Answer:
Given:
Sample Mean $ (x\bar)=13.8 $
Population Standard Deviation $ (\sigma)=2 $
Sample Size $ (n) = 17 $
Confidence Interval Level $ (CI) = 95\% $
Solution:
The level of significance $ (\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): $
$\text{CI} = \bar{x} \pm Z_c \cdot \frac{\sigma}{\sqrt{n}} $
$ = 13.8 \pm 1.96 \cdot \frac{2}{\sqrt{17}} $
$ = (12.840, 14.760) $
Final Answer:
The 95% confidence interval $ = (12.840, 14.760) $