Answer:
The smallest sample size required to obtain the desired margin of error is 44.
Step-by-step explanation:
I think there was a small typing error, we have that [tex]\sigma = 60[/tex] is the standard deviation of these weighs.
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]
Which of these is the smallest approximate sample size required to obtain the desired margin of error?
This sample size is n.
n is found when [tex]M = 15[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]15 = 1.645*\frac{60}{\sqrt{n}}[/tex]
[tex]15\sqrt{n} = 1.645*60[/tex]
Simplifying by 15
[tex]\sqrt{n} = 1.645*4[/tex]
[tex](\sqrt{n})^{2} = (1.645*4)^{2}[/tex]
[tex]n = 43.3[/tex]
Rounding up
The smallest sample size required to obtain the desired margin of error is 44.
