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Sagot :
Answer:
A sample size of 68 should be anticipated using.
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].
That is z with a pvalue of [tex]1 - 0.05 = 0.95[/tex], so Z = 1.645.
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.
Experience suggests that a reasonable estimate for the population standard deviation is 15
This means that [tex]\sigma = 15[/tex]
What minimum sample size should be anticipate using?
Margin of error at most 3, which means that the sample size is n when M = 3. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]3 = 1.645\frac{15}{\sqrt{n}}[/tex]
[tex]3\sqrt{n} = 1.645*15[/tex]
Simplifying both sides by 3:
[tex]\sqrt{n} = 1.645*5[/tex]
[tex](\sqrt{n})^2 = (1.645*5)^2[/tex]
[tex]n = 67.7[/tex]
Rounding up
A sample size of 68 should be anticipated using.
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