Question

A student researcher compares the heights of American students and non-American students from the student body of a certain college in order to estimate the difference in their mean heights. A random sample of 12 American students had a mean height of 69.6 inches with a standard deviation of 2.96 inches. A random sample of 18 non-American students had a mean height of 64.2 inches with a standard deviation of 1.64 inches. Determine the 95% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students. Assume that the population variances are equal and that the two populations are normally distributed. Find the point estimate, margin of error, and confidence interval.

Answer #1

For Sample 1 :

x̅1 = 69.6, s1 = 2.96, n1 = 12

For Sample 2 :

x̅2 = 64.2, s2 = 1.64, n2 = 18

Point estimate = (x̅1 - x̅2) = (69.6 - 64.2) = 5.4

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((12-1)*2.96² + (18-1)*1.64²) / (12+18-2) = 5.0750

At α = 0.05 and df = n1+n2-2 = 28, two tailed critical value, t-crit = T.INV.2T(0.05, 28) = 2.048

Margin of error, E = t-crit*√(S²p*(1/n1 +1/n2)) = 2.048*√(5.075*(1/12 + 1/18)) = 1.7198

95% Confidence interval :

Lower Bound = (x̅1 - x̅2) - E = 5.4 - 1.7198 = 3.6802

Upper Bound = (x̅1 - x̅2) + E = 5.4 + 1.7198 = 7.1198

**3.6802 < µ1 - µ2 < 7.1198**

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