A professor is interested in comparing the average amount spent on textbooks for

Question

A professor is interested in comparing the average amount spent on textbooks for freshmen and sophomores. A random sample of 12 freshmen yielded a sample mean amount of $1036 and a sample standard deviation of $68. A random sample of 8 sophomores yielded a sample mean amount of $1227 and a sample standard deviation of $321. Construct a 90% confidence interval for the difference between the mean amount spent on textbooks by freshmen and the mean amount spent by sophomores (ie. do freshmen minus sophomores). Since the sample standard deviations are wildly different, use a method which does not assume the populations have the same variance. Assume normality.

b) What is the lower confidence limit on the interval? Give your answer to two decimal places.(Answer is not -409.26)

Explanation / Answer

Here, x1 = 1036 , x2 = 1227 , s1 = 68 , s2 = 321 , n1 = 12 , n2 =8

At 90% confidence interval t value with df 18 = 1.734

CI = (x1 - x2) + /- t * sqrt ( s1^2/ n1 + s2^2/n2)

= ( 1036 - 1227) +/- 1.734 *sqrt( 68^2 / 12 + 321^2/8)

= ( - 390.72 , 8.72)

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