Question

7. A recent Gallop poll showed that 672 of 1019 adults nationwide think that marijuana shouldbe legalized.a) Find the margin of error that corresponds to a 95% confidence interval.c) Find the 95% confidence interval estimate of the proportion of adults nationwide whothink marijuana should be legalized.

Answer 1

Given: n = 1019

x = 672

`\hat{p}=(672)/(1019)=0.66`

z-score for 95% confidence interval, Z = 1.96

a) Margin of Error, MoE

`\begin{gathered} MoE=Z\sqrt[]{\frac{\hat{p}(1-\hat{p})}{n}}=1.96\cdot\sqrt[]{(0.66(1-0.66))/(1019)}=.0291 \\ \\ \end{gathered}`

b) 95% confidence interval estimate of the proportion, E

`\begin{gathered} E=\hat{p}\pm Z\sqrt[]{\frac{\hat{p}(1-\hat{p})}{n}}=0.66\pm1.96\cdot\sqrt[]{(0.66(1-0.66))/(1019)}=0.66\pm0.0291 \\ 0.66+0.0291=0.6891 \\ 0.66-0.0291=0.6309 \\ (0.6891,0.6309) \end{gathered}`

Answer:

a) MoE = 0.0291 = 2.91%

b) The 95% confidence interval is (0.6891, 0.6309)