Answer:
5 ± 9.74
Explanation:
Confidence interval, two - sample, t procedure ;
μ1 - μ2 = (x1 - x2) ± Tcritical*S
S = sqrt[(s1²/n1) + (s2²/n2)]
n1 = 9 ; n2 = 9 ; s1 = 8 ; s2 = 6 ; x1 = 30 ; x2 = 25
S = sqrt[(8^2/9) + (6^2/9)]
S = sqrt[11.111111]
S = 3.333
Tcritical at 99% confidence interval
df = 18 - 2= 16
Tcritical = 2.92
(30 - 25) ± 2.92*(3.333)
5 ± 9.74