Answer:
(-0.607 ; 1.757)
Step-by-step explanation:
Student: __ 1 ___ 2 ___ 3 ___ 4
Fall _____2.6__ 3.0__ 1.5 __1.4
Spring___ 1.7__ 2.2 __1.4 __0.90
Using calculator :
Fall:
Xbar1 = 2.125
Sample size, n1 = 4
Standard deviation, s1 = 0.8
Spring :
Xbar2 = 1.55
Sample size, n2 = 4
Standard deviation, s2 = 0.54
Confidence interval :
(xbar1 - xbar2) ± Tcritical * Sp* √1/n1 + 1/n2
Sp = √[(df1*s1² + df2*s2²) ÷ (n1 + n2 - 2)]
df = n - 1
(xbar1 - xbar2) = 2.125 -1.55 = 0.575
Hence,
Sp = √(((3*0.8^2) + (3*0.54^2)) / 6)
Sp= √2.7948 ÷ 6
Sp = √0.4658
Sp= 0.682
Tcritical at 95%, df = 4 = 2.45
Error margin = 2.45 * 0.682 * √1/4 + 1/4 = 1.182
Confidence Interval:
Lower boundary = 0.575 -1.182 = -0.607
Upper boundary = 0.575 + 1.182 = 1.757
(-0.607 ; 1.757)
