Question

College students are a unique population when it comes to drinking behavior, given that while in school, they often become able to drinking legally. We think that attitudes towards drinking may be different among different types of college students, particularly those that attend a traditional 4-year college and live on campus versus those that live off campus. The idea here is that students that are not around their peers as much (i.e., those that live off campus), are likely to be more conservative about drinking behaviors. We asked these students to tell use their average number of drinks per day (this is completely made up data). Using the information below, and α = 0.05, test whether students who live off campus drink less than their on campus counterparts. Live off Campus Live on Campus n = 12 n = 6 s_2=2.3 s_1=1.2 x ̅_2=0.5 x ̅_1=2 Would the alternative hypothesis for this test be directional or non-directional? Directional, greater than Directional, less than Non-directional What is the degrees of freedom for this test? 16 15 11 5 What is your obtained t? -1.981 -1.212 -1.483 -1.548 What is your critical value(s) for this test? - 2.947 - 1.761 - 1.746 - 2.306 What do you conclude? Reject H0 Fail to Reject H0

Live off Campus	Live on Campus
n = 12	n = 6
s2=2.3	s1=1.2
x2=0.5	x1=2

1. Would the alternative hypothesis for this test be directional or non-directional?
   1. Directional, greater than
   2. Directional, less than
   3. Non-directional

2. What is the degrees of freedom for this test?
   1. 16
   2. 15
   3. 11
   4. 5

3. What is your obtained t?
   1. -1.981
   2. -1.212
   3. -1.483
   4. -1.548

4. What is your critical value(s) for this test?
   1. - 2.947
   2. - 1.761
   3. - 1.746
   4. - 2.306

5. What do you conclude?
   1. Reject H₀
   2. Fail to Reject H₀

Answer 1

1)option B:Directional, less than

2)degree of freedom =n1+n2-2 =16

3)

Pooled Variance Sp2=((n1-1)s21+(n2-1)*s22)/(n1+n2-2)=				4.0869
Pooled Std dev Sp=((n1-1)s21+(n2-1)*s22)/(n1+n2-2)=				2.0216
Point estimate : x1-x2=			-1.5000
standard error se =Sp*√(1/n1+1/n2)=			1.011
test stat t =(x1-x2-Δo)/Se=		-1.483

4)

Decision rule :                   reject Ho if test statistic t<-1.746

5)

Fail to Reject H₀ (since test statistic is not less than crtiical value)