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Sagot :
We have that the poll had 705 participants, and one third keeps a dog for protection. We can write this information with the following variables:
[tex]\begin{gathered} n=705 \\ \text{Number of people that keeps a dog for protection:} \\ \frac{705}{3}=235 \\ \Rightarrow p=\frac{235}{705} \end{gathered}[/tex]since we want the 95% of confidence, we have to use the following z score:
[tex]z_{95\%}=1.96[/tex]then, we can find the margin of error with the following equation:
[tex]m=z_{\alpha}\cdot\sqrt[]{\frac{p(1-p)}{n}}[/tex]in this case, we have the following:
[tex]m=1.96(\sqrt[]{\frac{(\frac{235}{705})(1-\frac{235}{705})}{705}})=1.96(0.018)=0.04=4\%[/tex]therefore, the margin of error is 4% and the confidence interval is:
[tex]\begin{gathered} (p-m,p+m) \\ \Rightarrow(\frac{235}{705}-0.04,\frac{235}{705}+0.04)=(0.293,\text{0}.373) \end{gathered}[/tex]therefore, the proportion of americans that keep a dog for protection is between 29.3% and 37.3%
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