Respuesta :
Answer:
A sample size of 35 is needed.
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 width W as such
[tex]W = 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.
How large must the sample size be if the width of the 95% interval for mu is to be 1.0:
We need to find n for which W = 1.
We have that [tex]\sigma^{2} = 9[/tex], then [tex]\sigma = \sqrt{\sigma^{2}} = \sqrt{9} = 3[/tex]. So
[tex]W = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]1 = 1.96*\frac{3}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = 1.96*3[/tex]
[tex](\sqrt{n})^2 = (1.96*3)^{2}[/tex]
[tex]n = 34.57[/tex]
Rounding up
A sample size of 35 is needed.
