Answer:
Given,
degree of freedom = n1 + n2 - 2 = 40 + 40 - 2 = 78
alpha = 0.1
t(alpa/2 , df) = 1.665
Sp^2 = ((n1-1)s1^2 + (n2-1)s2^2)/(n1+n2-2)
substitute values
= ((40-1)2.32^2 + (40-1)1.62^2)/(40+40-2)
= 4.0034
Sp = sqrt(4.0034) = 2
90% CI = (x1 - x2) +/- t*Sp*sqrt(1/n1 + 1/n2)
substitute values
= (5.69 - 4.25) +/- 1.665*2*sqrt(1/40 + 1/40)
= 1.44 +/- 0.7446
= (0.695 , 2.185)
Here 90% confidence interval lies b/w 0.695 to 2.185 hrs.
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