Question

A professor is interested in comparing the average amount spent on textbooks for freshmen and sophomores. A random sample of 15 freshmen yielded a sample mean amount of $1016 and a sample standard deviation of $54. A random sample of 12 sophomores yielded a sample mean amount of $1236 and a sample standard deviation of $320. Construct a 95% 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.
Do not round down or take the next level of freedom.
1) What is the appropriate degrees of freedom in this case? Give your answer to four decimal places.  
2) What is the lower confidence limit on the interval? Give your answer to two decimal places.  
3) What is the upper confidence limit on the interval? Give your answer to two decimal places.  
4) Using this interval, is it reasonable to conclude that the average sophomore spends more than the average freshman?

a)No because the interval contains 0.

b)Yes because the upper endpoint on the interval is below 0.     

c)Yes because the interval contains 0.

d)Yes because the upper endpoint on the interval is above 0.

Answer 1

The statistical software output for this problem is:

Two sample T summary confidence interval:

μ₁ : Mean of Population 1
μ₂ : Mean of Population 2
μ₁ - μ₂ : Difference between two means
(without pooled variances)

95% confidence interval results:

Difference	Sample Diff.	Std. Err.	DF	L. Limit	U. Limit
μ1 - μ2	-220	93.422339	11.502206	-424.53112	-15.468877

Hence,

1) Degrees of freedom = 11.5022

2) Lower limit = -424.53

3) Upper limit = -15.47

4) Option B is correct.