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6.Suppose the Gallup Organization wants to estimate the population proportion of those who think there should be a law that would ban the possession of handguns. In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law. How large a sample size is needed to be 95% confident with a margin of error of E

Sagot :

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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