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 18 American students had a mean height of 69.9inches with a standard deviation of 2.79inches. A random sample of 12 non-American students had a mean height of 63.8 inches with a standard deviation of 2.31 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.

Answer 1

sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1 + 1/n2))
sp = sqrt((((18 - 1)*2.79^2 + (12 - 1)*2.31^2)/(18 + 12 - 2))*(1/18 + 1/12))
sp = 0.9734

Given CI level is 0.98, hence α = 1 - 0.98 = 0.02                  
α/2 = 0.02/2 = 0.01, zc = z(α/2, df) = 2.47                  
                  
Margin of Error                  
ME = zc * sp                  
ME = 2.47 * 0.9734                  
ME = 2.404                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (69.9 - 63.8 - 2.47 * 0.9734 , 69.9 - 63.8 - 2.47 * 0.9734                  
CI = (3.70 , 8.50)

Point estimate = 69.9 - 63.8 = 6.1