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Sagot :
Answer:
The minimum sample size needed is 39.
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 Ztable as such z has a pvalue 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.
Assuming that the standard deviation of the number of hours is 120 hour
This means that [tex]\sigma = 120[/tex]
The minimum sample size needed in order to construct a 99% confidence interval for the mean number of hours of flight time of all such pilots, to within 50 hours, is about?
The minimum sample size needed is n.
n is found when [tex]M = 50[/tex]
So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]50 = 2.575\frac{120}{\sqrt{n}}[/tex]
[tex]50\sqrt{n} = 2.575*120[/tex]
[tex]\sqrt{n} = \frac{2.575*120}{50}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.575*120}{50})^2[/tex]
[tex]n = 38.19[/tex]
Rounding up
The minimum sample size needed is 39.
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