Some college professors make bound lecture notes available to their classes in an effort to improve teaching effectiveness. A study of business student's opinions of lecture notes. Two groups of students were surveyed - 86 students enrolled in a promotional strategy class that required the purchase of lecture notes, and 35 students enrolled in a sales/retailing elective that did not offer lecture notes. At the end of the semester :"Having a copy of the lecture notes was helpful in understanding the material." Responses were measured on a nine-point semantic difference scale, where 1="strongly disagree" and 9=" strongly agree." A summary of the results is reported in the follow:
Classes Buying Lecture Notes Classes Not Buying Lecture Notes
n1=86 n2=35
X1=8.48 X2=7.80
S21=.94 S22=2.99
a. Describe the two populations involved in the comparison.
b. Do the samples provides sufficient evidence to conclude that there is a difference in the mean responses of the two groups of the students? Test using α=.01
c. Construct a 99% confidence interval for (μ1-μ2). Interpret the result.
d. Would a 95% confidence interval for (μ1-μ2) be narrow or wider than the one you found in part c? Why?

a) The number of students sampled in both populations are large. We can assume that the populations are normally distributed. The populations are also independent.

b) This is a test of 2 independent groups. Let μ1 be the mean responses of students buying lecture notes and μ2 be the mean responses of students not buying lecture notes.

The random variable is μ1 - μ2 = difference in the mean responses of students buying lecture notes and the mean responses of students not buying lecture notes.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

This is a two tailed test.

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

x1 = 8.48

x2 = 7.8

s1 = 0.94

s2 = 2.99

n1 = 86

n2 = 35

t = (8.48 - 7.8)/√(0.94²/86 + 2.99²/35)

t = 1.32

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.94²/86 + 2.99²/35]²/[(1/86 - 1)(0.94²/86)² + (1/35 - 1)(2.99²/35)²] = 0.0706/0.00192021883

df = 37

We would determine the probability value from the t test calculator. It becomes

p value = 0.195

c) Since alpha, 0.01 < than the p value, 0.195, then we would fail to reject the null hypothesis. Therefore, at 5% significance level, the samples do not provide sufficient evidence to conclude that there is a difference in the mean responses of the two groups of the students.

d) The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

For a 99% confidence interval, the z score is 1.2.58. This is determined from the normal distribution table.

x1 - x2 = 8.48 - 7.8 = 0.68

z√(s1²/n1 + s2²/n2) = 2.58√(0.94²/86 + 2.99²/35) = 1.33

The confidence interval is

0.68 ± 1.33

The upper boundary for the confidence interval is

0.68 + 1.01 = 2.01

The lower boundary for the confidence interval is

0.68 - 1.33 = - 0.65

We are confident that the difference in population means responses between the students buying lecture notes and the students not buying lecture notes is between - 0.65 and 2.01

d) For a 95% confidence interval, the z score is 1.96.

z√(s1²/n1 + s2²/n2) = 1.96√(0.94²/86 + 2.99²/35) = 1.01

The confidence interval is

0.68 ± 1.01

The upper boundary for the confidence interval is

0.68 + 1.01 = 1.69

The lower boundary for the confidence interval is

0.68 - 1.01 = - 0.33

Therefore, a 95% confidence interval for (μ1-μ2) would be narrower. This is seen in the values in both scenarios.