Answer:
(19229.11 ,20770.89)
Step-by-step explanation:
We are given the following information:
Sample size, n = 17
Sample mean = 20,000 pounds
Sample standard deviation = 1,500 pounds
Confidence level = 95%
Significance level = 5% = 0.05
95% Confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 16 and}~\alpha_{0.05} = \pm 2.119[/tex]
[tex]20000 \pm 2.119(\displaystyle\frac{1500}{\sqrt{17}} ) = 20000 \pm 770.89 = (19229.11 ,20770.89)[/tex]