Notes
Practice Problems
\(\textbf{1)}\) Find \(n\) if \( \sigma =7.8\) and error is \(2\) with \(95\%\) confidence interval.
\(n = 59 \)
\(\,\,\,\,\,\displaystyle n=\left(\frac{z_c \sigma}{E}\right)^2\)
\(\,\,\,\,\,\displaystyle n=\left(\frac{1.96 \cdot 7.8}{2}\right)^2\)
\(\,\,\,\,\,\displaystyle n=58.4\)
Round up to \(59\)
\(\textbf{2)}\) How large should the sample be if you want to do a \(90\%\) confidence level and error is .03 from p and there is no preliminary study.
The sample size should be \(1068\)
\(\,\,\,\,\,\displaystyle n=\frac{1}{4}\left(\frac{z_c}{E}\right)^2\)
\(\,\,\,\,\,\displaystyle n=\frac{1}{4}\left(\frac{1.96}{.03}\right)^2\)
\(\,\,\,\,\,\displaystyle n=1067.1\)
Round up to \(1068\)
\(\textbf{3)}\) How large should the sample be if you want to do an \(90\%\) confidence level and width of interval was \(.04\) and prelim study found that \(\hat{p}=.35\)
The sample size should be \(1540\)
\(\,\,\,\,\,\displaystyle n=\left(\hat{p}\right)\left(1-\hat{p}\right)\left(\frac{z_c}{E}\right)^2\)
\(\,\,\,\,\,\displaystyle n=\left(.35\right)\left(1-.35\right)\left(\frac{1.645}{.02}\right)^2\)
\(\,\,\,\,\,\displaystyle n=1539.1\)
Round up to \(1540\)
\(\textbf{4)}\) Find \(n\) if \( \sigma =3\) and error is \(.2\) with \(99\%\) confidence interval.
\(n = 1492 \)
\(\,\,\,\,\,\displaystyle n=\left(\frac{z_c \sigma}{E}\right)^2\)
\(\,\,\,\,\,\displaystyle n=\left(\frac{2.575 \cdot 3}{.2}\right)^2\)
\(\,\,\,\,\,\displaystyle n=1491.9\)
Round up to \(1492\)
\(\textbf{5)}\) How large should the sample be if you want to do a \(80\%\) confidence level and error is .02 from p and there is no preliminary study.
The sample size should be \(1024\)
\(\,\,\,\,\,\displaystyle n=\frac{1}{4}\left(\frac{z_c}{E}\right)^2\)
\(\,\,\,\,\,\displaystyle n=\frac{1}{4}\left(\frac{1.28}{.02}\right)^2\)
\(\,\,\,\,\,\displaystyle n=1024\)
\(\textbf{6)}\) How large should the sample be if you want to do an \(95\%\) confidence level and width of interval was \(.16\) and prelim study found that \(\hat{p}=.4\)
The sample size should be \(145\)
\(\,\,\,\,\,\displaystyle n=\left(\hat{p}\right)\left(1-\hat{p}\right)\left(\frac{z_c}{E}\right)^2\)
\(\,\,\,\,\,\displaystyle n=\left(.4\right)\left(1-.4\right)\left(\frac{1.96}{.08}\right)^2\)
\(\,\,\,\,\,\displaystyle n=144.06\)
Round up to \(145\)
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