Notes
Effects on Margin of Error
If confidence level increases, the margin of error increases.
If sample size increases, the margin of error decreases.
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 \(59\)
\(\,\,\,\,\,\displaystyle n=\left(\frac{z_c \sigma}{E}\right)^2\)
\(\,\,\,\,\,\displaystyle n=\left(\frac{1.96 \cdot 7.8}{2}\right)^2\)
\(\,\,\,\,\,\displaystyle n=58.4\)
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 \(1068\)
\(\,\,\,\,\,\displaystyle n=\frac{1}{4}\left(\frac{z_c}{E}\right)^2\)
\(\,\,\,\,\,\displaystyle n=\frac{1}{4}\left(\frac{1.96}{.03}\right)^2\)
\(\,\,\,\,\,\displaystyle n=1067.1\)
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 \(1540\)
\(\,\,\,\,\,\displaystyle n=\left(\hat{p}\right)\left(1-\hat{p}\right)\left(\frac{z_c}{E}\right)^2\)
\(\,\,\,\,\,\displaystyle n=\left(.35\right)\left(1-.35\right)\left(\frac{1.645}{.02}\right)^2\)
\(\,\,\,\,\,\displaystyle n=1539.1\)
Round up to \(1540\)