Question

If p = .8 and n = 50, then we can conclude that the sampling distribution of pˆ p ^ is approximately a normal distribution.

Answer 1

Expected value of the sampling distribution of P^:

E(p^) = p=0.8

Standard Error(SE) of the Sample Proportion:

√ (p(1-p) / n)=√ (0.8(1-0.8) / 50)=0.0693

Explanation:

If you have a large enough sample size, you can use the normal distribution for the sampling distribution of p^.

How large is “large enough”? Use these formulas for a general guideline:

nPp^≥5

n(1-p^)≥5.

Here

50*0.8=40>5

50*(1-0.8)=10>5

Both are above 5, so we can use the normal distribution.

Expected value of the sampling distribution of P^:

E(p^) = p=0.8

Standard Error(SE) of the Sample Proportion:

√ (p(1-p) / n)=√ (0.8(1-0.8) / 50)=0.0693

Hope this will be helpful. Thank and God Bless You :)