Respuesta :
Answer:
a) A sample of 198 must be drawn.
b) Smaller, because the confidence level is smaller.
Step-by-step explanation:
Question a:
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 - 0.999}{2} = 0.0005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.0005 = 0.9995[/tex], so Z = 3.291.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
How large a sample must be drawn so that a 99.9% confidence interval for u will have a margin of error equal to 4.1?
This is n for which M = 4.1. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]4.1 = 3.291\frac{17.5}{\sqrt{n}}[/tex]
[tex]4.1\sqrt{n} = 3.291*17.5[/tex]
[tex]\sqrt{n} = \frac{3.291*17.5}{4.1}[/tex]
[tex](\sqrt{n})^2 = (\frac{3.291*17.5}{4.1})^2[/tex]
[tex]n = 197.3[/tex]
Rounding up:
A sample of 198 must be drawn.
(b) If the required confidence level were 95%, would the necessary sample size be larger or smaller?
We have that:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
Solving for n
[tex]\sqrt{n} = \frac{z\sigma}{M}[/tex]
That is, n and z are directly proportion, meaning that a higher value of z(higher confidence level) leads to a higher sample size needed.
95% < 99.9%, so a smaller confidence interval.
Smaller, because the confidence level is smaller.
