Answer: C
Step-by-step explanation:
Given:
Sample size (n) = 50
x = 12
[tex]\widehat{\mathbf{p}}=\frac{\mathbf{x}}{n}=\frac{12}{50}=0.24[/tex]
Confidence level = 90%
α = 1 − 0.90 = 0.10
α/2 = 0.05
[tex]\text { Critical value }\left(z_{c}\right)=z_{\frac{\alpha}{2}}=z_{0.05}=1.6449[/tex]
(from standard normal table)
90% Confidence interval is,
[tex]\begin{aligned}&\text { Confidence interval }=\widehat{\mathbf{p}} \pm z_{c} \times \sqrt{\frac{\hat{\mathbf{p}}(1-\hat{\mathbf{p}})}{n}} \\&\text { C. I }=0.24 \pm 1.6449 \times \sqrt{\frac{0.24(1-0.24)}{50}} \\&\text { C. I }=0.24 \pm 1.65 \times \sqrt{\frac{0.24(1-0.24)}{50}}\left\end{aligned}[/tex]
Therefore, 90% confidence interval for the true proportion of sophomores who favour the adoption of uniforms is C