Out of a sample of 60 people, 25 individuals are smokers. Compute a 95% confidence interval for the proportion of individuals who are smokers.

Answer:
Given:
Favorable Cases $(x)=25$

Sample Size $(n)=60$

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

Solution:
The Sample Proportion $(\hat{p}):$
$\hat{p}=\frac{25}{60}=0.4167$

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 $(CI):$
$$CI=\hat{p}\pm z_c\times \sqrt{\frac{\hat{p}(1–\hat{p})}{n}}$$ $$=0.4167\pm 1.96\times \sqrt{\frac{0.4167(1–0.4167)}{60}}$$ $$=(0.292,0.541)$$

Final Answer:
The 95% confidence interval $=(0.292,0.541)$