Statistics and Data
Analysis
Professor William Greene
Stern School of Business
IOMS Department
Department of Economics
1/18
Part 15: Hypothesis Tests
Statistics and Data Analysis
Part 15 – Hypothesis
Tests: Part 3
2/18
Part 15: Hypothesis Tests
A Test of Independence




3/18
In the credit card example, are Own/Rent and
Accept/Reject independent?
Hypothesis: Prob(Ownership) and Prob(Acceptance)
are independent
Formal hypothesis, based only on the laws of
probability:
Prob(Own,Accept) = Prob(Own)Prob(Accept)
(and likewise for the other three possibilities.
Rejection region: Joint frequencies that do not look like
the products of the marginal frequencies.
Part 15: Hypothesis Tests
A Contingency Table Analysis
4/18
Part 15: Hypothesis Tests
Independence Test
Step 2: Expected proportions assuming independence: If the
factors are independent, then the joint proportions should equal
the product of the marginal proportions.
[Rent,Reject]
[Rent,Accept]
[Own,Reject]
[Own,Accept]
5/18
Hypothetical
0.54404 x 0.21906 = 0.11918
0.54404 x 0.78094 = 0.42486
0.45596 x 0.21906 = 0.09988
0.45596 x 0.78094 = 0.35606
(Actual)
(.13724)
(.40680)
(.08182)
(.37414)
Part 15: Hypothesis Tests
Comparing Actual to Expected
The statistic is N times the sum over the four cells
(Observed-Expected)2
 = N ×  Rows  Columns
Expected
If this is large (because the observed proportions don't
2
look like the expected ones) then rej ect the hypothesis.
(This is a "chi squared statistic.")
 (0.13724  0.11918)2 (0.40680  0.42486) 2 



0.11918
0.42486

2  13,444 
2
2
 (0.08182  0.09988)
(0.37414  0.35608) 



0.09988
0.35608


= 103.33013
6/18
Part 15: Hypothesis Tests
When is Chi Squared Large?
For a 2x2 table, the critical chi squared
value for α = 0.05 is 3.84.
 (Not a coincidence, 3.84 = 1.962)
 Our 103.33 is large, so the hypothesis of
independence between the acceptance
decision and the own/rent status is
rejected.

7/18
Part 15: Hypothesis Tests
Computing the Critical Value
For an R by C Table, D.F. = (R-1)(C-1)
CalcProbability
Distributions  Chisquare
The value reported is
3.84146.
8/18
Part 15: Hypothesis Tests
Analyzing Default



9/18
Do renters default
more often (at a
different rate) than
owners?
To investigate, we
study the cardholders
(only)
We have the raw
observations in the
data set.
OWNRENT
0
DEFAULT
0
1
All
4854
615
5469
46.23 5.86
52.09
1
4649
44.28
381
3.63
5030
47.91
All
9503
90.51
996
9.49
10499
100.00
Part 15: Hypothesis Tests
10/18
Part 15: Hypothesis Tests
11/18
Part 15: Hypothesis Tests
 [.4623  (.9051 .5209)]2
[.0586  (.0949  .5209)]2 



(.9051 .5209)
(.0949  .5209)
2


  10499
 [.4428  (.9051 .4791)]2
[.0363  (.0949  .4791)]2 



(.9051 .4791)
(.0949  .4791)


12/18
Part 15: Hypothesis Tests
Hypothesis Test
13/18
Part 15: Hypothesis Tests
In my sample of 210 travelers between Sydney and Melbourne, it appears
that there is a relationship between income and the decision whether to fly or
not. Do the data suggest that the mode choice and income are
independent?
14/18
Part 15: Hypothesis Tests
Treatment Effects in Clinical Trials


Does Phenogyrabluthefentanoel (Zorgrab)
work?
Investigate: Carry out a clinical trial.



N+0 = “The placebo effect”
N+T – N+0 = “The treatment effect”
Is N+T > N+0 (significantly)?
Placebo
15/18
Drug Treatment
No Effect
N00
N0T
Positive Effect
N+0
N+T
Part 15: Hypothesis Tests
16/18
Part 15: Hypothesis Tests
Confounding Effects
17/18
Part 15: Hypothesis Tests
What About Confounding Effects?
Normal Weight
Obese
Nonsmoker
Smoker
Age and Sex are usually relevant as well. How can all these
factors be accounted for at the same time?
18/18
Part 15: Hypothesis Tests