Answer :
Answer:
At least 98 people need to be sampled. It is not a huge number, that is, it is not difficult to sample 98 people, so it is a reasonable sample size for a real world calculation.
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 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 question:
We need a sample size of at least n.
n is found when M = 3. We have that [tex]\sigma = 18[/tex].
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]3 = 1.645*\frac{18}{\sqrt{n}}[/tex]
[tex]3\sqrt{n} = 1.645*18[/tex]
Simplifying by 3
[tex]\sqrt{n} = 1.645*6[/tex]
[tex](\sqrt{n})^{2} = (1.645*6)^{2}[/tex]
[tex]n = 97.41[/tex]
Rounding up,
At least 98 people need to be sampled. It is not a huge number, that is, it is not difficult to sample 98 people, so it is a reasonable sample size for a real world calculation.