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Sagot :
Given:
Confidence level = 98%
Margin of error, E = 2% = 0.02
Let's find the number of students the instructor should sample.
Here, we are to find the sample size (n).
We have:
[tex]P^{^{\prime}}=1-P^{\prime}=0.5[/tex]At 98% confidence level, we have:
[tex]\alpha=1-\text{ 98\%= 2\% = 0.02}[/tex][tex]\frac{\alpha}{2}=\frac{0.02}{2}=0.01[/tex]Using the z-score table, we have:
[tex]Z_{\frac{\alpha}{2}}=Z_{0.01}=2.33[/tex]To find the sample size(number of students) n, we have:
[tex]n=(\frac{Z_{\frac{\alpha}{2}}}{E})^2\ast P^{\prime}\ast(1-P^{\prime})[/tex]Thus, we have:
[tex]\begin{gathered} n=(\frac{2.33}{0.02})^2\ast0.5\ast0.5 \\ \\ n=(116.5)^2\ast0.5\ast0.5 \\ \\ n=13572.25\ast0.5\ast0.5 \\ \\ n=3393.06\approx3393 \end{gathered}[/tex]Therefore, the statistics instructor should sample 3393 students.
ANSWER:
3393
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