Question

Suppose 48H% of the population has a retirement account. If a random sample of size 632632 is selected, what is the probability that the proportion of persons with a retirement account will differ from the population proportion by greater than 3%3%

Answer 1

Answer: 0.1312

Explanation:

Given: The proportion of population has a retirement account : p = 0.48

Sample size : n = 632

Let q be th sample proportion.

The probability that the proportion of persons with a retirement account will differ from the population proportion by greater than 3% will be :-

`P(|q-p|>0.03)=1-P((|q-p|\leq 0.03)\\\\=1-P(-0.03<q-p<0.03)\\\\=1-P(\frac{-0.03}{\sqrt{((0.48)(1-0.48))/(632)}}<\frac{q-p}{\sqrt{(p(1-p))/(n)}}<\frac{0.03}{\sqrt{((0.48)(1-0.48))/(632)}})\\\\=1-P(-1.5096<z<1.5095) \ \ \ \ [\ Z=\frac{q-p}{\sqrt{(p(1-p))/(n)}}\ ]\\\\=1-(2P(Z<1.5095)-1)\ \ \ \ [P(-z<Z<z)=2(Z<z)-1]\\\\=2-2P(Z<1.5095)=2-2( 0.9344)\ \ \ [\text{by p-value table}]\\\\=0.1312`

Hence, the required probability = 0.1312