Chapter 12 Inference about two probabilities -...

Chapter 12

Inference about twoprobabilities

In this chapter, we discuss inferences about proportions in two populationsin two ways:

1. inferences about independence

2. inferences about two populations - difference in probabilities

12.1 Contingency tables -Tests of indepen-

dence

These tests occur with two measurements on the same person, for example,smoking; respiratory infection (including chest cold). Both measurementsare binary, in this case Y and N

The question is: Are these measurements related?

In general, the results are reported as a two-by-two table.They are sometimes called a fourfold table, a cross-tabulation tables or,most often, a contingency table, where one measurement is contingent (de-pendent) on the other.

1

2 CHAPTER 12. INFERENCE ABOUT TWO PROBABILITIES

The general form of the contingency table is

Table 12.1: A typical contingency table

Characteristic BCharacteristic A Yes No TotalYes a b r1No c d r2Total c1 c2 T

As an example consider a study looking at the relationship between smok-ing and respiratory illness, referred to as a cold.

Table 12.2: A study of smoking and colds

Had a cold this yearSmoked this year Yes No TotalYes 15 8 23No 10 12 22Total 25 20 45

The traditional null and alternative hypotheses are quite vagueHo : smoking and colds are not related (12.1)that is, smoking and colds are independent

HA : smoking and colds are related

Despite the vagueness of the null hypothesis, we can, under this hypoth-esis, compute expected probabilities, for example,

Pr(Smoked this year AND had a cold) = Pr(smoked this year)Pr(had a cold this year)

=23

45

25

45

=23

81= 0.2839

12.1. CONTINGENCY TABLES -TESTS OF INDEPENDENCE 3

We can also compute the expected number of smokers with a cold

45

(

23

45

25

45

)

=23(25)

45= 12.7777

In general we have Eij =ricjT

Table 12.3: Resulting table of Expected values

study of smoking and coldsHad a cold this year

Yes 12.78 10.22 23No 12.22 9.78 22Total 25 20 45

The test statistic is Pearsons test of association SP (after Karl Pearson,1900)

SP =2∑

i=1

2∑

j=1

(Oij − Eij)2

Eij

=(15− 12.7778)2

12.7778+

(8− 10.2222)2

10.2222

+(10− 12.2222)2

12.2222+

(12− 9.7778)2

9.7778= 4.9382(0.3602)

= 1.7786

pvalue = Pr(SP > 1.7786)

= 2Pr(ZN > 1.3336)

= 2(.0912)

= 0.1824


There is a continuity-corrected version suggested by Frank Yates (1934)

SY =2∑

i=1

2∑

j=1

(|Oij − Eij| − 0.5)2

Eij

=(|15− 12.78| − 0.5)2

12.78+

(|8− 10.22| − 0.5)2

10.22

+(|10− 12.22| − 0.5)2

12.22+

(|12− 9.78| − 0.5)2

9.78

=2.9660

0.3602= 1.0668

pvalue = Pr(SY > 1.0683)

= 2Pr(ZN > 1.0336)

= 2(.1506)

= 0.3012


12.1.1 alternative presentation of statistics

These are called the epidemiological version of the test statistics:

1. Pearson’s test

SP = T(ad− bc)2

c1c2r1r2

= 45(15(12)− 8(10))2

23(22)(25)(20)

= 45

(

1002

253000

)

=450

253= 1.7786

2. Yates’ version

SY = T(|ad− bc| − T/2)2

c1c2r1r2

= 45(|15(12)− 8(10)| − 22.5)2

23(22)(25)(20)

= 45

(

77.52

253000

)

= 1.0683


12.1.2 exact test

Meanwhile, in 1922, Ronald Fisher had suggested an exact test:

• fix column and row totalsthese are denoted in Table 12.1 as c1, c2, r1, r2, T

• given we have c1 items of the total T in column 1calculate the probability of gettinga of r1 in row 1c of r2 in row 2this comes from the hypergeometric distribution, and is written as

(

r1a

)(

r2c

)

(

Tc1

)

• calculate for observed data

(

2315

)(

2210

)

(

4525

)

= 0.100023

• repeat for all tableswith smaller probability!!!

For this particular data, p-value = 0.2361889, that is, 0.2362


12.1.3 Summary of contingency table tests

This table gives the summary of the results of three possible statistical testsof Independence:

Method p-value commentFisher 0.2362 exactPearson 0.1824 anticonservativeYates 0.3012 conservative

This table demonstrates the method of testing hypotheses like (12.1)

1. where possible, use Fisher’s exact test;

2. when necessary, use SY , Yates’s continuity-corrected test statistic

12.1.4 Hypotheses about differences in probabilities

If we selected a 23 smokers and 22 non-smokers, and followed them for ayear to see who got colds, we would be using the study design called a cohort

study by epidemiologists. Here are the results of such a study:

Table 12.4: cohort study of smoking and colds

Had a cold during yearSmoker at beginning of year Yes No TotalYes 15 8 23No 10 12 22Total 25 20 45

As with the hypothesis test methodology already discussed, we can think ofthis data as coming from two populations, smokers and non-smokers, eachwith probability, πi, of getting a cold

Here are the possible null and alternative hypotheses for this situation:Ho : π1 = π2(12.2)HA : π1 6= π2


Any of the approaches to hypothesis testing would use a statistic of theform

p1 − p2√

p1(1−p1)n1

+ p2(1−p2)n2

(12.3)

It can be shown that :

1. the square of the test statistic (12.3) is equivalent to SP , the Pearsontest statistic described in the previous section(this equivalence is proved at the end of this chapter)

2. there is a continuity-corrected version of (12.3), and it can the shownthat the square of this statistic is equivalent to SY , the Yates teststatistic.

Hence SP and/or SY can be used to test hypotheses given in (12.2), andthere is no need for use of the statistic (12.3).

Hence for testing hypotheses such as those given in (12.2)

1. use Fisher’s exact test, when available;

2. otherwise, use SY , Yates continuity-corrected test

WHAT about ESTIMATION

12.2. CONFIDENCE INTERVALS FORDIFFERENCES IN PROBABILITIES9

12.2 Confidence Intervals for differences in

probabilities

1. for large samplesnipi > 5, i = 1, 2

(p1 − p2)± zα/2

√

(

p1(1− p1)

n1

+p2(1− p2)

n2

)

(12.4)

2. small samples Iuse Continuity correction for each sample

(p1 − p2)±(

12r1

+ 12r2

)

+ zα/2

√

(

p1(1−p1)n1

+ p2(1−p2)n2

)

3. small samples IIusing intervals from each sample, according to either WIlson/Score orWald Adjusted methods, then combining them in a special way;

4. small samples IIIalternative due to Zou and Donner (2004)

12.2.1 Newcombe’s (1998) method

1. use Wilson or Wald Adjusted to get(l1, u1) for π1

(l2, u2) for π2

2. compute

(p1 − p2)− zα/2

√

l1(1− l1)

n1

+u2(1− u2)

n2

(12.5)

(p1 − p2) + zα/2

√

u1(1− u1)

n1

+l2(1− l2)

n2

(12.6)

or

(p1 − p2)−√

(p1 − l1)2 + (u2 − p2)2(12.7)

(p1 − p2) +√

(p2 − l2)2 + (u1 − p1)2(12.8)


12.2.2 example

12.2.2.1 Continuity correction

1. since this is a small sample,we don’t use the large sample method, butinstead, use the Continuity correctionFirst wee need point estimates:p1 = 15/23 = 0.6522, p2 = 10/22 = 0.4545

2. construct the 95% confidence interval

(0.6522− 0.4545)

±[

1

46+

1

44+ 1.96

√

0.6522(0.3478)

23+

0.4545(0.5455)

22

]

= (0.1977)± (0.0445 + 1.96√0.009862 + 0.011270)

= 0.1977± [0.0445 + 1.96(0.1454)]

= 0.1977± 0.3294

that is, (-0.132,0.527), a very wide interval.

12.2.2.2 using adjusted wald

1. need point estimatesp1 = (15 + 2)/(23 + 4) = 0.6296p2 = (10 + 2)/(22 + 4) = 0.4615

2. construct first 95% CI

0.6296± 1.96

√

0.6296(0.3704)

27

= 0.6296± 1.96√0.00864

= 0.6296± 1.96(0.09293) = 0.6296± 0.1821

= (0.4475, 0.8118)


3. construct second 95% confidence interval

0.4615± 1.96

√

0.4615(0.5385)

26

= 0.4615± 1.96√0.009558

= 0.4615± 1.96(0.09776) = 0.6296± 0.1916

= (0.2699, 0.6532)

(p1 − p2)−√

(p1 − l1)2 + (u2 − p2)2

= (0.6522− 0.4545)

−√

(0.6522− 0.4475)2 + (0.6532− 0.4545)2

= 0.1977−√0.20472 + 0.19872

= 0.1977−√0.08138378 = 0.1977− 0.2853 = −0.0876

(p1 − p2) +√

(p2 − l2)2 + (u1 − p1)2

= (0.6522− 0.4545)

−√

(0.4545− 0.2699)2 + (0.8118− 0.6522)2

= 0.1977 +√0.18462 + 0.15962

= 0.1977 +√0.05954932 = 0.1977 + 0.2440 = 0.4417

Hence interval is (-0.0876,0.4417)


12.2.2.3 Advantages of Newcombe’s method

1. narrower than Continuity corrected interval

2. shifted away from 0

3. asymmetric around point estimate 0.1977

12.2.2.4 The origins of Newcombe’s formulae

The lower bound of the confidence interval for the difference in probabilitiesis give by

(p1 − p2)− zα/2

√

σ2p1+ σ2

p2(12.9)

In the usual (large sample) method, the ”overall” estimates of σ2p1

and σp2

are used, that is, p1(1−p1)n1

and p2(1−p2)n2

, respectivelyThese are used in equation (12.9) to get equation (12.4).

However, as Newcombe suggested, at the lower and upper bounds of the in-terval, there are better estimates of the variances

At the lower bound, we want the smallest value of π̂1 − π̂2; this occurs whenwe use l1 and u2.For example, if the 95%CI for π1 is (0.4,0.5) and the 95% Ci for π2 is (0.2,0.3),the the difference π̂1 − π̂2 if minimized if we use l1 = 0.4 and u2 = 0.3 to get0.4− 0.2 = 0.1Similarly, the difference is maximized it we use u1 and l2, for example,0.5− 0.1 = 0.4.


Returning to the formula, when we plug l1 and u2 into (12.4) or (12.9) toget (12.5), namely,

(p1 − p2)− zα/2

√

l1(1− l1)

n1

+u2(1− u2)

n2

Similarly, plug plug u1 and l2 into (12.4) or (12.9) to get (12.6) namely,

(p1 − p2) + zα/2

√

u1(1− u1)

n1

+l2(1− l2)

n2

Now for the other equations:l1 is the lower bound of the interval estimate of π1. It is calculated asl1 = p1 − zα/2σMoving terms around = sign, this becomeszα/2σ = p1 − l1Thus p1 − l1 is an estimate of zα/2σp1

Similarly u2 − p2 is an estimate of zα/2σp2

Plug p1 − l1 and u2 − p2 into (12.4) above to get (12.7)

Similarly at upper bound we plug u1−p1 and p2− l2 into (12.4) to get (12.8).


12.3 Using SAS for contingency tables

Now we look at Proc FREQ and its use with count data (rather than rawdata) to produce the analysis of contingency tables.

12.3.1 Simple SAS program

Here is a sample SAS program for contingency table analyses:

title ’simple contingency table’;

proc format;

yesno 0=’yes’ 1=’no’;

DATA marj;

INPUT r o freq;

format r yesno. o yesno.;

DATALINES;

0 0 15

0 1 8

1 0 10

1 1 12

;

The SAS commands for contingency table analysis have to a repeatedRISKDIFF for each method:

PROC FREQ;

WEIGHT freq;

TABLES r*o/ NOROW NOCOL NOPERCENT CHISQ RISKDIFF;

TABLES r*o/ NOROW NOCOL NOPERCENT RISKDIFF(CORRECT);

TABLES r*o/ NOROW NOCOL NOPERCENT

RISKDIFF(METHOD=NEWCOMBE);

quit;

12.3. USING SAS FOR CONTINGENCY TABLES 15

Here is the output, first the contingency table:

simple contingency table

The FREQ Procedure

Table of r by o

Frequency

| yes| no| Total

---------+--------+--------+

yes| 15 | 8 | 23 |

---------+--------+--------+

no| 10 | 12 | 22 |

---------+--------+--------+

Total 25 20 45

and next the tests of the null hypothesis:

Statistic DF Value Prob

-------------------------------------------------

Chi-Square 1 1.7787 0.1823

Likelihood Ratio Chi-Square 1 1.7900 0.1809

Continuity Adj. Chi-Square 1 1.0683 0.3013

Mantel-Haenszel Chi-Square 1 1.7391 0.1872

Phi Coefficient 0.1988

Contingency Coefficient 0.1950

Cramer’s V 0.1988

Fisher’s Exact Test

----------------------------------

Cell (1,1) Frequency (F) 15

Left-sided Pr <= F 0.9493

Right-sided Pr >= F 0.1507

Table Probability (P) 0.1000

Two-sided Pr <= P 0.2362


Here are the estimates of risk difference:

Column 1 Risk Estimates

(Asymptotic)95\% (Exact)95\%

Risk ASE Confide Limits Confide Limits

-----------------------------------------------------

Row 1 0.6522 0.0993 0.4575 0.8468 0.4273 0.8362

Row 2 0.4545 0.1062 0.2465 0.6626 0.2439 0.6779

Total 0.5556 0.0741 0.4104 0.7007 0.4000 0.7036

Difference 0.1976 0.1454 -0.0873 0.4825

Difference is (Row 1 - Row 2)

The next page of output gives ”Column 2 Risk Estimates” which you maychoose to ignore.

Here are the Risk estimates with continuity correction:

Column 1 Risk Estimates

(Asymptotic) 95% (Exact) 95%

Risk ASE Confide Limits Confide Limits

-----------------------------------------------------

Row 1 0.6522 0.0993 0.4358 0.8686 0.4273 0.8362

Row 2 0.4545 0.1062 0.2238 0.6853 0.2439 0.6779

Total 0.5556 0.0741 0.3993 0.7118 0.4000 0.7036

Difference 0.1976 0.1454 -0.1318 0.5270

Difference is (Row 1 - Row 2)

The asymptotic confidence limits include

a continuity correction.

This agrees with our hand calculations done previously in this chapter.SAS also gives a similar table for ”Column 2 Risk Estimates”, which,

again, you may wish to ignoreNote, although we requested a Newcombe Method calculation of confi-

dence interval for the difference in proportions, it DOES NOT work in SAS9.2


12.3.2 Creating pretty tables

In this section, the emphasis is on creating tables in a suitable, readable for-mat, but the statistical analyses are still the same. In the previous example,we used the default for creation of a table, that is, the numerical value of thelevels of a discrete variables.

In this particular example,

1. the level with ”0” is first level

2. the level with ”1” is second level

and we created formats appropriately.In this case, because we wanted the observed count of ”15”, in the first

row and first column, and we know that the the values of ”0” are the firstlevel for each variable, we label the ”0”’s as YES, and, similarly, the ”1”’s asNO.

However we may want formats defined differently, for example, we maywant

• ”1” to be formatted as YES

• ”0” to be formatted as NO.

Hence the variables values of (1,1) have a count of 15. In order to do this,we create data with ”1 1 15” in one line, but we put this value first after theDATALINES command. We tell SAS to use this information by adding theoption ”ORDER=DATA” on the PROC FREQ line.

In other words

1. create formats so that columns and rows of the table are correctlylabelled;

2. input table in order a, b, c, d;

3. say PROC FREQ ORDER=DATA;


Here is a SAS program which does this:

title1 ’simple contingency table’;

title2 ’with different formats’;

PROC FORMAT; value yesno 1=’yes’ 0=’no’;

DATA marj;

INPUT r o freq; FORMAT r yesno. o yesno.;

DATALINES;

1 1 15

1 0 8

0 1 10

0 0 12

;

PROC FREQ ORDER=DATA;

WEIGHT freq;

TABLES r*o/NOROW NOCOL NOPERCENT ;

And here is output with the new new formats. Note that the output is thesame, but we have ”1” as YES and ”0” as NO, which we may wish to use inlater analyses.

simple contingency table

with different formats

The FREQ Procedure

Table of r by o

Frequency|yes |no | Total

---------+--------+--------+

yes | 15 | 8 | 23

---------+--------+--------+

no | 10 | 12 | 22

---------+--------+--------+

Total 25 20


12.3.3 Dealing with Raw data - not counts

In this case, the data consists of records of individuals, who have discretevalues (in this case, ”1” or ”2”) for each variable, and we have to use SASto sum over individuals to create counts, and hence tables.

Here is a typical program, where the values of the discrete variables arepre-defined, and we merely code those values and we create a contingencytable:[12:46am mario]

title ’ analysis from raw data’;

filename two ’two.dat’;

proc format;

value gender 1=’Male’ 2=’Female’;

value smoking 1=’No’ 2=’Yes’;

data second;

infile two;

input id age sex hr smoking;

label sex=’Sex of Subject’

smoking=’Smoking status’;

format sex gender. smoking smoking.;

proc freq ;

tables sex*smoking/norow nocol nopercent;

Here is the data:

1 22 1 60 1

2 25 1 80 2

3 24 1 75 1

4 23 1 65 1

5 27 1 90 2

6 26 1 70 1

7 27 2 55 1

8 26 2 65 1

9 23 2 70 1

10 30 2 75 2

11 25 2 85 2

12 24 2 90 2


We know that SAS will put the count for values (1,1) in the upper leftcell of the table, that is, the males who said No:

analysis from raw data

The FREQ Procedure

Table of sex by smoking

sex(Sex of Subject)

smoking(Smoking status)

Frequency|No |Yes | Total

---------+--------+--------+

Male | 4 | 2 | 6

---------+--------+--------+

Female | 3 | 3 | 6

---------+--------+--------+

Total 7 5 12

We actually want the YES outcome in the first column (and are not worriedabout rows)

12.3.3.1 Getting tables in correct order

There are many ways to do this:

1. SORT on the variables defining row and column of the table; in thisexample, we would only have to sort on the row defining SMOKING,and use the command

Proc FREQ ORDER=DATA;

The problem with this method is that it depends on the physical orderof data. You don’t want to create a permanent dataset that depends onthe order of the data, because it can be changed by a SORT on anothervariable. You should use a SORT to create a temporary dataset for aparticular analysis.Because of these problems, nothing further will be shown about thismethod


2. use ORDER=FORMATTED to get the rows and columns of the tablein the alphabetic order of formats. The problem with this is that bothvariables will be done this way; in this particular case, the Rows willchanges from Males first to Females first, and the columns will remainwith No first, then Yes (see below);

3. Use a trick on the names of the formats to get the formats in the correctorder. You may have to do this with both variables (see below);

4. Define a new version of variablewith values in the correct order. Two examples are given below.

12.3.3.2 order defined by formats

This requires a particular option on the Proc FREQ line:

Proc FREQ ORDER=FORMATTED;

Here is the output:

analysis from raw data - use formats


sex(Sex of Subject)


Frequency|No |Yes | Total

---------+--------+--------+

Female | 3 | 3 | 6

---------+--------+--------+

Male | 4 | 2 | 6

---------+--------+--------+

Total 7 5 12

Note that the order of columns and rows is the the alphabetical order ofthe formats; hence ”No” still comes before ”Yes” which we don’t want, and”Females” comes before ”Males”, which his change, but is ok.


12.3.3.3 Redefine formats

In order to get the Yes column first, we can define one level as ’Yes’ and theother as ’Zno’; hence

proc format;

value smoking 1=’Zno’ 2=’Yes’;

and

Proc FREQ ORDER=FORMATTED;

Now, the first column will be the smokers, even though one strange format(Zno) is used, and Females come before Males, as before

The FREQ Procedure


sex(Sex of Subject)


Frequency|Yes |Zno | Total

---------+--------+--------+

Female | 3 | 3 | 6

---------+--------+--------+

Male | 2 | 4 | 6

---------+--------+--------+

Total 5 7 12


12.3.3.4 Redefine variable(s)

In this case, we create a new variable which has its values in the order wewish, AND we create a new format to go along the new values. Here is partof a SAS program to do this:

Proc FORMAT;

smokb value 2=’Yes’ 3=’No’;

DATA ...;

...

smok2 = smoking;

if (smoking=1) then (smok2=3);

format smok2 smokb;

label smok2=’Smoking status 2’;

...

PROC FREQ DATA=...;

TABLES sex*smok2 ...

SAS uses the default numerical order, but the values of the new SMOKINGvariable are 2 and 3, where the lower value of 2 corresponds to the outcomeYES. Hence in this contingency table, ”Yes” comes before ”No”, which isgood, and ”Males” come before ”Females”, which is ok:

analysis from raw data

The FREQ Procedure\su

Table of sex by smok2

sex(Sex of Subject)

smok2(Smoking status 2)

Frequency|Yes |No | Total

---------+--------+--------+

Male | 3 | 3 | 6

---------+--------+--------+

Female | 2 | 4 | 6

---------+--------+--------+

Total 5 7 12


12.3.3.5 Alternative redefinition

Another way of redefining the variable is to switch, in a new variable, thevalues of 1 and 2; here is sample code:

Proc FORMAT;

smokb value 1=’Yes’ 2=’No’;

DATA ...;

...

smok2 = smoking;



format smok2 smokb;

label smok2=’Smoking status 2’;

...

PROC FREQ DATA=;

TABLES sex*smok2 ...

The output is exactly like the preceding output (but the values of the newvariables are different, as are their formats).

12.4. DESIGNING STUDIES 25

12.4 Designing studies

When planning a study which compares a binary outcome between two pop-ulations, there are two possible computations to be made when trying to plana study:

1. sample sizeHow big a sample size do we need to detect the difference in two pro-portions (probabilities);

2. powerwhat is the chance (probability) of detecting the effect of the new treat-ment

Both these calculations use the effect size, that is,

1. in a population study, the difference in the proportions in two popula-tions,

2. in a clinical study, the change in proportion (population probability)between the new treatment and the old one.

12.4.1 Sample size

We are interested in the change in probability δπ = π1 − π2

Since the variance depends on mean, for this calculation we have to specifyπ1, π2

The simple formula from previous work on difference in means becomes morecomplex

n =

(

√

2π(1− π)zα/2 +√

π1(1− π1) + π2(1− π2)zβES

)2

(12.1)where π = π1+π2

2. The effect size, ES, is δπ.

The σ, which was included in δ in the earlier formula, does not appear inthis ES; it does appear in the numerator of the equation above.

usually α = 0.05, and β = 0.20 (that is, the power is 80%)so that zα/2 = z0.025 = 1.960 and zβ = z0.20 = 0.842


12.4.1.1 Sample size calculation example

In a clinical trial, where the usual success rate is 50%, we hope to improvethis by 10% to a new rate of 60%.Mathematically, this may be stated as π2 = 0.5 and π1 = 0.60, δ = 0.10 = ES(We label the two groups to that ES is positive and π = 0.50+0.60

2= 0.55

Substituting into the preceding equation, we get

n =

(

√

2(0.55)(0.45)(1.96) +√

0.50(0.50) + 0.60(0.40)(0.842)

0.10

)2

=

(

[√0.495]1.96 + [

√0.49]0.842

0.10

)2

=

(

1.3789 + 0.5894

0.10

)2

=

(

1.9683

0.10

)2

= 387.4528

that is 388 (close enough to 400)


12.4.1.2 An approximation

It is possible to assume that π1 and π2 are close enough to π that they canbe replaced in equation (12.1), which becomes

n =

(

√

2π(1− π)(zα/2 + zβ)

ES

)2

(12.2)

Using numbers from the previous example, equation (12.2) becomes

n =

(

√

2(0.55)(0.45)(1.96 + 0.842)

0.10

)2

= 388.6

which we round up to 389. This seems to be a conservative approximation.


12.4.2 Power for difference in proportions

Recall, that, for the difference in means this was calculated as

Pr

(

ZN > zα/) −|δA|σD̄

)

For proportions, this becomes

Pr

(

ZN > zα/) −|δπ|σD̄p

)

(12.3)

whereδπ = π1 − π2

and

σD̄p=

√

π1(1− π1)

n1

+π2(1− π2)

n2

which, for equal sample size, becomes

=

√

π1(1− π1) + π2(1− π2)

n

12.4.2.1 example: power for difference in proportions

Say, for the smoking-cold example,π1 = 0.65, π2 = 0.45, n1 = n2 = n = 25.Hence δπ = π1 − π2 = 0.65− 0.45 = 0.20

σD̄p=

√

0.65(0.35) + 0.45(0.55)

25= 0.1378

so that power is

Pr

(

ZN > z0.025 −0.20

0.1378

)

= Pr(ZN > 1.960− 1.4510) = Pr(ZN > 0.509)

= z0.509 = 0.305

Hence the chance of finding a difference of 0.20 with samples of size 25 is30%


12.4.2.2 Approximation

If, for the purposes of calculating σD̄p, we are prepared to assume that π1 =

π2 = π, then

σD̄p=

√

2(0.55(0.45)

25= 0.1407

so that the power is approximately

Pr

(

ZN > z0.025 −0.20

0.1407

)

= Pr(ZN > 1.960− 1.421) = Pr(ZN > 0.539)

= z0.539 = 0.2950

This is smaller than the more exact calculation, and is thus conservative(although this may not always be so).


12.5 One-sided alternatives

If we nave a one-side alternative, we can use the text baseed on p1− p2, thatis,

p1 − p2√

p(1− p)(

1n1

+ 1n2

)

where p1 and p2 are the proportions in the two samples, and p is the overallproportion

p =(X1 +X2)

(n1 + n2)

12.5.1 example of one-sided test

Suppose we have an alternative

HA : π1 > π2

ie, smokers are more likely to get colds than non-smokers.We have observed p1 = (15/23) = 0.65217; p2 = (10/22) = 0.45455and p = 25/45 = 0.55555so that

p = Pr

ZN >0.65217− 0.45455

√

0.55555(0.44444)(

123

+ 122

)

= Pr

(

ZN >0.19762

√

0.24691(0.08893)

)

= Pr

(

ZN >0.19762

0.14818

)

= Pr(ZN > 1.3336) = z1.3336 = 0.0912

Hence, at α = 0.05, cannot reject Ho.

12.5. ONE-SIDED ALTERNATIVES 31

12.5.2 Continuity correction

In this case, the test statistic is

(p1 − p2)−(

12n1

+ 12n2

)

√

p(1− p)(

1n1

+ 1n2

)

and

p = Pr

ZN >0.65217− 0.45455−

(

146

+ 144

)

√

0.55555(0.44444)(

123

+ 122

)

= Pr

(

ZN >0.15316

0.14818

)

= Pr(ZN > 1.0336) = z1.0336 ≈ 0.1506

Hence, at α = 0.05, we cannot reject Ho.


12.5.3 use of SY for one-sided alternatives

Recall that SY was developed for two-sided alternatives. In calculating thep-value, we used the formula

p = Pr(χ21 > x) = 2Pr(ZN >

√x)(12.7)

we actually wanted

Pr(χ21 > x) = Pr(ZN >

√x) + Pr(ZN < −

√x)

but becausePr(ZN >

√x) = Pr(ZN < −

√x)

we get equation (12.7)

For one-sided alternativesuse only one of

Pr(ZN >√x)

andPr(ZN < −

√x)

for example, for one-sided alternative, use

p = Pr(ZN >√

SY )

12.5.3.1 example: use of SY for one-sided test

For the two-sided alternative

pvalue = Pr(SY > 1.0668) = 2Pr(ZN > 1.0336)

but for the one-sided alternative

pvalue == 2Pr(ZN >√1.0668)

= Pr(ZN > 1.0336)

which is equivalent to the result using statistic p1 − p2...

12.6. PROOF OF EQUIVALENCE OF TESTS 33

12.6 Proof of Equivalence of tests

12.6.1 difference in proportions and SP

For testing hypotheses about the difference in probabilities (population pro-portions) we can use the test based on the difference in proportions, or wecan use SP .These two test statistics can be shown to be mathematically equivalent.Take the test based on the difference in proportions and square it.

(p1 − p2)2

p(1− p)(

1r1+ 1

r2

)

Rewrite this using the notation from contingency table

=

(

ar1− c

r2

)2

(

c1T

c2T

)

(

1r1+ 1

r2

)

=

(

ar2−cr1r1r2

)2

(

c1c2T 2

)

(

Tr1r2

)

=(ar2 − cr1)

2T

c1c2r1r2


A term in numerator can be written

ar2 − cr1 = a(c+ d)− c(a+ b)

= ac+ ad− ac− bc = ad− bc

so test statistic is

(ad− bc)2T

c1c2r1r2= SP

QED

12.6. PROOF OF EQUIVALENCE OF TESTS 35

12.6.2 Two versions of SP

We want to show the equivalence of usual version of SP to the epidemiological

version

SP =2∑

i=1

2∑

j=1

(

(Oij − Eij)2

Eij

)

=

(

a− r1c1T

)2

r1c1T

+

(

b− r1c2T

)2

r1c2T

+

(

c− r2c1T

)2

r2c1T

+

(

d− r2c2T

)2

r2c2T

=1

T

(aT − r1c1)2

r1c1+

(bT − r1c2)2

r1c2

+(cT − r2c1)

2

r2c1+

(dT − r2c2)2

r2c2

In numerator of first term

aT − r1c1 = a(r1 + r2)− r1(a+ c) = ar2 − cr1

= a(c+ d)− c(a+ b) = ad− bc

similarly for numerator of all the other termsNow we have

(ad− bc)2

T

(

1

r1c1+

1

r1c2+

1

r2c1+

1

r2c2

)


(ad− bc)2

T

(

r2c2 + r2c1 + r1c2 + r1c1r1r2c1c2

)

The numerator of second factor becomes

r2(c2 + c1) + r1(c2 + c1) = (r2 + r1)(c2 + c1) = T 2

hence(ad− bc)2T

r1r2c1c2

1. We have demonstrated the ”unadjusted” version of all three statisticsare mathematically equivalent

2. We could also show that the ”continuity correction unadjusted” versionof all three statistics are mathematically equivalentUse whichever of the three you find easiest for your calculations

3. this applies to two-sided andone-sided alternative hypotheses

12.7. USING SAS WITH 2X3 TABLES 37

12.7 Using SAS with 2x3 tables

For 2x2 tables,the exact test (Fisher’s) comes automatically; however, is ispossible to get exact probabilities for tables larger than 2X2 . One has toinsert the command EXACT into Proc FREQ

title "get exact pvalue for 2x3 table";

options ls=64 ps=24;

data exact;

input r c n;

datalines;

1 1 147

1 2 435

1 3 256

2 1 147

2 2 392

2 3 323

;

proc freq ORDER=DATA;

weight n;

table r*c/CHISQ NOROW NOCOL NOPERCENT EXACT;


Here is the output to this calculation:

get exact pvalue for 2x3 table

The FREQ Procedure

Table of r by c

r c

Frequency| 1| 2| 3| Total

---------------------------------------------

1 | 147 | 435 | 256 | 838

---------------------------------------------

2 | 147 | 392 | 323 | 862

----------------------------------------------

Total | 294 | 827 | 579 | 1700

Statistics for Table of r by c

Statistic DF Value Prob

------------------------------------------------

Chi-Square 2 9.6519 0.0080

Likelihood Ratio Chi-Square 2 9.6684 0.0080

Mantel-Haenszel Chi-Square 1 4.8042 0.0284

Phi Coefficient 0.0753

Contingency Coefficient 0.0751

Cramer’s V 0.0753

Fisher’s Exact Test

----------------------------------

Table Probability (P) 1.771E-05

Pr <= P 0.0080

Sample Size = 1700

12.8 Questions

see end of next chapter

Chapter 12 Inference about two probabilities -...

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