Answer:
Given:
Sample mean $(\bar{x}) = 8.2$
Population standard deviation $(\sigma) = 0.7$
Sample size $(n) = 40$
Confidence interval level $= 95\%$
Solution:
The significance level $(\alpha):$
$\therefore (\alpha) = 1 – 0.95 = 0.05$
The critical value $(z_c):$
The critical value at 0.05 significance level $= 1.96$
The confidence interval:
$CI = \bar{x} \pm z_c \cdot \frac{\sigma}{\sqrt{n}} = 8.2 \pm 1.96 \cdot \frac{0.7}{\sqrt{40}} = (7.9844,8.4156) \approx (7.98, 8.42)$
Final answer:
The 95\% confidence interval $= (7.98, 8.42)$