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 70.5 inches with a standard deviation of 2.26 inches. A random sample of 12 non-American students had a mean height of 65.8 inches with a standard deviation of 2.44 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 2 of 3 :

Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places

Step 3: Calculate the 98% confidence interval. Round to three decimal places.

Answer #1

Step 2

Margin of Error = t(α/2 , n1+n2-2) SP √( (1/n1) + (1/n2)) =
2.144401

Step 3

Confidence interval is :-

( X̅1 - X̅2 ) ± t( α/2 , n1+n2-2) SP √( (1/n1) + (1/n2))

t(α/2, n1 + n1 - 2) = t( 0.02/2, 18 + 12 - 2) = 2.467

( 70.5 - 65.8 ) ± t(0.02/2 , 18 + 12 -2) 2.3324 √ ( (1/18) +
(1/12))

Lower Limit = ( 70.5 - 65.8 ) - t(0.02/2 , 18 + 12 -2) 2.3324 √(
(1/18) + (1/12))

Lower Limit = 2.556

Upper Limit = ( 70.5 - 65.8 ) + t(0.02/2 , 18 + 12 -2) 2.3324 √(
(1/18) + (1/12))

Upper Limit = 6.844

**98% Confidence Interval is ( 2.556 , 6.844 )**

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