Respuesta :
Answer:
The minimum sample size is 158.
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{1-0.75}{2} = 0.125[/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]1-0.125 = 0.875[/tex], so [tex]z = 1.15[/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.
In this question:
The minimum sample size is n.
n is found when M = 30.
We have that [tex]\sigma = 327.8[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]30 = 1.15*\frac{327.8}{\sqrt{n}}[/tex]
[tex]30\sqrt{n} = 1.15*327.8[/tex]
[tex]\sqrt{n} = \frac{1.15*327.8}{30}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.15*327.8}{30})^{2}[/tex]
[tex]n = 157.9[/tex]
Rounding up
The minimum sample size is 158.
