Chapter 8.2, Problem 49UYK

(a)

To determine

To find: The formula for the mean and standard deviation of ${\widehat{p}}_1\text{and }{\widehat{p}}_2$.

Answer to Problem 49UYK

Solution: The formula for mean of ${\widehat{p}}_1$ is as follows:

${\mu }_{{\widehat{p}}_1}=p_1$

Similarly, the formula for mean of ${\widehat{p}}_2$ is as follows:

${\mu }_{{\widehat{p}}_2}=p_2$

The formula for standard deviation of ${\widehat{p}}_1$ is as follows:

${\sigma }_{{\widehat{p}}_1}=\sqrt{\frac{p_1\left(1-p_1\right)}{n_1}}$

Similarly, the formula for standard deviation of ${\widehat{p}}_2$ is as follows:

${\sigma }_{{\widehat{p}}_2}=\sqrt{\frac{p_2\left(1-p_2\right)}{n_2}}$

Explanation of Solution

Calculation: The formula for mean of population proportion $\left(\widehat{p}\right)$ is as follows:

${\mu }_{\widehat{p}}=p$

The formula for standard deviation of population proportion $\left(\widehat{p}\right)$ is as follows:

${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

So, the mean for ${\widehat{p}}_1$ and ${\widehat{p}}_2$ will be $p_1$ and $p_2$, and the standard deviation of ${\widehat{p}}_1$ and ${\widehat{p}}_2$ will be

${\sigma }_{{\widehat{p}}_1}=\sqrt{\frac{p_1\left(1-p_1\right)}{n_1}}$ and ${\sigma }_{{\widehat{p}}_2}=\sqrt{\frac{p_2\left(1-p_2\right)}{n_2}}$.

(b)

To determine

To find: The mean of difference $D={\widehat{p}}_1-{\widehat{p}}_2$.

Answer to Problem 49UYK

Solution: The mean of D is $\underset{\_}{p_1-p_2}$.

Explanation of Solution

Calculation: The mean of the sum of two random variables is the sum of the individual means of those random variables.

The formula for mean of the sum of two random variables X and Y is as follows:

$\left({\mu }_{X+Y}\right)={\mu }_X+{\mu }_Y$

where ${\mu }_X$ and ${\mu }_Y$ represent the individual mean of the random variables X and Y, respectively.

So, the mean of D is calculated as follows:

$\begin{array}{c}{\mu }_D={\mu }_{{\widehat{p}}_1-{\widehat{p}}_2}\\ ={\mu }_{{\widehat{p}}_1}+{\mu }_{-{\widehat{p}}_2}\\ ={\mu }_{{\widehat{p}}_1}-{\mu }_{{\widehat{p}}_2}\end{array}$

From part (a), the value of ${\mu }_{{\widehat{p}}_1}=p_1$ and ${\mu }_{{\widehat{p}}_2}=p_2$.

On substituting the values,

$\begin{array}{c}{\mu }_D={\mu }_{{\widehat{p}}_1}-{\mu }_{{\widehat{p}}_2}\\ =p_1-p_2\end{array}$

Hence, the required mean of the difference is $p_1-p_2$.

(c)

To determine

To find: The variance of D.

Answer to Problem 49UYK

Solution: The variance of D is $\underset{\_}{\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_2\left(1-p_2\right)}{n_2}}$.

Explanation of Solution

Calculation: The variance of the sum of two random variables will be equal to the sum of the individual variances and twice the covariance between the variables. The formula for variance of the sum of two random variables X and Y is as follows:

${\sigma }_{X-Y}^2={\sigma }_X^2+{\sigma }_Y^2-2\times Cov\left(X,Y\right)$

where ${\sigma }_X^2$ and ${\sigma }_Y^2$ represent the variance of random variables X and Y, respectively, and $Cov\left(X,Y\right)$ is the covariance of X and Y. So,

${\sigma }_{{\widehat{p}}_1-{\widehat{p}}_2}^2={\sigma }_{{\widehat{p}}_1}^2+{\sigma }_{{\widehat{p}}_2}^2-2\times Cov\left({\widehat{p}}_1,{\widehat{p}}_2\right)$

But the variables are independent, so $Cov\left({\widehat{p}}_1,{\widehat{p}}_2\right)=0$.

$\begin{array}{c}{\sigma }_{{\widehat{p}}_1-{\widehat{p}}_2}^2={\sigma }_{{\widehat{p}}_1}^2+{\sigma }_{{\widehat{p}}_2}^2\\ =\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_2\left(1-p_2\right)}{n_2}\end{array}$

Hence, the required variance of the difference is $\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_2\left(1-p_2\right)}{n_2}$.