Chapter 14 Tests of Hypotheses Based on Count Data

14.2 Tests concerning proportions (large samples)

14.3 Differences between proportions 14.4 The analysis of an r x c table

14.2 Tests concerning proportions (large samples)

np>5; n(1-p)>5 n independent trials; X=# of successes p=probability of a success Estimate:

n

Xp ˆ

Tests of Hypotheses

Null H0: p=p0

Possible Alternatives:

HA: p<p0

HA: p>p0

HA: pp0

Test Statistics

Under H0, p=p0, and

Statistic:

is approximately standard normal under H0 .

Reject H0 if z is too far from 0 in either direction.

n

pp

n

ppp

)1()1( 00ˆ

npp

ppppz

p )1(

ˆˆ

00

0

ˆ

0

Rejection Regions

Alternative Hypotheses

HA: p>p0 HA: p<p0 HA: pp0

Rejection Regions

z>z z<-z z>z/2 or

z<-z/2

Equivalent Form:

0 0

0 0 0 0

ˆ*

(1 ) (1 )

p p X npnz

np p np p

n

Example 14.1

H0: p=0.75 vs HA: p0.75

=0.05 n=300 x=206 Reject H0 if z<-1.96 or z>1.96

68667.0300

206ˆ

n

Xp

Observed z value

Conclusion: reject H0 since z<-1.96 P(z<-2.5 or z>2.5)=0.0124<reject H0.

0.68667 0.752.5

0.75(1 0.75)300

206 2252.5

300(0.75)(1 0.75)

z

or

z

Example 14.2

Toss a coin 100 times and you get 45 heads Estimate p=probability of getting a head

Is the coin balanced one? =0.05

Solution:

H0: p=0.50 vs HA: p0.50

45.0100

45ˆ p

Enough Evidence to Reject H0?

Critical value z0.025=1.96 Reject H0 if z>1.96 or z<-1.96

Conclusion: accept H0

15

5

)50.01)(50.0(100

)50.0(10045

z

Another example

The following table is for a certain screening test

91010Results Negative

80140Result Positive

Cancer Absent

Cancer Present

FNA status

Truth = surgical biopsy

Total

220

920

Total 150 990 1140

True positive 140sensitivity 0.93

True Positives False Negatives 150

Test to see if the sensitivity of the screening test is less than 97%.

Hypothesis

Test statistic

0 0

0

: .97

: .97

H p p

Ha p p

0 0

ˆ 0 0

estimated proportion-prestated proportion

standard error of the estimated proportion

ˆ ˆ 140 150 .972.6325

(1 ) .97 (1 .97)150

p

z

p p p p

SE p p

n

Check p-value when z=-2.6325, p-value = 0.004

Conclusion: we can reject the null hypothesis at level 0.05.

What is the conclusion?

One word of caution about sample size: If we decrease the sample size by a factor of 10,

911Results Negative

814Result Positive

Cancer Absent

Cancer Present

FNA status

Truth = surgical biopsy

Total

22

92

Total 15 99 114

True positive 14sensitivity 0.93

True Positives False Negatives 15

And if we try to use the z-test,

0 0

ˆ 0 0

estimated proportion-prestated proportion

standard error of the estimated proportion

ˆ ˆ 14 15 .970.8324

(1 ) .97 (1 .97)15

p

z

p p p p

SE p p

n

P-value is greater than 0.05 for sure (p=0.2026). So we cannot reach the same conclusion.

And this is wrong!

So for test concerning proportions

We want

np>5; n(1-p)>5

14.3 Differences Between Proportions

Two drugs (two treatments) p1 =percentage of patients recovered after

taking drug 1 p2 =percentage of patients recovered after

taking drug 2 Compare effectiveness of two drugs

Tests of Hypotheses

Null H0: p1=p2 (p1-p2 =0)

Possible Alternatives:

HA: p1<p2

HA: p1>p2

HA: p1p2

Compare Two Proportions

Drug 1: n1 patients, x1 recovered

Drug 2: n2 patients, x2 recovered

Estimates: Statistic for test:

If we did this study over and over and drew a histogram of the resulting values of , that histogram or distribution would have standard deviation

1 2ˆ ˆp p

2

22

1

11 ˆ;ˆ

n

xp

n

xp

21 ˆˆ

21 ˆˆ

pp

ppz

1 2ˆ ˆp p

Estimating the Standard Error

Under H0, p1=p2=p. So

Estimate the common p by

)11

)(1(21

)1()1(2ˆ

2ˆ

2ˆˆ 2

22

1

11

2121

nnpp

npp

npp

pppp

21

21

21

2211 ˆˆˆ

nn

xx

nn

pnpnp

So put them together

1 2

1 2

ˆ ˆ

1 1ˆ ˆ(1 )( )

p pz

p pn n

Example 12.3

Two sided test: H0: p1=p2 vs HA: p1p2

n1=80, x1=56

n2=80, x2=38

7.0ˆ1 p 475.0ˆ 2 p 5875.08080

3856ˆ

p

Two Tailed Test

Observed z-value:

Critical value for two-tailed test: 1.96 Conclusion: Reject H0 since |z|>1.96

88.2078.0

225.0

)801

801

(4125.05875.0

475.07.0

z

Rejection Regions

Alternative Hypotheses

HA: p1>p2

HA: p1<p2

HA: p1p2

Rejection Regions

z>z z<-z z>z/2 or

z<-z/2

P-value of the previous example

P-value=P(z<-2.88)+P(z>2.88)=2*0.004

So not only we can reject H0 at 0.05 level, we can also reject at 0.01 level.

14.4 The analysis of an r x c table

Recall Example 12.3 Two sided test; H0: p1=p2 vs HA: p1p2

n1=80, x1=56 n2=80, x2=38

We can put this into a 2x2 table and the question now becomes is there a relationship between treatment and outcome? We will come back to this example after we introduce 2x2 tables and chi-square test.

Recover Not Rec

Treat 1 56 24 | 80

Treat 2 38 42 | 80

94 66 | 160

2x2 Contingency Table

The table shows the data from a study of 91 patients who had a myocardial infarction (Snow 1965). One variable is treatment (propranolol versus a placebo), and the other is outcome (survival for at least 28 days versus death within 28 days).

Treatment

1729

738

OUTCOME

Survival for at least 28 days Death

Propranolol

Placebo

Total

45

46

91Total 67 24

Hypotheses for Two-way TablesHypotheses for Two-way Tables

The hypotheses for two-way tables are very “broad stroke”.

• The null hypothesis H0 is simply that there is no association between the row and column variable.

• The alternative hypothesis Ha is that there is an association between the two variables. It doesn’t specify a particular direction and can’t really be described as one-sided or two-sided.

Hypothesis statement in Our Example

Null hypothesis: the method of treating the myocardial infarction patients did not influence the proportion of patients who survived for at least 28 days.

The alternative hypothesis is that the outcome (survival or death) depended on the treatment, meaning that the outcomes was the dependent variable and the treatment was the independent variable.

Calculation of Expected Cell Count

To test the null hypothesis, we compare the observed cell counts (or frequencies) to the expected cell counts (also called the expected frequencies)

The process of comparing the observed counts with the expected counts is called a goodness-of-fit test. (If the chi-square value is small, the fit is good and the null hypothesis is not rejected.)

TotalStudy

TotalColumnTotalRowE

111,1

Observed cell counts

Treatment1729

738

OUTCOMESurvival for at least 28 days

Death

Propranolol

Placebo

Total

45

4691Total 67 24

Treatment

12.1333.87

11.8733.13

OUTCOMESurvival for at least 28 days

Death

Propranolol

Placebo

Total

45

4691Total 67 24

Expected cell counts

The Chi-Square ( The Chi-Square ( 22) Test Statistic) Test Statistic

The chi-square statistic is a measure of how much the observed cell counts in a two-way table differ from the expected cell counts. It can be used for tables larger than 2 x 2, if the average of the expected cell counts is > 5 and the smallest expected cell count is > 1; and for 2 x 2 tables when all 4 expected cell counts are > 5. The formula is:

2 = (observed count – expected count)2/expected count

Degrees of freedom (df) = (r –1) x (c – 1)

Where “observed” is an observed sample count and “expected” is the computed expected cell count for the same cell, r is the number of rows, c is the number of columns, and the sum () is over all the r x c cells in the table (these do not include the total cells).

The Chi-Square ( The Chi-Square ( 22) Test Statistic) Test Statistic

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Example: Patient Compliance w/ RxExample: Patient Compliance w/ Rx

TreatmentCompliance 10 mg Daily 5 mg bid Total

95%+ 46 40 86

< 95% 4 10 14

Total 50 50 100

In a study of 100 patients with hypertension, 50 were randomly allocated to a group prescribed 10 mg lisinopril to be taken once daily, while the other 50 patients were prescribed 5 mg lisinopril to be taken twice daily. At the end of the 60 day study period the patients returned their remaining medication to the research pharmacy. The pharmacy then counted the remaining pills and classified each patient as < 95% or 95%+ compliant with their prescription. The two-way table for Compliance and Treatment was:

Example: Patient Compliance w/ RxExample: Patient Compliance w/ Rx

TreatmentCompliance 10 mg Daily 5 mg bid Total95%+ 460 400 860< 95% 40 100 140Total 500 500 1000

TreatmentCompliance 10 mg Daily 5 mg bid Total95%+ 430 430 860< 95% 70 70 140Total 500 500 1000

The expected cell counts were:

2 = 29.9, df = (2-1)*(2-1) = 1, P-value <0.001

If we use the two sample test for proportion

1 2

1 2

1 2

2

2 2(1)

460 400 860ˆ ˆ ˆ, ,

500 500 1000ˆ ˆ

5.468121 1

ˆ ˆ(1 )( )

5.46812 29.9

p p p

p pz

p pn n

z

The The 22 and and zz Test Statistics Test StatisticsThe comparison of the proportions of “successes” in two populations leads to a 2 x 2 table, so the population proportions can be compared either using the 2 test or the two-sample z test .

For a 2-sided test, it really doesn’t matter, because • they always give exactly the same result, because the 2 is equal

to the square of the z statistic and • the chi-square with one degree of freedom 2

(1) critical values are equal to the squares of the corresponding z critical values.

• A P-value for the 2 x 2 2 can be found by calculating the square root of the chi-square, looking that up in Table for P(Z > z) and multiplying by 2, because the chi-square always tests the two-sided alternative.

The The 22 and and zz Test Statistics Test Statistics

• For a 2 x 2 table with a one-sided alternative:

• The 1-sided two-sample z statistic could to be used.

• The chi-square p-value could be modified as•1-sided p-value = 0.5*(2-sided p-value) if the observed difference is in the direction of the alternative

•HA: p1< p2 and

•1-sided p-value = 1 - 0.5*(2-sided p-value) if the observed difference is away from the alternative, e.g.

• The chi-square is the one most often seen in the literature.

21 ˆˆ pp

21 ˆˆ pp

Summary: Computations for Two-way TablesSummary: Computations for Two-way Tables

1. Create the table, including observed cell counts, column and row totals.

2. Find the expected cell counts.

• Determine if a 2 test is appropriate.

• Calculate the 2 statistic and number of degrees of freedom.

3. Find the approximate P-value

• Use Table III chi-square table to find the approximate P-value.

• Or use z-table and find the two-tailed p-value if it is 2 x 2.4. Draw conclusions about the association between the row and

column variables.

Yates Correction for Continuity

The chi-square test is based on the normal approximation of the binomial distribution (discrete), many statisticians believe a correction for continuity is needed.

It makes little difference if the numbers in the table are large, but in tables with small numbers it is worth doing.

It reduces the size of the chi-square value and so reduces the chance of finding a statistically significant difference, so that correction for continuity makes the test more conservative.

E

EOYates

22 )5.0|(|

What do we do if the expected values in any of the cells in a 2x2

table is below 5?For example, a sample of teenagers might be divided into male and female on the one hand, and those that are and are not currently dieting on the other. We hypothesize, perhaps, that the proportion of dieting individuals is higher among the women than among the men, and we want to test whether any difference of proportions that we observe is significant. The data might look like this:

diet no diet total

men 1 11 12

women 9 3 12

totals 10 14 24

The question we ask about these data is: knowing that 10 of these 24 teenagers are dieters, what is the probability that these 10 dieters would be so unevenly distributed between the girls and the boys? If we were to choose 10 of the teenagers at random, what is the probability that 9 of them would be among the 12 girls, and only 1 from among the 12 boys?

--Hypergeometric distribution!

--Fisher’s exact test uses the hypergeometric distribution to calculate the “exact” probability of obtaining such set of the values.

Fisher’s exact test

Before we proceed with the Fisher test, we first introduce some notation. We represent the cells by the letters a, b, c and d, call the totals across rows and columns marginal totals, and represent the grand total by n. So the table now looks like this:

men women total

dieting a b a + b

not dieting

c d c + d

totals a + c b + d n

Fisher showed that the probability of obtaining any such set of values was given by the hypergeometric distribution:

                                                                     

diet no diet total

men a b a + b

women c d c + d

totals a + c b + d n

In our example

10!14!12!12!0.00134

24!1!9!11!3!p

HA: pM= pW HA: pM< pW

Recall that p-value is the probability of observing data as extreme or more extreme if the null hypothesis is true. So the p-value is this problem is 0.00137.

10!14!12!12!0.00003

24!0!10!12!2!p

241410totals

1239women

12111men

totalno dietdiet

241212totals

12210women

12120men

totalno dietdiet

As extreme

as observed

More extreme than

observed

Two Sided Test

HA: pM= pW HA: pM≠ pW

Extreme observations include either mostly women dieters or mostly men.

Since the numbers of men and women are equal, probabilities remain the same if we interchange men and women. So the p-value is this problem is 2*0.00137.

241410totals

12women

121, 11men

totalno dietdiet

241212totals

12women

120, 12men

totalno dietdiet

As extreme

as observed

More extreme than

observed

The Fisher Exact Probability Test Used when one or more of the expected counts in a contingency

table is small (<2). Fisher's Exact Test is based on exact probabilities from a

specific distribution (the hypergeometric distribution). There's really no lower bound on the amount of data that is

needed for Fisher's Exact Test. You can use Fisher's Exact Test when one of the cells in your table has a zero in it. Fisher's Exact Test is also very useful for highly imbalanced tables. If one or two of the cells in a two by two table have numbers in the thousands and one or two of the other cells has numbers less than 5, you can still use Fisher's Exact Test.

Fisher's Exact Test has no formal test statistic and no critical value, and it only gives you a p-value.

Does pregnancy affect outcome of methadone maintenance treatment? Does pregnancy affect outcome of methadone

maintenance treatment?

Journal of Substance Abuse Treatment, June 2004, 295-303

51 pregnant

51 non-pregnant

Pick 3 randomly for 3 Axis 1 Disorder Cases

2426.02

1213.0100

49

101

50

102

51)3(

1213.0100

49

101

50

102

51)0(

#

valuepsided

XP

XP

pregnantX

0 1 2 3

0.0

0.1

0.2

0.3

0.1213

0.3787 0.3787

0.1213

Observed X=0

Adinoma Tumors in Mice Genetic basis of variation in adenoma multiplicity in ApcMin/+

Mom1S mice. Proceedings of the National Academy of Science, February 22, 2005, 2868-2873

“The results (Table 5) indicated that the frequency of allele loss was significantly higher in line V (96%) than in line I (77%)

P = 0.003, Fisher's exact test).”

Table 5. Frequency of allele loss of WT Apc in lines

Type Number Percent     

Line MinI/+I 46/60 77%     

Line MinV/+V 52/54 96%

Loss No Loss

Line I 46 14 | 60

Line V 52 X=2 | 54

98 16 | 114X P(X)0 0.000012 1 0.000223 2 0.001925 3 0.009938 4 0.034317

12 0.012970 13 0.002941 14 0.000445 15 0.000040 16 0.000002

P 0.002647

More extreme values are values with smaller probability.

Observed count

0 1 2 3 4 etc etc 12 13 14 15 16

0.0

0.01

0.02

0.03