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 70.7 inches with a standard deviation of 2.41 inches. A random sample of 17 non-American students had a mean height of 62.7 inches with a standard deviation of 3.07 inches. Determine the 98% 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.

Step 1 of 3 : Find the point estimate that should be used in constructing the confidence interval.

Step 2 of 3: Find the margin of error to be used in constructing the confidence interval. Round to six decimal places

Step 3 of 3: Construct the 98% confidence interval. Round to two decimal places

Answer 1

Pooled Variance
sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1 + 1/n2))
sp = sqrt((((12 - 1)*2.41^2 + (17 - 1)*3.07^2)/(12 + 17 - 2))*(1/12 + 1/17))
sp = 1.0632

point estimate = 70.7 - 62.7 = 8

Given CI level is 0.98, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.473

Margin of Error
ME = tc * sp
ME = 2.473 * 1.0632
ME = 2.629294

CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)
CI = (70.7 - 62.7 - 2.473 * 1.0632 , 70.7 - 62.7 - 2.473 * 1.0632
CI = (5.37 , 10.63)