We have to calculate a 90% confidence interval for the proportion that chose the answer "Spruce".
The sample proportion is p = 0.869:
[tex]p=\frac{X}{n}=\frac{272}{313}\approx0.869[/tex]The standard error of the proportion is:
[tex]\begin{gathered} \sigma_p=\sqrt{\frac{p(1-p)}{n}} \\ \\ \sigma_p=\sqrt{\frac{0.869\cdot0.131}{313}} \\ \\ \sigma_p\approx\sqrt{0.0003637} \\ \sigma_p\approx0.019 \end{gathered}[/tex]The critical z-value for a 90% confidence interval is z = 1.645.
The margin of error (MOE) can be calculated as:
[tex]MOE=z\cdot\sigma_p=1.645\cdot0.019\approx0.031[/tex]Then, the lower and upper bounds of the confidence interval are:
[tex]\begin{gathered} LL=p-z\sigma_p=0.869-0.031=0.838 \\ UL=p+z\sigma_p=0.869+0.031=0.900 \end{gathered}[/tex]Answer: The 90% confidence interval for the population proportion is (0.838, 0.900).