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 $= 95\%$

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 $=(0.292, 0.541)$

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