1. Two Sample Hypothesis Tests: Answers to Practice Problems
#8.4 (2/3e)/ 9.4 (1e)
#8.8 (2/3e)/ 9.8 (1e)
#8.12a (2/3e)/ 9.12a (1e)

2. #8.4 (2/3e)/ 9.4 (1e) Problem information: Use the 5-step method….
s1=3.9 s2=2.0
n1=97 n2=100
Use the 5-step method….
Steps 1 - 3
2 random samples, Interval-ratio
Sample is large n1+n2 ≥ 100, normal
z-test
Question: Are employment equity employees different?”
2-tailed test, α=.05, zcr=±1.96
Ho: µ1= µ2 H1: µ1≠ µ2

3. #8.4 (2/3e)/ 9.4 (1e) cont. Step 4: Step 5:
Zobtained<Zcritical therefore fail to reject Ho
The difference is not significant and is random chance
The employment equity employees were not significantly different from the regular employees.

4. #8.8 (2/3e)/ 9.8 (1e) Problem information: Use the 5-step method….
s1=2.7 s2=4.2
n1=40 n2=55
Use the 5-step method….
Steps 1 - 3
2 random samples, Interval-ratio
Sample is small n1+n2 < 100, t-distribution
t-test
Question: Has membership reduced delinquency?”
1-tailed test, α=.05, df = n1+n2 -2 = 93, tcr= (reduced)
Ho: µ1= µ2 H1: µ1< µ2

5. #8.8 (2/3e)/ 9.8 (1e) cont. Step 4: Step 5:
tobtained>tcritical (further out in tail) therefore reject Ho
The difference is significant and is not random chance
Membership has significantly reduced delinquency (tobt=-2.61, df=93, α=.05)

6. #8.12a (2/3e)/ 9.12a (1e) Problem information: Use the 5-step method….
Ps1 = Ps2 = 0.59
n1 = n2 = 82
Use the 5-step method….
Steps 1 - 3:
1 random sample, Nominal
Sample is large n1 + n2 ≥ 100, normal
z-distribution (z-test)
Question asks “did the special program work?”
1-tailed test, α=.05, z=-1.65 (special has lower divorce rate)
Ho: Pu1 = Pu2 H1: Pu1 < Pu2

7. #8.12a (2/3e)/ 9.12a (1e) Step 4: Step 5:
Zobtained<Zcritical therefore fail to reject Ho
The difference is not significant and is random chance
The special program did not reduce divorce significantly.