n1 = 171, n2 = 162, x1-bar = 10.5, x2-bar = 12.4, s1 = 1.9, s2 = 2.4
Ho: μ1 = μ2
Ha: μ1 ≠ μ2
α = 0.10
Df = 171 + 161 - 2 = 330
Pooled SD, s = √[{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)]
= √(((171 - 1) * 1.9^2 + (161 - 1) * 2.4^2)/(171 + 161 -2))
= 2.157
SE = s * √{(1 /n1) + (1 /n2)}
= 2.157 * √((1/171) + (1/161))
= 0.2368
t = (x1-bar -x2-bar)/SE = -8.022
p- value = 0
Since the p- value < 0.10, there is a significant difference between the two groups on time spent towards campus life
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