Answer:
(0.7044 ; 3.0956)
Step-by-step explanation:
Given:
GROUP A:
n1 = 25
x1 = 12.7
s1 = 2.2
GROUP B :
n2 = 25
x2 = 10.8
s2= 2.0
The obtain the confidence interval assuming unequal population variance :
(x1 - x2) ± tα/2[√(s1²/n1 + s2²/n2)]
The degree of freedom :
df = (s1²/n1 + s2²/n2)² ÷ (s1²/n1)²/n1-1 + (s2²/n2)²/n2-1
The degree of freedom :
(2.2²/25 + 2²/25)² ÷ (2.2²/25)²/24 + (2²/25)²/24
df = 0.12503296 ÷ (0.0015617 + 0.0010666)
df = 47.57 ;
df = 48
Tcritical value ; α = 95% ; df = 48
Tcritical = 2.0106
C.I = (12.7 - 10.8) ± 2.0106[√(2.2²/25 + 2²/25)]
C.I = 1.9 ± (2.0106 * 0.5946427)
C.I = 1.9 ± 1.1955887
C. I = (0.7044 ; 3.0956)