We have to find the 95% confidence interval for the proportion.
We first need the sample proportion, the standard error and the critical value of z.
The sample proportion is p=0.38.
[tex]p=X/n=133/350=0.38[/tex]The standard error of the proportion is:
[tex]\begin{gathered} \sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.38*0.62}{350}} \\ \sigma_p=\sqrt{0.000673}=0.0259 \end{gathered}[/tex]The critical z-value for a 95% confidence interval is z=1.96.
Now we can can calculate the margin of error (MOE) as:
[tex]MOE=z\cdot\sigma_p=1.96\cdot0.0259=0.0509[/tex]Then, the lower and upper bounds of the confidence interval are:
[tex]\begin{gathered} LL=p-z\cdot\sigma_p=0.38-0.0509=0.3291 \\ UL=p+z\cdot\sigma_p=0.38+0.0509=0.4309 \end{gathered}[/tex]The 95% confidence interval for the population proportion is (0.3291, 0.4309).
When expressed as percentage, we can conclude that the 95% confidence interval is (33%, 43%).
Answer: the 95% confidence interval is (33%, 43%).
