Chapter 8.6, Problem 68E

a.

To determine

Obtain the variance of $Y_1-Y_2$.

Answer to Problem 68E

The variance of $Y_1-Y_2$ is $n{p}_1\left(1-p_1\right)+n{p}_2\left(1-p_2\right)+2n{p}_1{p}_2$.

Explanation of Solution

It is given that $Y_1,Y_2,Y_3,Y_4$ have multinomial distribution with n trials and $p_1,p_2,p_3,p_4$ are the probabilities of the four cells. The pdf is given as follows:

$f\left(x_1,\dots ,x_4,p_1,\dots ,p_4\right)=\{\frac{n!}{x_1,!\dots x_4!}{p}_1^{x_1}\cdot \dots \cdot p_4^{x_4},n={\displaystyle \sum _{i=1}^4{n}_i}$

The mean and variance of multinomial distribution are given as follows:

$\begin{array}{l}E\left(Y_i\right)=n\cdot p_i\\ V\left(Y_i\right)=n\cdot p_i\left(1-p_i\right),i=1\text{to}\text{4}\end{array}$

Consider the following:

 $I_k^{\left(i\right)}=\{\begin{array}{l}1,k=i\\ 0,\text{else}\end{array}$.

The expectation of the products of Y_i and Y_j  (i≠j) is as follows:

$\begin{array}{c}E\left(Y_i{Y}_j\right)=E\left({\displaystyle \sum _{k=1}^n{I}_k^{\left(i\right)}}{\displaystyle \sum _{m=1}^n{I}_l^{\left(j\right)}}\right)\\ =E\left({\displaystyle \sum _{k=l}^n{I}_k^{\left(i\right)}{I}_k^{\left(j\right)}}+{\displaystyle \sum _{k\ne l}^n{I}_k^{\left(i\right)}{I}_l^{\left(j\right)}}\right)\\ =E\left({\displaystyle \sum _{k\ne l}^n{I}_k^{\left(i\right)}{I}_l^{\left(j\right)}}\right)\left[\text{since }{\displaystyle \sum _{k=l}^n{I}_k^{\left(i\right)}{I}_k^{\left(j\right)}}=0\right]\\ ={\displaystyle \sum _{k\ne l}^{n}E\left(I_k^{\left(i\right)}{I}_l^{\left(j\right)}\right)}\\ ={\displaystyle \sum _{k\ne l}^{n}E\left(I_k^{\left(i\right)}\right)E\left(I_l^{\left(j\right)}\right)}\\ ={\displaystyle \sum _{k\ne l}^n{p}_i{p}_j}\\ =n\left(n-1\right)p_i{p}_j\\ E\left(Y_1{Y}_2\right)=n\left(n-1\right)p_1{p}_2\end{array}$

The variance of $Y_1-Y_2$ is computed as follows:

$\begin{array}{c}V\left(Y_1-Y_2\right)=V\left(Y_1\right)+V\left(Y_2\right)-2E\left(Y_1{Y}_2\right)+2E{Y}_{1}E{Y}_2\\ =n{p}_1\left(1-p_1\right)+n{p}_2\left(1-p_2\right)-2n\left(n-1\right)p_1{p}_2+2n^2{p}_1{p}_2\\ =n{p}_1\left(1-p_1\right)+n{p}_2\left(1-p_2\right)-2n^2{p}_1{p}_2+2n{p}_1{p}_2+2n^2{p}_1{p}_2\\ =n{p}_1\left(1-p_1\right)+n{p}_2\left(1-p_2\right)+2n{p}_1{p}_2\end{array}$

Thus, the variance of $Y_1-Y_2$ is $n{p}_1\left(1-p_1\right)+n{p}_2\left(1-p_2\right)+2n{p}_1{p}_2$.

b.

To determine

Construct the 95% confidence interval for the difference between the population proportion favoring complete protection and the population proportion favoring destruction by wildlife officers.

Answer to Problem 68E

The 95% confidence interval for the difference between the population proportion favoring complete protection and the population proportion favoring destruction by wildlife officers is $\left(-0.14,-0.06\right)$.

Explanation of Solution

It is known that ${\widehat{p}}_1=0.06,{\widehat{p}}_2=0.16,\text{and }n=500$.

Since Y_i’s is dependent, the formula for confidence interval is given as follows:

$\left(\begin{array}{l}\left({\widehat{p}}_1-{\widehat{p}}_2\right)-z_{\frac{\alpha }{2}}\sqrt{\frac{1}{n}\left({\widehat{p}}_1\left(1-{\widehat{p}}_1\right)+{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)+2{\widehat{p}}_1{\widehat{p}}_2\right)},\\ \left({\widehat{p}}_1-{\widehat{p}}_2\right)+z_{\frac{\alpha }{2}}\sqrt{\frac{1}{n}\left({\widehat{p}}_1\left(1-{\widehat{p}}_1\right)+{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)+2{\widehat{p}}_1{\widehat{p}}_2\right)}\end{array}\right)$

From Table 4: Normal Curve Areas, the z value corresponding to 0.025$\left(=\frac{1-0.95}{2}\right)$) is 1.96.

The 95% confidence interval for the difference between the population proportion favoring complete protection and the population proportion favoring destruction by wildlife officers is computed as follows:

$\begin{array}{c}\left({\widehat{p}}_1-{\widehat{p}}_2\right)-z_{\frac{\alpha }{2}}\sqrt{\frac{1}{n}\left({\widehat{p}}_1\left(1-{\widehat{p}}_1\right)+{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)+2{\widehat{p}}_1{\widehat{p}}_2\right)}=\{\begin{array}{c}\left(0.06-0.16\right)-\\ \left(1.96\right)\sqrt{\frac{1}{500}\left(\left(0.06\left(1-0.06\right)\right)+\left(0.16\left(1-0.16\right)\right)+2\left(0.06\right)\left(0.16\right)\right)}\end{array}\\ \left({\widehat{p}}_1-{\widehat{p}}_2\right)+z_{\frac{\alpha }{2}}\sqrt{\frac{1}{n}\left({\widehat{p}}_1\left(1-{\widehat{p}}_1\right)+{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)+2{\widehat{p}}_1{\widehat{p}}_2\right)}=\{\begin{array}{c}\left(0.06-0.16\right)+\\ \left(1.96\right)\sqrt{\frac{1}{500}\left(\left(0.06\left(1-0.06\right)\right)+\left(0.16\left(1-0.16\right)\right)+2\left(0.06\right)\left(0.16\right)\right)}\end{array}\\ =\left(-0.14,-0.06\right)\end{array}$

Therefore, the 95% confidence interval for the difference between the population proportion favoring complete protection and the population proportion favoring destruction by wildlife officers is $\left(-0.14,-0.06\right)$.

Since the interval does not contain 0, it implies that the two probabilities are not the same.