Suppose 43% of the population has a retirement account. If a random sample of size 774 is selected, what is the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 3%?
Answer:
the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 3% is 0.9082
Step-by-step explanation:
Given that:
sample size n = 774
Let P be the population proportion for having a retirement account = 0.43
Also
Let consider [tex]\hat p[/tex] be the sample proportion of having a retirement account.
However; as n is > 30 ; we can say:
[tex]\mathbf{\mu_{\hat p} = 0.43}[/tex] ;
[tex]\mathbf{\sigma_{\hat p^2} = \dfrac{p(1-p)}{n}}[/tex]
⇒ [tex]\mathbf{\sigma_{\hat p^2} = \dfrac{0.43(1-0.43)}{774}}[/tex]
⇒ [tex]\mathbf{\sigma_{\hat p^2} = \dfrac{0.43(0.57)}{774}}[/tex]
So; we need P( the sample proportion will differ from 'p' by less than 3% i.e 0.03)
[tex]=P(| \hat p- p|< 0.03)[/tex]
[tex]=P(| \hat p- \mu _p|< 0.03)[/tex]
[tex]= P ( |\dfrac{\hat P - \mu_p}{\sigma_{\hat p}}|< \dfrac{0.03}{\sqrt{ \dfrac{0.43*0.57}{774} }})[/tex]
[tex]= P(|Z|<1.6859)\ \ \ \ [Z=(\dfrac{\hat P - \mu_{\hat P}}{\sigma_{\hat P}}) \sim N(0,1)][/tex]
[tex]= P(-1.6859 <Z<1.6859) \\ \\ = \Phi(1.6859)- \Phi (-1.6859) \\ \\ = \Phi (1.6859) - (1- \Phi(1.6859) \\ \\ = 2 \Phi (1.6859)-1[/tex]
From Normal Cumulative Distribution Function Table
[tex]= 2*0.9541 -1[/tex]
= 1.9082 - 1
= 0.9082
Thus; the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 3% is 0.9082
