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Sagot :
Answer:
The minimum sample size which should be used is 315.
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:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
Previous data estimates the proportion of adults who are fully literate in Colombia to be 0.92.
This means that [tex]\pi = 0.92[/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].
What is the MINIMUM sample size which should be used to construct this interval if the margin of error for the interval is to be 0.03 or less?
The minimum sample size that should be used is n for which M = 0.03. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.92*0.08}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96\sqrt{0.92*0.08}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.92*0.08}}{0.03}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.92*0.08}}{0.03})^2[/tex]
[tex]n = 314.2[/tex]
Rounding up
The minimum sample size which should be used is 315.
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