We have to find the 80% confidence interval for a population proportion.
The sample size is n = 362 and the number of successes is X = 54.
Then, the sample proportion is p = 0.149171.
[tex]p=\frac{X}{n}=\frac{54}{362}\approx0.149171[/tex]The standard error of the proportion is:
[tex]\begin{gathered} \sigma_s=\sqrt{\frac{p(1-p)}{n}} \\ \sigma_s=\sqrt{\frac{0.149171*0.850829}{362}} \\ \sigma_s=\sqrt{0.000351} \\ \sigma_s=0.018724 \end{gathered}[/tex]The critical z-value for a 80% confidence interval is z = 1.281552.
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=p-z\cdot\sigma_s=0.149171-1.281552\cdot0.018724\approx0.1492-0.0240=0.1252[/tex][tex]UL=p+z\cdot\sigma_s=0.1492+0.0240=0.1732[/tex]As the we need to express it as a trilinear inequality, we can write the 80% confidence interval for the population proportion (π) as:
[tex]0.125<\pi<0.173[/tex]Answer: 0.125 < π < 0.173
