Respuesta :
Answer:
B. 97
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].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find 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.
How many students should be sampled to be within 1 drink of the population mean with 95% probability?
This is n when [tex]M = 1[/tex]
We know from preliminary studies that the standard deviation is around 5, so [tex]\sigma = 5[/tex]
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]1 = 1.96*\frac{5}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = 5*1.96[/tex]
[tex](\sqrt{n})^{2} = (5*1.96)^{2}[/tex]
[tex]n = 96.04[/tex]
We round up, so the correct answer is:
B. 97
The correct option is an option (B).
Margin Error:
The margin of error formula is.
[tex]ME=Z_{\frac{\alpha}{2} } \times \frac{\sigma}{\sqrt{n} }[/tex]
Given that,
The margin error is 1
Then by the above formula, we get,
[tex]ME=Z_{\frac{\alpha}{2} } \times \frac{\sigma}{\sqrt{n} }\\1=Z_{0.0250}\times \frac{5}{\sqrt{n} } \\{\sqrt{n} }=1.96\times 5\\n=(9.8)^2\\n=96.04\\n\approx97[/tex]
Learn more about Margin Error:
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