a. The $t_{\alpha/2}$ at 80% confidence level is equal to:
b. Find the best point estimate of the population mean.
Answer:
Given Data:
- Sample mean $ (\bar{x}) = 5.5 $
- Sample standard deviation $ (s) = 0.77 $
- Sample size $ (n) = 14 $
- Confidence interval level $ = 80% $
Solution:
a) The $ t_{\frac{\alpha}{2}} $ at the $ 80% $ confidence level:
The significance level is calculated as:
$ \alpha = 1 – 0.8 = 0.2 $
The degree of freedom is calculated as:
$ df = n – 1 = 14 – 1 = 13 $
The critical value is calculated as:
$ t_c = t_{\frac{\alpha}{2}, df} = t_{0.2/2, 13} \approx 1.35 , \text{[use Excel function]} $
The confidence interval is calculated as:
$ CI = \bar{x} \pm t_c \times \frac{s}{\sqrt{n}} $
$ = 5.5 \pm 1.35 \times \frac{0.77}{\sqrt{14}} $
$ = (5.22, 5.78) $
