Respuesta :
Answer:
a) 25
b) 67
c) 97
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 $0.10.
This is n when M = 0.1. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.1 = 1.96*\frac{0.25}{\sqrt{n}}[/tex]
[tex]0.1\sqrt{n} = 1.96*0.25[/tex]
[tex]\sqrt{n} = \frac{19.6*0.25}{0.1}[/tex]
[tex](\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}[/tex]
[tex]n = 24.01[/tex]
Rounding up to the nearest whole number, 25.
(b) The desired margin of error is $0.06.
This is n when M = 0.06. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.06 = 1.96*\frac{0.25}{\sqrt{n}}[/tex]
[tex]0.06\sqrt{n} = 1.96*0.25[/tex]
[tex]\sqrt{n} = \frac{19.6*0.25}{0.06}[/tex]
[tex](\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}[/tex]
[tex]n = 66.7[/tex]
Rounding up, 67
(c) The desired margin of error is $0.05.
This is n when M = 0.05. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.05 = 1.96*\frac{0.25}{\sqrt{n}}[/tex]
[tex]0.05\sqrt{n} = 1.96*0.25[/tex]
[tex]\sqrt{n} = \frac{19.6*0.25}{0.05}[/tex]
[tex](\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}[/tex]
[tex]n = 96.04[/tex]
Rounding up, 97
