{1}}-E(\hat{p_{1}})]^{2}+E[(\hat{p_{2}}-E(\hat{p_{2}})]^{2}+0$

$=V(\hat{p_{1}})+V(\hat{p_{2}})$

$=\frac{n_{1}p_{1}(1-p_{1})}{n_{1}^{2}}+\frac{n_{2}p_{2}(1-p_{2})}{n_{2}^{2}}\rightarrow (1)$

{ samples and independent and mean deviantion about mean is 0}

(2)

Central limit theorem states that if we have population with mean $\mu$ and standard deviation $\sigma ,$ if we take sufficiently large random samples the distribution of samples mean will be approximately distributed as $\mu$ ,$\sigma_{x} /\sqrt{n}$ and normally distribution

$(\hat{p_{1}}-\hat{p_{2}})=p_{1}-p_{2}$$Var(\hat{p_{1}}-\hat{p_{2}})=\frac{p_{1}(1-p_{1})}{n_{1}}+\frac{p_{2}(1-p_{2})}{n_{2}}$

Hence, proved that

$\hat{p_{1}}-\hat{p_{1}}\sim N\left ( p_{1} -p_{2} ,\frac{p_{1} (1-p_{1} )}{n_{1}}+\frac{p_{2} (1-p_{2} )}{n_{2}}\right )$