A professor is interested in comparing the average amount spent on textbooks for freshmen and sophomores. A random sample of 10 freshmen yielded a sample mean amount of $1023 and a sample standard deviation of $68. A random sample of 14 sophomores yielded a sample mean amount of $1257 and a sample standard deviation of $338. Construct a 99% 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. a) What is the appropriate degrees of freedom in this case? Give your answer to four decimal places. 14.4480 Correct: Your answer is correct. b) What is the lower confidence limit on the interval? Give your answer to two decimal places. Incorrect: Your answer is incorrect. c) What is the upper confidence limit on the interval? Give your answer to two decimal places. d) Using this interval, is it reasonable to conclude that the average sophomore spends more than the average freshman? Yes because the upper endpoint on the interval is above 0. Yes because the interval contains 0. No because the interval contains 0. Yes because the upper endpoint on the interval is below 0.
For Freshman, sample 1 :
x̅1 = 1023, s1= 68, n1 =
10
For sophmore sample 2 :
x̅2 = 1257, s2 = 338, n2 =
14
Population standard deviation does not have same variance.
a) Degree of freedom:
At α = 0.01 and df =14, two tailed critical value, tc = T.INV.2T( 0.01 , 14 ) = 2.977
b) Lower interval = (x̅1 - x̅2) - tc*√(s1²/n1
+s2²/n2) = - 510.425
c) Upper interval = (x̅1 - x̅2) + tc*√(s1²/n1 +s2²/n2) =
42.425
d) No, because the interval contains 0.
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