Respuesta :
Answer:
The minimum sample size required to create the specified confidence interval is 1804.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
Minimum sample size for a margin of error of 0.06:
This sample size is n.
n is found when [tex]M = 0.06, \sigma = 1.3[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.06 = 1.96*\frac{1.3}{\sqrt{n}}[/tex]
[tex]0.06\sqrt{n} = 1.96*1.3[/tex]
[tex]\sqrt{n} = \frac{1.96*1.3}{0.06}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*1.3}{0.06})^{2}[/tex]
[tex]n = 1803.41[/tex]
Rounding up to the next integer
The minimum sample size required to create the specified confidence interval is 1804.
