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Construct a 95% confidence interval of the population proportion using the given information x=180, n = 300 the lower bound is the upper bound isRound to three decimal places as needed

Sagot :

Given;

[tex]x=180,n=300[/tex]

Then, we can find the point estimation as;

[tex]\begin{gathered} \hat{p}=\frac{x}{n} \\ \hat{p}=\frac{180}{360}=0.60 \end{gathered}[/tex][tex]Z_{\frac{\alpha}{2}}=Z_{0.05}=1.96[/tex]

Thus, the margin of error E is;

[tex]\begin{gathered} E=Z_{\frac{\alpha}{2}}\sqrt[]{\frac{\hat{p}(1-\hat{p})}{n}} \\ E=1.96\sqrt[]{\frac{0.60(0.40)}{300}} \\ E=1.96\sqrt[]{0.0008} \\ E=0.055 \end{gathered}[/tex]

A 95% confidence interval for population proportion p is;

[tex]\hat{p}\pm E=0.60\pm0.055[/tex]

The lower bound is;

[tex]0.60-0.055=0.545[/tex]

The upper bound is;

[tex]0.60+0.055=0.655[/tex]

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