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Sagot :
Answer:
A sample of 2017 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 0.30.
This means that [tex]\pi = 0.3[/tex]
Confidence level:
Not given, so I will use 95%.
This means that [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].
a. [2.5 pts] How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of 0.02?
A sample of n should be taken.
n is found fo M = 0.02. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.96\sqrt{\frac{0.3*0.7}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.96\sqrt{0.3*0.7}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.3*0.7}}{0.02}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.3*0.7}}{0.02})^2[/tex]
[tex]n = 2016.8[/tex]
Rounding up:
A sample of 2017 should be taken.
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