Question

Find the probability that $\widehat{p_1}-\widehat{p_2}$is less than or equal to $0.02$. Show your work.

Short answer

The required probability is $0.2061$.

Step-by-step solution

Given Information

$Mean\left({\mu }_{{\hat{p}}_1-{\hat{p}}_2}\right)=-0.10$.

Standard deviation $\left({\sigma }_{{\hat{p}}_1-{\hat{p}}_2}\right)=0.097$.

Explanation

The probability that ${\hat{p}}_1-{\hat{p}}_2$is less than or equal to $-0.02$can be computed as:

localid="1650444373982" $$\begin{gathered} P\left({\hat{p}}_1-{\hat{p}}_2\ge -0.02\right)=P\left(\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\left({\sigma }_{{\hat{p}}_1}-{\hat{p}}_2\right)}\ge \frac{-0.02-\left(p_1-p_2\right)}{\left({\sigma }_{{\hat{p}}_1}-{\hat{p}}_2\right)}\right) \\ =P\left(Z\ge \frac{-0.02-(-0.1)}{0.097}\right) \\ =P(Z>0.82) \\ =0.2061 \end{gathered}$$