Question

Literacy A researcher reports that $80\%$of high school graduates, but only $40\%$of high school dropouts, would pass a basic literacy test. Assume that the researcher’s claim is true. Suppose we give a basic literacy test to a random sample of $60$high school graduates and a separate random sample of $75$high school dropouts.${\hat{p}}_G,{\hat{p}}_D$be the sample proportions of graduates and dropouts, respectively, who pass the test.

a. What is the shape of the sampling distribution of ${\hat{p}}_G-{\hat{p}}_D$. 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 is approximately normal.

b. ${\mu }_{{\hat{p}}_G-{\hat{p}}_D}=0.40$

c.${\sigma }_{{\hat{p}}_G-{\hat{p}}_D}=0.07745$

Step-by-step solution

Given Information

It is given that $n_G=60$

$n_D=75$

$p_G=0.80$

$p_D=0.40$

Shape of p^G-p^D

Assuming shape of ${\hat{p}}_G-{\hat{p}}_D$is normal.

Conditions are:

$n_G{p}_G\ge 10$

$n_G\left(1-p_G\right)\ge 10$

$n_{DPD}\ge 10$

$n_D\left(1-p_D\right)\ge 10$

$n_G{p}_G=\left(60\right)(0.80)=48$

$n_G\left(1-p_G\right)=\left(60\right)(1-.80)=\left(60\right)(.20)=12$

$n_D{p}_D=\left(75\right)(0.40)=30$

$n_D\left(1-p_D\right)=\left(75\right)(1-0.40)=\left(75\right)(0.60)=45$

As all four conditions are satisfied, the shape of${\hat{p}}_C-{\hat{p}}_A$is approximately normal.

Mean of Sampling Distribution

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

$=0.80-0.40=0.40$

The mean is$0.40$

Standard Deviation

High School graduates, $\left(60\right)<10\%\text{of all high school graduates}$

Dropouts, $\text{(75)}<10\%\text{of all high school graduates}$

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

$=\sqrt{\frac{0.80(1-0.80)}{60}+\frac{0.40(1-0.40)}{75}}$

$=\sqrt{\frac{0.80(0.20)}{50}+\frac{0.40(0.60)}{75}}\approx 0.07745$