Respuesta :
Answer:
A sample of 1015 should be taken.
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 of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of 0.12.
This means that [tex]\pi = 0.12[/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]. I used 95% because the confidence level was not given.
How large a sample should be taken to estimate the actual proportion of smokers with a margin of error of 0.02?
This is n for which M = 0.02. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.96\sqrt{\frac{0.12*0.88}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.96\sqrt{0.12*0.88}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.12*0.88}}{0.02}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.12*0.88}}{0.02})^2[/tex]
[tex]n = 1014.2[/tex]
Rounding up:
A sample of 1015 should be taken.
