Answer:
(19228.737 ; 20771.263)
Explanation:
Given that:
Sample size, n = 17
Sample mean, m = 20,000
Sample standard deviation, s = 1500
α = 95%
The confidence interval,
Mean ± margin of error
Margin of Error : Tcritical * s/sqrt(n)
Tcritical at α = 0.05 ; df = n-1 = 17 - 1 = 16 ; 2 - tailed test = 2.120
Margin of Error = 2.120 * 1500/sqrt(17) = 771.263
Lower boundary :
20,000 - 771.263 = 19228.737
Upper boundary :
20,000 + 771.263 = 20771.263
Confidence interval :
(19228.737 ; 20771.263)