Answer:
- 18.377
Explanation:
n1 = 17 ; xbar1 = 78.4 ; s1² = 90.3
n2 = 13 ; x2bar = 92 ; s2² = 107.8
Confidence interval :
(x1 - x2) ± Tcritical * sqrt[(sp²/n1 + sp²/n2)]
Pooled variance = Sp² = (df1*s1² + df2*s2²) ÷ (n1 + n2 - 2)
df = n - 1
Sp² = (16*90.3 + 12*107.8) / 28
Sp² = 97.8
Tcritical, 90%, df = n1 + n2 - 2 = 28 (1 - tail)
Tcritical = 1.313
(78.4-92)±1.313*sqrt[(sp²/n1 + sp²/n2)]
Margin of Error = Tcritical * sqrt[(sp²/n1 + sp²/n2)] Margin of Error = (1.313 * √(97.8/17 + 97.8/13)) = 4.7771473
Lower boundary :
(78.4-92) - 4.7771473
-13.6 - 4.7771473
= −18.3771473
= - 18.377