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 1818 American students had a mean height of 68.968.9 inches with a standard deviation of 2.712.71 inches. A random sample of 1212 non-American students had a mean height of 65.765.7 inches with a standard deviation of 2.172.17 inches. Determine the 90%90% 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 critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 2 of 3: Find the standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer to two decimal places.

Step 3 of 3: Construct the 90%90% confidence interval. Round your answers to two decimal places.

Answer 1

Solution :

Step 1 of 3 : Critical value t = 1.701

Step 2 of 3 : Standard error Se = 0.94 (rounded)

Step 3 of 3 : The 90% confidence interval is (1.61,4.79) (rounded)

Explanation :-

Given that, for American students : X1-bar = 68.9 , s1 = 2.71 , n = 18

for Non-American students : X2-bar = 65.7 , s1 = 2.17 , n = 12