Question

Suppose 56V% of politicians are lawyers. If a random sample of size 787787 is selected, what is the probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by greater than 4%4%

Answer 1

Answer: 0.0238

Explanation:

Given : Proportion of politicians are lawyers: p= 0.56

Sample size : n= 787

Let
`\hat{p}` be th sample proportion.

The the probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by greater than 4% will be :-

`P(|\hat{p}-0.56|>0.04)=P(-0.04>\hat{p}-0.56>0.04)=\\\\ P(-0.04+0.56>\hat{p}>0.04+0.56)\\\\=P(0.52>\hat{p}>0.60)\\\\=P(\frac{0.52-0.56}{\sqrt{(0.56(1-0.56))/(787)}}>\frac{\hat{p}-p}{\sqrt{(p(1-p))/(n)}}>\frac{0.60-0.56}{\sqrt{(0.56(1-0.56))/(787)}})\\\\=P(-2.26>z>2.26)\ \ \ [Z=\frac{\hat{p}-p}{\sqrt{(p(1-p))/(n)}}]\\\\=1-P(-2.26<z<2.26)\\\\=1-(2P(Z > 2.26)-1)\ \ \ [P(-z<Z<z)=2P(Z > |z|)-1]\\\ =2-2P(Z > 2.26)\\\\=2-2(0.9881)=0.0238`

Hence, the required probability = 0.0238