Answer:
b. 98
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{1-0.9}{2} = 0.05[/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]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]
Now, find 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, we have that:
[tex]M = 0.5, \sigma = 3[/tex]
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.5 = 1.645*\frac{3}{\sqrt{n}}[/tex]
[tex]0.5\sqrt{n} = 4.935[/tex]
[tex]\sqrt{n} = 9.87[/tex]
[tex]n = 97.42[/tex]
So a sample of at least 98 is required.
The correct answer is:
b. 98
