Respuesta :
Answer:
The minimum sample size required to create the specified confidence interval is 295.
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 Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
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.
Variance of 0.49:
This means that [tex]\sigma = \sqrt{0.49} = 0.7[/tex]
They want to construct a 95% confidence interval with an error of no more than 0.08. What is the minimum sample size required to create the specified confidence interval?
The minimum sample size is n for which M = 0.08. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.08 = 1.96\frac{0.7}{\sqrt{n}}[/tex]
[tex]0.08\sqrt{n} = 1.96*0.7[/tex]
[tex]\sqrt{n} = \frac{1.96*0.7}{0.08}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*0.7}{0.08})^2[/tex]
[tex]n = 294.1[/tex]
Rounding up:
The minimum sample size required to create the specified confidence interval is 295.
