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Sagot :
Answer:
The minimum sample size is 239.
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 - 0.99}{2} = 0.005[/tex]
Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.005 = 0.995[/tex], so Z = 2.575.
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.
Population standard deviation is equal to 1.5
This means that [tex]\sigma = 1.5[/tex]
Margin of error of 0.25
This means that [tex]M = 0.25[/tex]
What's the minimum size of the sample?
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.25 = 2.575\frac{1.5}{\sqrt{n}}[/tex]
[tex]0.25\sqrt{n} = 2.575*1.5[/tex]
[tex]\sqrt{n} = \frac{2.575*1.5}{0.25}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.575*1.5}{0.25})^2[/tex]
[tex]n = 238.7[/tex]
Rounding up:
The minimum sample size is 239.
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