Respuesta :
Answer:
a) Sample size of 96 or higher
b) Sample size of 196 or higher
c) Sample size of 385 or higher
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.
In this problem, [tex]\sigma = 0.25[/tex]
a.The desired margin of error is $.10.
Sample size of n or higher when [tex]M = 0.1[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.1 = 1.96*\frac{0.5}{\sqrt{n}}[/tex]
[tex]0.1\sqrt{n} = 1.96*0.5[/tex]
[tex]\sqrt{n} = \frac{1.96*0.5}{0.1}[/tex]
[tex]\sqrt{n} = 9.8[/tex]
[tex]\sqrt{n}^{2} = (9.8)^{2}[/tex]
[tex]n = 96[/tex]
b.The desired margin of error is $.07.
Sample size of n or higher when [tex]M = 0.07[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.07 = 1.96*\frac{0.5}{\sqrt{n}}[/tex]
[tex]0.07\sqrt{n} = 1.96*0.5[/tex]
[tex]\sqrt{n} = \frac{1.96*0.5}{0.07}[/tex]
[tex]\sqrt{n} = 14[/tex]
[tex]\sqrt{n}^{2} = (14)^{2}[/tex]
[tex]n = 196[/tex]
c.The desired margin of error is $.05.
Sample size of n or higher when [tex]M = 0.05[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.05 = 1.96*\frac{0.5}{\sqrt{n}}[/tex]
[tex]0.05\sqrt{n} = 1.96*0.5[/tex]
[tex]\sqrt{n} = \frac{1.96*0.5}{0.05}[/tex]
[tex]\sqrt{n} = 19.6[/tex]
[tex]\sqrt{n}^{2} = (19.6)^{2}[/tex]
[tex]n = 384.1[/tex]
Rounding up, sample size of 385 or higher
