Question

I want red!A candy maker offers Child and Adult bags of jelly beans with

different color mixes. The company claims that the Child mix has $30\%$red jelly beans, while the Adult mix contains $15\%$red jelly beans. Assume that the candy maker’s claim is true. Suppose we take a random sample of $50$jelly beans from the Child mix and a separate random sample of $100$jelly beans from the Adult mix. Let ${\hat{p}}_C$and ${\hat{p}}_A$be the sample proportions of red jelly beans from the Child and

Adult mixes, respectively.

a. What is the shape of the sampling distribution of ${\hat{p}}_C-{\hat{p}}_A$? Why?

b. Find the mean of the sampling distribution.

c. Calculate and interpret the standard deviation of the sampling distribution.

Short answer

a. The shape in normal.

b. ${\mu }_{{\hat{p}}_C-{\hat{p}}_A}=0.15$

c.${\sigma }_{{\hat{p}}_C-{\hat{p}}_A}=0.07399$

Step-by-step solution

Given Information

It is given that $n_C=50$

$n_A=100$

$p_C=0.30$

$p_A=0.15$

Shape of p^C-p^A

Assuming ${\hat{p}}_C-{\hat{p}}_A$is normal.

Condition is:

$n_C{p}_C\ge 10$

$n_C\left(1-p_C\right)\ge 10$

$n_A{p}_A\ge 10$

$n_C{p}_C=\left(50\right)(0.30)=15$

$n_A{p}_A=\left(100\right)(0.15)=15$

$n_A\left(1-p_A\right)=\left(100\right)(1-0.15)=\left(100\right)(0.85)=85$

The shape is approximately normal as it satisfies all the four conditions.

Mean of Sampling Distribution

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

$=0.30-0.15=0.15$

Mean is$0.17$

Standard Deviation

$\text{Child jelly}\left(50\right)<10\%\text{of all child jelly beans}$and

$\text{Adult jelly}\left(100\right)<10\%\text{of all adult jelly beans}$

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

$=\sqrt{\frac{0.30(1-0.30)}{50}+\frac{0.15(1-0.15)}{100}}$

$=\sqrt{\frac{0.30(0.70)}{50}+\frac{0.15(0.85)}{100}}\approx 0.07399$