Answer:
Given:
Sample mean $(x\bar) = 7.1 $
Population standard deviation $ (σ)=0.5 $
Sample size $ (n)=50 $
Confidence interval level $=98% $
Solution:
The significance level $(\alpha): $
$ \therefore (\alpha) = 1-0.98 = 0.02 $
The critical value $(z_c): $
The critical value at 0.02 significance level = 2.33
The confidence interval:
$CI = \bar{x} \pm z_c \cdot \frac{\sigma}{\sqrt{n}} = 7.1 \pm 2.33 \cdot \frac{0.5}{\sqrt{50}} = (6.9369, 7.2631) \approx (6.94, 7.26)$
Final answer:
The 98% confidence interval $= (6.94,7.26)$
