Chapter 10.1, Problem 2E

(a)

To determine

To Find: the shape of ${\widehat{p}}_G-{\widehat{p}}_D$ and why?

Explanation of Solution

Given:

  $n_G$ =60

  $n_D$ =75

  $p_G$ =0.80

  $p_D$ =0.40

Calculation:

Let that the sampling distribution ${\widehat{p}}_G-{\widehat{p}}_D$ is approximately normal

Condition

  $\begin{array}{l}{n}_G{p}_G\ge 10,\\ n_G(1-p_G)\ge 10,\\ n_D{p}_D\ge 10,\\ n_D(1-p_D)\ge 10\end{array}$

  $\begin{array}{c}{n}_G{p}_G=\left(60\right)\left(0.80\right)=48\\ n_G(1-p_G)=\left(60\right)\left(1-.80\right)=\left(60\right)\left(.20\right)=12\\ n_D{p}_D=\left(75\right)\left(0.40\right)=30\\ n_D(1-p_D)=\left(75\right)\left(1-0.40\right)=\left(75\right)\left(0.60\right)=45\end{array}$

The shape of ${\widehat{p}}_C-{\widehat{p}}_A$ is approximately normal because it is satisfying all condition four conditions which is $n_G{p}_G\ge 10,n_G(1-p_G)\ge 10,n_D{p}_D\ge 10,n_D(1-p_D)\ge 10$

(b)

To determine

To Calculate: the mean of sampling distribution

Answer to Problem 2E

  ${\mu }_{{\widehat{p}}_G-{\widehat{p}}_D}=0.40$

Explanation of Solution

Formula used:

  ${\mu }_{{\widehat{p}}_G-{\widehat{p}}_D}=p_G-p_D$

Calculation:

  $\begin{array}{c}{\mu }_{{\widehat{p}}_G-{\widehat{p}}_D}=p_G-p_D\\ =0.80-0.40\\ =0.40\end{array}$

(c)

To determine

To Calculate: the standard deviation of sampling distribution

Answer to Problem 2E

  ${\sigma }_{{\widehat{p}}_G-{\widehat{p}}_D}=0.07745$

Explanation of Solution

Formula used:

  ${\sigma }_{{\widehat{p}}_G-{\widehat{p}}_D}=\sqrt{\frac{p_G\left(1-p_G\right)}{n_G}+\frac{p_D\left(1-p_D\right)}{n_D}}$

Calculation:

High school graduates (60) <10% of all high school graduates

High school dropouts (75) <10% of all high school graduates

Standard deviation given as

  ${\sigma }_{{\widehat{p}}_G-{\widehat{p}}_D}=\sqrt{\frac{p_G\left(1-p_G\right)}{n_G}+\frac{p_D\left(1-p_D\right)}{n_D}}$

  $\begin{array}{c}=\sqrt{\frac{0.80\left(1-0.80\right)}{60}+\frac{0.40\left(1-0.40\right)}{75}}\\ =\sqrt{\frac{0.80\left(0.20\right)}{50}+\frac{0.40\left(0.60\right)}{75}}\\ \approx 0.07745\end{array}$

Standard deviation is 0.07745 and mean is 0.40 so proportions of graduates and dropouts vary about 0.07745 from 0.40