Respuesta :
Answer:
Sample size of 586 or higher.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
What sample size could be 95% confident that the estimated (sample) proportion is within 4 percentage points of the true population proportion?
Sample size of at least n when [tex]M = 0.04[/tex]
42.1% of women-owned businesses provided retirement plans contributions, which means that [tex]p = 0.421[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.96\sqrt{\frac{0.421*0.579}{n}}[/tex]
[tex]0.04\sqrt{n} = 0.9677[/tex]
[tex]\sqrt{n} = \frac{0.9677}{0.04}[/tex]
[tex]\sqrt{n} = 24.19[/tex]
[tex]\sqrt{n}^{2} = (24.19)^(2)[/tex]
[tex]n = 585.2[/tex]
We need a sample size of 586 or higher.
