Question

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 1

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

Explanation:

Given that:

sample size n = 774

Let P be the population proportion for having a retirement account = 0.43

Also

Let consider
`\hat p` be the sample proportion of having a retirement account.

However; as n is > 30 ; we can say:

`\mathbf{\mu_(\hat p) = 0.43}` ;

`\mathbf{\sigma_(\hat p^2) = (p(1-p))/(n)}`

⇒
`\mathbf{\sigma_(\hat p^2) = (0.43(1-0.43))/(774)}`

⇒
`\mathbf{\sigma_(\hat p^2) = (0.43(0.57))/(774)}`

So; we need P( the sample proportion will differ from 'p' by less than 3% i.e 0.03)

`=P(| \hat p- p|< 0.03)`

`=P(| \hat p- \mu _p|< 0.03)`

`= P ( |(\hat P - \mu_p)/(\sigma_(\hat p))|< \frac{0.03}{\sqrt{ (0.43*0.57)/(774) }})`

`= P(|Z|<1.6859)\ \ \ \ [Z=((\hat P - \mu_(\hat P))/(\sigma_(\hat P))) \sim N(0,1)]`

`= P(-1.6859 <Z<1.6859) \\ \\ = \Phi(1.6859)- \Phi (-1.6859) \\ \\ = \Phi (1.6859) - (1- \Phi(1.6859) \\ \\ = 2 \Phi (1.6859)-1`

From Normal Cumulative Distribution Function Table

`= 2*0.9541 -1`

= 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