AMS 572 Lecture Notes#11
October31st, 2011
Ch.9. Categorical Data Analysis
Exact test on one population proportion:
Data: sample of n, X are # of “Successes”, n-X are # of “Failures”
,
(1)
(2)
(3)
.
Inference on Two Population Proportions: ,
Independent samples, both are large
e.g. Suppose we wish to compare the proportions of smokers among male and female students in SBU.
Two large independent samples:
Population 1: Population 2:
Sample 1: Sample 2:
(,)
①Point estimator:
②By CLT,
Two samples are independent
③P.Q. for :
not P.Q.
P.Q.
④100(1-)% large samples CI for :
Here,
⑤Test
General case:
p-value=P(Z≥)
At the significance level , we reject if and p-value<α.
When =0, one often uses the following test statistic
here
Example 1. A random sample of Democrats and a random sample of Republicans were polled on an issue. Of 200 Republicans, 90 would vote yes on the issue; of 100 democrats, 58 would vote yes. Let p1 and p2 denote respectively the proportions of all Democrats or all Republicans who would vote yes on this issue.
(a)Construct a 95% confidence interval for (p1 - p2)
(b)Can we say that more Democrats than Republicans favor the issue at the 1% level of significance? Please report the p-value.
(c)Please write up the entire SAS program necessary to answer question raised in (b). Please include the data step.
Solution:
(a)Democrats:
Republicans:
The 100(1-α)% confidence interval for (p1 - p2) is
After plugging in Z0.025 = 1.96 etc., we found the 95% CI to be [0.01, 0.25]
(b)Hypotheses are v.s .
.
.
We cannot reject at . Therefore, Democrats favor the issue as same as Republicans at the 1% significance level.
(c) SAS code:
Data Poll;
Input Party $ outcome $ count;
Datalines;
Republican yes 90
Republican no 110
Democrats yes 58
Democrats no 42
;
Run;
Procfreqdata=poll;
Tables party*outcome/chisq;
Weight count;
Run;
Output:
The SAS System
The FREQ Procedure
Table of Party by outcome
Party outcome
Frequency‚
Percent ‚
Row Pct ‚
Col Pct ‚no ‚yes ‚ Total
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Democrat ‚ 42 ‚ 58 ‚ 100
‚ 14.00 ‚ 19.33 ‚ 33.33
‚ 42.00 ‚ 58.00 ‚
‚ 27.63 ‚ 39.19 ‚
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Republic ‚ 110 ‚ 90 ‚ 200
‚ 36.67 ‚ 30.00 ‚ 66.67
‚ 55.00 ‚ 45.00 ‚
‚ 72.37 ‚ 60.81 ‚
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Total 152 148 300
50.67 49.33 100.00
Statistics for Table of Party by outcome
Statistic DF Value Prob
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Chi-Square 1 4.5075 0.0337
Likelihood Ratio Chi-Square 1 4.5210 0.0335
Continuity Adj. Chi-Square 1 4.0024 0.0454
Mantel-Haenszel Chi-Square 1 4.4924 0.0340
Phi Coefficient -0.1226
Contingency Coefficient 0.1217
Cramer's V -0.1226
Fisher's Exact Test
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Cell (1,1) Frequency (F) 42
Left-sided Pr <= F 0.0226
Right-sided Pr >= F 0.9877
Table Probability (P) 0.0103
Two-sided Pr <= P 0.0377
Sample Size = 300
Inference on several proportions—the Chi-square test (large sample)
Def. Multinomial Experiment.
We have a total of n trials (sample size=n)
①For each trial, it will result in 1 of k possible outcomes.
②The probability of getting outcome i is , and =1
③These trials are independent.
Example2. Previous experience indicates that the probability of obtaining 1 healthy calf from a mating is 0.83. Similarly, the probabilities of obtaining 0 and 2 healthy calves are 0.15 and 0.02 respectively. If the farmer breeds 3 dams from the herd, find the probability of getting exact 3 health calves.
Def. Multinomial Distribution
Let be the number of trials resulted in i-th category out of a total of n trials and be the probability of getting i-th category outcome, then
Solution:
P(exact 3 health calves)=+
=0.015+0.572=0.59
*Relations to the Binomial Distribution (k=2)
Category / 1 / 2Probability / =p / =1-p
# trials / =x / =n-x
Chi-square goodness of fit test
Example3. Gregor Mendel (1822-1884) was an Austrian monk whose genetic theory is one of the greatest scientific discovery of all time. In his famous experiment with garden peas, he proposed a genetic model that would explain inheritance. In particular, he studied how the shape (smooth or wrinkled) and color (yellow or green) of pea seeds are transmitted through generations. His model shows that the second generation of peas from a certain ancestry should have the following distribution.
wrinkled-green / wrinkled-yellow / smooth-green / smooth-yellowTheoretical probabilities / / / /
n=556
General test:
Test whether the theoretical probability is correct
T.S
where is the observed number of observations in category i
is the expected count of the i-th category ,
At the significance level α, reject iff
Solution:
wrinkled-green / wrinkled-yellow / smooth-green / smooth-yellowTheoretical probabilities / / / /
Observed count out of 556 / =31 / =102 / =108 / =315
Expected counts / =34.75 / =104.25 / =104.25 / =312.75
T.S
=7.815
At significance level 0.05, we cannot reject
SAS Code:
DATA GENE;
INPUT @1 COLOR $13. @15 NUMBER 3.;
DATALINES;
YELLOWSMOOTH 315
YELLOWWRINKLE 102
GREENSMOOTH 108
GREENWRINKLE 31
;
* HYPOTHESIZING A 9:3:3:1 RATIO;
PROCFREQDATA=GENE ORDER=DATA; WEIGHT NUMBER;
TITLE3'GOODNESS OF FIT ANALYSIS';
TABLES COLOR / CHISQNOCUMTESTP=(0.56250.18750.18750.0625);
RUN;
The SAS System
GOODNESS OF FIT ANALYSIS
The FREQ Procedure
Test
COLOR Frequency Percent Percent
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YELLOWSMOOTH 315 56.65 56.25
YELLOWWRINKLE 102 18.35 18.75
GREENSMOOTH 108 19.42 18.75
GREENWRINKLE 31 5.58 6.25
Chi-Square Test
for Specified Proportions
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Chi-Square 0.6043
DF 3
Pr > ChiSq 0.8954
Sample Size = 556
Example 4. A classic tale involves four car-pooling students who missed a test and gave as an excuse of a flat tire. On the make-up test, the professor asked the students to identify the particular tire that went flat. If they really did not have a flat tire, would they be able to identify the same tire?
To mimic this situation, 40 other students were asked to identify the tire they would select.
The data are:
Tire / Left front / Right front / Left rear / Right rearFrequency / 11 / 15 / 8 / 6
At α=0.05, please test whether each tire has the same chance to be selected?
Solution:
n=40, =n=10
Fail to reject .
The chi-square goodness of fit test is an extension of the Z-test for one population proportion.
Data: sample size n, x: successes with probability p
n-x: failures with probability 1-p
TS.
At α, reject iff
Success / FailureExpected / /
Observed / =x /
=
=
Recall: If , then .
When k=1,
Let Z~N(0,1), then
=
The two tests are identical.
Exact Tests for Inference on Two Population Proportions
- Fisher’s exact test:
A little bit of history (Fisher’s Tea Drinker): R.A. Fisher described the following experiment. Muriel Bristol, his colleague, claimed that when drinking tea, she could distinguish whether milk or tea was added to the cup first (she preferred milk first). To test her claim, Fisher designed an experiment with 8 cups of tea – 4 with milk added first and 4 with tea added first. Muriel was told that there were 4 cups of each type, and was asked to predict which four had the milk added first. The order of presenting the cups to her was randomized. It turned out that Muriel correctly identified 3 from each type. Now we are testing the null hypothesis that she did so by pure guessing, versus the alternative hypothesis that she could do better than pure guessing. The p-value of the test is derived as follows:
Since the p-value is large, we could not reject the null hypothesis. It means that it is possible that she chose 3 correctly by pure guessing.
Inference on 2 population proportions, 2 independent samples
Example 5. The result of a randomized clinical trial for comparing Prednisone and Prednisone+VCR drugs, is summarized below. Test if the success and failure probabilities are the same for the two drugs.
Drug / Success / Failure / Row totalPred
PVCR / 14
38 / 7
4 / =21
=42
m=52 / n-m=11 / n=63
General setting:
“S” / “F” / TotalSample1 / x / -x /
Sample2 / y / -y /
m=x+y / n-m / n
Solution:
<0.05
Reject
SAS code:
Data trial;
input drug $ outcome$ count;
datalines;
pred S 14
pred F 7
PVCR S 38
PVCR F 4
;
run;
procfreqdata=trial;
tables drug*outcome/chisq;
weight count;
run;
- McNemar’s test
Inference on 2 population proportions- paired samples
Example6. A preference poll of a panel of 75 voters was conducted before and after a TV debate during the campaign for the 1980 presidential election between Jimmy Carter and Ronald Reagan. Test whether there was a significant shift from Carter as a result of the TV debate.
Preferencebefore / Preference after
Carter / Reagan
Carter / 28 / 13
Reagan / 7 / 27
General setting:
Condition1response / Condition2 response
Yes / No
Yes / A=a, / B=b,
No / C=c, / D=d,
+++=1, A+B+C+D=n, (A, B, C, D)~Multinomial
,
P(B=k| B+C=m)~Bin(m,p=)
Under , P(B=k| B+C=m)~Bin(m,p=1/2)
①
②
③
Solution:
SAS code:
Data election;
input before $ after $ count;
datalines;
Carter Carter 28
Carter Reagan 13
Reagan Reagan 27
Reagan Carter 7
;
run;
procfreqdata=election;
exactagree;
tables before*after/agree;
weight count;
run;
The SAS System 21:15 Saturday, November 6, 2010 2
The FREQ Procedure
Table of before by after
before after
Frequency‚
Percent ‚
Row Pct ‚
Col Pct ‚Carter ‚Reagan ‚ Total
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Carter ‚ 28 ‚ 13 ‚ 41
‚ 37.33 ‚ 17.33 ‚ 54.67
‚ 68.29 ‚ 31.71 ‚
‚ 80.00 ‚ 32.50 ‚
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Reagan ‚ 7 ‚ 27 ‚ 34
‚ 9.33 ‚ 36.00 ‚ 45.33
‚ 20.59 ‚ 79.41 ‚
‚ 20.00 ‚ 67.50 ‚
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Total 35 40 75
46.67 53.33 100.00
Statistics for Table of before by after
McNemar's Test
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Statistic (S) 1.8000
DF 1
Asymptotic Pr > S 0.1797
Exact Pr >= S 0.2632 (= 2*0.1316)
Simple Kappa Coefficient
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Kappa (K) 0.4700
ASE 0.1003
95% Lower Conf Limit 0.2734
95% Upper Conf Limit 0.6666
Test of H0: Kappa = 0
ASE under H0 0.1140
Z 4.1225
One-sided Pr > Z <.0001
Two-sided Pr > |Z| <.0001
Exact Test
One-sided Pr >= K 3.614E-05
Two-sided Pr >= |K| 5.847E-05
Sample Size = 75
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