Question

Suppose that .06 of each of two populations possess a given characteristic. Samples of size 400 are randomly drawn from each population. The probability that the difference between the first sample proportion which possess the given characteristic and the second sample proportion which possess the given characteristic being more than .03 is _______.

Answer 1

Answer:

The correct answer to the following question will be "0.0367".

Step-by-step explanation:

The given values are:

$p1=p2=0.06$

$q1=q2=1-p1=0.94$

$n1=n2=400$

As we know,

$E(p1-p2)=p1-p2=0$

$SE(p1-p2)=\sqrt{\frac{p1q1}{n1}+\frac{p2q2}{n2}}$

On putting the given values in the above expression, we get

                   $= \sqrt{p1q1(\frac{1}{400}+\frac{1}{400})}$

                   $=0.0168$

Now, consider

$P(p1-p2>0.03)=P[\frac{(p1-p2)-E(p1-p2)}{SE(p1-p2)}>\frac{0.03-0}{0.0168}]$

                            $=P(Z>1.7857)$

                            $=P(Z>1-79)$

                            $=0.036727$

Therefore, "0.0367" is the right answer.