Respuesta :
Answer:
b. false
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.
The sample size they should take is.
n, when [tex]\sigma = 8, M = 2[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]2 = 1.96*\frac{8}{\sqrt{n}}[/tex]
[tex]2\sqrt{n} = 1.96*8[/tex]
Dividing by 2
[tex]\sqrt{n} = 4*1.96[/tex]
[tex](\sqrt{n})^{2} = (4*1.96)^2[/tex]
[tex]n = 61.46[/tex]
We have to round up, so the needed sample size is 62, and the problem's statement is false.
