Question

Find the mean and standard deviation of the sampling distribution. Show your work.

Short answer

The required mean and standard deviation are $-0.10$ and $0.097$ respectively.

Step-by-step solution

Given Information

Proportion of red crackers in Bag $1\left(p_1\right)=25\%$

$=0.25$.

Proportion of red crackers in Bag $2\left(p_2\right)=35\%$

$=0.35$.

Number of crackers in Bag $1\left(n_1\right)=50$.

Number of crackers in Bag $2\left(n_2\right)=40$.

Explanation

The formulas to compute the mean and standard deviation are:

$Mean\left({\mu }_{{\hat{p}}_1-{\hat{p}}_2}\right)=p_1-p_2$$\text{Standard deviation}\left({\sigma }_{{\hat{p}}_1-{\hat{p}}_2}\right)=\sqrt{\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_2\left(1-p_2\right)}{n_2}}$

The mean and standard deviation can be computed as:

$Mean\left({\mu }_{{\hat{p}}_1-{\hat{p}}_2}\right)=p_1-p_2$

$=0.25-0.35$

$=-0.10$

Now,

$\text{Standard deviation}\left({\sigma }_{{\hat{p}}_1-{\hat{p}}_2}\right)=\sqrt{\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_2\left(1-p_2\right)}{n_2}}$

$=\sqrt{\frac{0.25(1-0.25)}{50}+\frac{0.35(1-0.65)}{40}}$

$=0.097$