Chapter 10.1, Problem 1E

(a)

To determine

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

Explanation of Solution

Given:

  $n_C$ =50

  $n_A$ =100

  $p_C$ =0.30

  $p_A$ =0.15

Calculation:

Let that the sampling distribution ${\widehat{p}}_C-{\widehat{p}}_A$ is approximately normal

Condition

  $\begin{array}{l}{n}_C{p}_C\ge 10,\\ n_C(1-p_C)\ge 10,\\ n_A{p}_A\ge 10,\\ n_A(1-p_A)\ge 10\end{array}$

  $\begin{array}{c}{n}_C{p}_C=\left(50\right)\left(0.30\right)=15\\ n_C(1-p_C)=\left(50\right)\left(1-.30\right)=\left(50\right)\left(.70\right)=35\\ n_A{p}_A=\left(100\right)\left(0.15\right)=15\\ n_A(1-p_A)=\left(100\right)\left(1-0.15\right)=\left(100\right)\left(0.85\right)=85\end{array}$

The shape of ${\widehat{p}}_C-{\widehat{p}}_A$ is approximately normal because it is satisfying all condition four conditions which is $n_C{p}_C\ge 10,n_C(1-p_C)\ge 10,n_A{p}_A\ge 10,n_A(1-p_A)\ge 10$

(b)

To determine

To Calculate: the mean of sampling distribution and explain it

Answer to Problem 1E

  ${\mu }_{{\widehat{p}}_C-{\widehat{p}}_A}=0.15$

Explanation of Solution

Formula used:

  ${\mu }_{{\widehat{p}}_C-{\widehat{p}}_A}=p_C-p_A$

Calculation:

  $\begin{array}{c}{\mu }_{{\widehat{p}}_C-{\widehat{p}}_A}=p_C-p_A\\ =0.30-0.15\\ =0.15\end{array}$

(c)

To determine

To Calculate: the standard deviation of sampling distribution

Answer to Problem 1E

  ${\sigma }_{{\widehat{p}}_C-{\widehat{p}}_A}=0.07399$

Explanation of Solution

Formula used:

  ${\sigma }_{{\widehat{p}}_C-{\widehat{p}}_A}=\sqrt{\frac{p_C\left(1-p_C\right)}{n_C}+\frac{p_A\left(1-p_A\right)}{n_A}}$

Calculation:

Child jelly (50) <10% of all child jelly beans

Adult jelly (100) <10% of all adult jelly beans

Standard deviation given as

  ${\sigma }_{{\widehat{p}}_C-{\widehat{p}}_A}=\sqrt{\frac{p_C\left(1-p_C\right)}{n_C}+\frac{p_A\left(1-p_A\right)}{n_A}}$

  $\begin{array}{c}=\sqrt{\frac{0.30\left(1-0.30\right)}{50}+\frac{0.15\left(1-0.15\right)}{100}}\\ =\sqrt{\frac{0.30\left(0.70\right)}{50}+\frac{0.15\left(0.85\right)}{100}}\\ \approx 0.07399\end{array}$

Standard deviation is 0.07399 and mean is 0.155 so proportion of red jelly bean vary about 0.07399 from mean 0.15