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Sagot :
Answer:
A sample size of 21 is needed.
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 - 0.95}{2} = 0.025[/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 - 0.025 = 0.975[/tex], so Z = 1.96.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
Using an estimated standard deviation of $11,605
This means that [tex]\sigma = 11605[/tex]
What sample size do you need to have a margin of error equal to $5000 with 95% confidence
A sample size of n is needed. n is found when M = 5000. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]5000 = 1.96\frac{11605}{\sqrt{n}}[/tex]
[tex]5000\sqrt{n} = 1.96*11605[/tex]
[tex]\sqrt{n} = \frac{1.96*11605}{5000}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*11605}{5000})^2[/tex]
[tex]n = 20.69[/tex]
Rounding up,
A sample size of 21 is needed.
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