Respuesta :
Answer:
We need a sample of at least 2090 families if we want to be 95% confident that our estimate of p is within 0.02 of the true value.
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 interval [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 of the interval is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In this problem, we have that:
[tex]n = 500, x = 340, p = \frac{340}{500} = 0.68[/tex]
95% confidence interval
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].
How large a sample is required if we want to be 95% confident that our estimate of p is within 0.02 of the true value?
The margin of error decreases when the sample size increases. So we need a sample of at least n people when M = 0.02 for the estimate of p being within 0.02 of the value.
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.96\sqrt{\frac{0.68*0.32}{n}}[/tex]
[tex]0.02\sqrt{n} = 0.9143[/tex]
[tex]\sqrt{n} = \frac{0.9143}{0.02}[/tex]
[tex]\sqrt{n} = 45.71[/tex]
[tex]\sqrt{n}^{2} = (45.71)^{2}[/tex]
[tex]n = 2089.83[/tex]
We need a sample of at least 2090 families if we want to be 95% confident that our estimate of p is within 0.02 of the true value.
