B. Develop a 90%,95%,98%,99% confidence interval estimate of the true mean difference between the two lines.
solution:-
a. point estimate = (x1-x2) = (25.08-25.26) = -0.18
B. degrees of freedom df = (n1+n2)-2 = (18+23)-2 = 39
90% confidence
the value of 90% confidence with df from t table is t = 1.685
confidence interval formula
=> point estimate +/- t * sqrt(s1^2/n1 + s2^2/n2)
=> -0.18 +/- 1.685 * sqrt((0.05^2/18) + (0.08^2/23))
=> (-0.2144 , -0.1456)
95% confidence
the value of 95% confidence with df from t table is t = 2.022
confidence interval formula
=> point estimate +/- t * sqrt(s1^2/n1 + s2^2/n2)
=> -0.18 +/- 2.022 * sqrt((0.05^2/18) + (0.08^2/23))
=>(-0.2213 , -0.1387)
98% confidence
the value of 98% confidence with df from t table is t = 2.426
confidence interval formula
=> point estimate +/- t * sqrt(s1^2/n1 + s2^2/n2)
=> -0.18 +/- 2.426 * sqrt((0.05^2/18) + (0.08^2/23))
=> (-0.2295 , -0.1305)
99% confidence
the value of 99% confidence with df from t table is t = 2.708
confidence interval formula
=> point estimate +/- t * sqrt(s1^2/n1 + s2^2/n2)
=> -0.18 +/- 2.708 * sqrt((0.05^2/18) + (0.08^2/23))
=> (-0.2353, -0.1247) (rounded to four decimal places)
B. Develop a 90%,95%,98%,99% confidence interval estimate of the true mean difference between the two lines. A pet fo...
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