Answer:
Given Data:
- Sample mean $ (\bar{x}) = 100,000 $
- Sample standard deviation $ (s) = 42,000 $
- Sample size $ (n) = 49 $
- Confidence interval level $ = 90% $
Solution:
The significance level:
$ \alpha = 1 – 0.90 = 0.10 $
The degrees of freedom:
$ df = n – 1 = 49 – 1 = 48 $
The critical value:
$ t_c = t_{\frac{\alpha}{2}, df} = t_{0.05, 48} \approx 1.677$
The confidence interval:
$ CI = \bar{x} \pm t_c \times \frac{s}{\sqrt{n}} $
$ = 100,000 \pm 1.677 \times \frac{42,000}{\sqrt{49}} $
$ = (89,938, 110,062) $