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Answer:
A sample of [tex]n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2[/tex] is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.
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]
In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law.
This means that [tex]n = 1012, \pi = \frac{374}{1012} = 0.3696[/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].
How large a sample size is needed to be 95% confident with a margin of error of E?
A sample size of n is needed, and n is found when M = E.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]E = 1.96\sqrt{\frac{0.3696*0.6304}{n}}[/tex]
[tex]E\sqrt{n} = 1.96\sqrt{0.3696*0.6304}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.3696*0.6304}}{E}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2[/tex]
[tex]n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2[/tex]
A sample of [tex]n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2[/tex] is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.
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