Out of a sample of 45 people, 20 individuals are obese. Compute a 95% confidence interval for the proportion of individuals who are obese.

Answer:
Given:
Favorable Cases $(x)=20$

Sample Size $(n)=45$

Confidence Interval Level $=95\mathrm{\%}$

Solution:
The Sample Proportion $(\hat{p}):$
$\hat{p}=\frac{20}{45}=0.4444$

The Significance Level $(\alpha ):$
$\alpha =1–0.95=0.05$

The critical value $(z_c):$
$z_c=Z_{\alpha /2}=Z_{0.05/2}=1.96$

The confidence interval $(CI):$
$CI=\hat{p}\pm z_c\times \sqrt{\frac{\hat{p}(1–\hat{p})}{n}}=0.4444\pm 1.96\times \sqrt{\frac{0.4444(1–0.4444)}{45}}=(0.299,0.590)$

Final Answer:
The 95% confidence interval $=(0.299,0.590)$