Biostat 200 Lecture 8 1. Hypothesis testing recap Hypothesis testing – Choose a null hypothesis,...

Biostat 200 Lecture 8

1

Hypothesis testing recap

• Hypothesis testing– Choose a null hypothesis, one-sided or two sided

test– Set , significance level, to set the probability of a

Type I error ( P(reject H0 | H0 )

– For a given test, a test statistic is calculated, e.g. for a two-sample t-test the test statistic is:

2

11

)()(

212

2121

nns

xxt

p

stat

• For an ANOVA, the test statistic is:

• For a Wilcoxon rank sum/Mann Whitney test, the test statistic is:

3

2

2

W

B

s

sF

12

)1(

2

)1(

LsLsW

LssW

W

Ww

nnnnand

nnnwhere

Wz

• These test statistics are derived to follow the corresponding theoretical distribution (tstat follows the t distribution, F statistic follows the F distribution, zw follows the Standard Normal) if certain assumptions are met.

• These assumptions are: – For ttest and ANOVA, the underlying distribution of the

random variable being measured (X) should be approximately normal

• In reality the t-test is rather robust, so with large enough sample size and without very large outliers, it is ok to use the t-test

– For the ANOVA, the variance of the subgroups should be approximately equal (Barlett’s test on Stata output)

– For the Wilcoxon Rank Sum Test the underlying distributions must have the same basic shape 4

• One hypothesis test will be “more conservative” than another if that test is less likely to reject the null– A test with a lower level of is more conservative,

e.g. =0.01, sometimes used in clinical trials– A two-sided test is more conservative than a one-

sided test, because even though you are using the same total level, it is divided between the two tails

– If the assumptions of a parametric test are met or are not grossly violated, then a non-parametric test is more conservative than the corresponding parametric test

5

ANOVA and t-test for 2 groups. ttest extot, by( sex)

Two-sample t test with equal variances------------------------------------------------------------------------------ Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]---------+-------------------------------------------------------------------- male | 295 114.9458 7.258138 124.6626 100.6613 129.2303 female | 237 152.1498 11.27012 173.5014 129.9469 174.3527---------+--------------------------------------------------------------------combined | 532 131.5197 6.478136 149.419 118.7938 144.2457---------+-------------------------------------------------------------------- diff | -37.20403 12.94578 -62.63536 -11.77269------------------------------------------------------------------------------ diff = mean(male) - mean(female) t = -2.8738Ho: diff = 0 degrees of freedom = 530

Ha: diff < 0 Ha: diff != 0 Ha: diff > 0 Pr(T < t) = 0.0021 Pr(|T| > |t|) = 0.0042 Pr(T > t) = 0.9979

. oneway extot sex

Analysis of Variance Source SS df MS F Prob > F------------------------------------------------------------------------Between groups 181902.478 1 181902.478 8.26 0.0042 Within groups 11673228.1 530 22024.9586------------------------------------------------------------------------ Total 11855130.5 531 22326.0462

Bartlett's test for equal variances: chi2(1) = 28.7299 Prob>chi2 = 0.0006

When there are 2 groups, the F-statistic equals the t-statistic squared

Wilcoxon rank sum Kruskal Wallis. ranksum extot, by(sex)

Two-sample Wilcoxon rank-sum (Mann-Whitney) test

sex | obs rank sum expected-------------+--------------------------------- male | 295 74838.5 78617.5 female | 237 66939.5 63160.5-------------+--------------------------------- combined | 532 141778 141778

unadjusted variance 3105391.25adjustment for ties -37529.45 ----------adjusted variance 3067861.80

Ho: extot(sex==male) = extot(sex==female) z = -2.158 Prob > |z| = 0.0310

. kwallis extot, by(sex)

Kruskal-Wallis equality-of-populations rank test

+-------------------------+ | sex | Obs | Rank Sum | |--------+-----+----------| | male | 295 | 74838.50 | | female | 237 | 66939.50 | +-------------------------+

chi-squared = 4.599 with 1 d.f.probability = 0.0320

chi-squared with ties = 4.655 with 1 d.f.probability = 0.0310

7

When there are two groups, the chi-square statistic is equal to the z statistic squared (here slightly different because of ties)

More on categorical outcomes

• With the exception of the proportion test, all the previous tests were for comparing continuous outcomes and categorical predictors– E.g., CD4 count by alcohol consumption– Minutes of exercise by sex

• We often have dichotomous outcomes and predictors– E.g. Had at least one cold in the prior 3 months by

sex8

• We can make tables of the number of observations falling into each category

• These are called contingency tables• E.g. At least one cold by sex

. tab coldany sex

At least | one cold, | Biological sex at prior 3 | birth months | male female | Total-----------+----------------------+---------- no | 131 100 | 231 yes | 166 140 | 306 -----------+----------------------+---------- Total | 297 240 | 537

9

Contingency tables• Often summaries of counts of disease versus no disease and

exposed versus not exposed• Frequently 2x2 but can generalize to n x k

– n rows, k columns• Note that Stata sorts on the numeric value, so for 0-1

variables the disease state will be the 2nd row

Exposure

+ - Total

Disease + a b a+b

- c d c+d

Total a+c b+d n=a+b+c+d

Pagano and Gavreau, Chapter 15 10

Contingency tables• Contingency tables

are usually summaries of data that originally looked like this.

Example of data set

Obs. Exposure (1=yes; 0=no)

Disease (1=yes; 0=no)

1 1 1

2 1 0

3 1 1

4 0 0

5 1 1

6 1 0

7 0 0

… … …

n 0 0

Pagano and Gavreau, Chapter 15 11

. list coldany sex

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

| coldany sex |

|------------------|

1. | yes male |

2. | no male |

3. | yes female |

4. | yes female |

5. | no male |

|------------------|

6. | no male |

7. | no male |

8. | yes male |

9. | yes male |

10. | yes male |

|------------------|

11. | no female |

12. | yes male |

13. | no male |

14. | yes female |

15. | no female |

|------------------|

16. | yes female |

. list coldany sex, nolabel

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

| coldany sex |

|---------------|

1. | 1 0 |

2. | 0 0 |

3. | 1 1 |

4. | 1 1 |

5. | 0 0 |

|---------------|

6. | 0 0 |

7. | 0 0 |

8. | 1 0 |

9. | 1 0 |

10. | 1 0 |

|---------------|

11. | 0 1 |

12. | 1 0 |

13. | 0 0 |

14. | 1 1 |

15. | 0 1 |

|---------------|

16. | 1 1 |

12

• We want to know whether the incidence of colds varies by gender.

• We could test the null hypothesis that the cumulative incidence of ≥1 cold in males equals that of females. The cumulative incidence is a proportion.

H0: pmales= pfemales HA: pmales≠ pfemales

13

. prtest coldany, by(sex)

Two-sample test of proportion male: Number of obs = 297

female: Number of obs = 240

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

Variable | Mean Std. Err. z P>|z| [95% Conf. Interval]

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

male | .5589226 .0288108 .5024545 .6153907

female | .5833333 .0318234 .5209605 .6457061

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

diff | -.0244108 .0429278 -.1085476 .0597261

| under Ho: .042973 -0.57 0.570

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

diff = prop(male) - prop(female) z = -0.5680

Ho: diff = 0

Ha: diff < 0 Ha: diff != 0 Ha: diff > 0

Pr(Z < z) = 0.2850 Pr(|Z| < |z|) = 0.5700 Pr(Z > z) = 0.7150

14

• There are other methods to do this (chi-square test)

• Why?– These methods are more general – can be used

when you have more than 2 levels in either variable

• We will start with the 2x2 example however

15

• Overall, the cumulative incidence of least one cold in the prior 3 months is 306/537=.569. This is the marginal probability of having a cold

• There were 297 males and 240 females• Under the null hypothesis, the expected

cumulative incidence in each group is the overall cumulative incidence

• So we would expect 297*.569=169.2 with at least one cold in the males, and 240*.569=136.8 with at least one cold in the females

16


• We can also calculate the expected number with no colds under the null hypothesis of no difference– Males: 297*(1-.569) = 127.8– Females: 240*(1-.569) = 103.2

• We can make a table of the expected counts

17

Observed data


EXPECTED COUNTS UNDER THE NULL HYPOTHESIS

At least | one cold, | Biological sex at prior 3 | birth months | male female | Total-----------+----------------------+---------- no | 127.8 103.2 | 231 yes | 169.2 136.8 | 306 -----------+----------------------+---------- Total | 297 240 | 537

• Generically

18

Expected counts

Exposure

+ - Total

Disease + (a+b)(a+c)/n (a+b)(b+d)/n a+b

- (c+d)(a+c)/n (c+d)(b+d)/n c+d


• The Chi-square test compares the observed frequency (O) in each cell with the expected frequency (E) under the null hypothesis of no difference

• The differences O-E are squared, divided by E, and added up over all the cells

• The sum of this is the test statistic and follows a chi-square distribution

19

Chi-square test of independence

• The chi-square test statistic (for the test of independence in contingency tables) for a 2x2 table (dichotomous outcome, dichotomous exposure)

• i is the index for the cells in the table – there are 4 cells• This test statistic is compared to the chi-square distribution

with 1 degree of freedom

4

1

221

)(i

i

ii

E

EO

Pagano and Gavreau, Chapter 1520


• The chi-square test statistic for the test of independence in an nxk contingency table is

• This test statistic is compared to the chi-square distribution• The degrees of freedom for the this test are (n-1)*(k-1), so for

a 2x2 there is 1 degree of freedom– n=the number of rows; k=the number of columns in the nxk table– The chi-square distribution with 1 degree of freedom is actually the

square of a standard normal distribution• Expected cell sizes should all be >1 and <20% should be <5• The Chi-square test is for two sided hypotheses


nk

ii

iikn E

EO1

22

)1(*)1(

)(

Chi-square distribution

22

Chi-square distribution

23


• For the example, the chi-square statistic for our 2x2 is (131-127.8)2 /127.8 + (100-103.2)2 /103.2 + (166-169.2)2 /169.2 + (140-136.8)2 /136.8 = .323

• There is 1 degree of freedom• Probability of observing a chi-square value with 1

degree of freedom of .323 is .570. di chi2tail(1,.323).56981031

• Fail to reject the null hypothesis of independence


If you want to see the row or column percentages, use row or col options

tab coldany sex, row col chi+-------------------+| Key ||-------------------|| frequency || row percentage || column percentage |+-------------------+

At least | one cold, | Biological sex at prior 3 | birth months | male female | Total-----------+----------------------+---------- no | 131 100 | 231 | 56.71 43.29 | 100.00 | 44.11 41.67 | 43.02 -----------+----------------------+---------- yes | 166 140 | 306 | 54.25 45.75 | 100.00 | 55.89 58.33 | 56.98 -----------+----------------------+---------- Total | 297 240 | 537 | 55.31 44.69 | 100.00 | 100.00 100.00 | 100.00

Pearson chi2(1) = 0.3227 Pr = 0.570

26

• Because we using discrete cell counts to approximate a chi-squared distribution, for 2x2 tables some use the Yates correction

• Not computed in Stata

27

4

1

221

)5.0|(|i

i

ii

E

EO

Lexicon

• When we talk about the chi-square test, we are saying it is a test of independence of two variables, usually exposure and disease.

• We also say we are testing the “association” between the two variables.

• If the test is statistically significant (p<0.05), we often say that the two variables are not independent or we say the association is statistically significant.

28

Test of independence• For small cell sizes in 2x2 tables, use the Fisher exact test• It is based on a discrete distribution called the hypergeometric

distribution• For 2x2 tables, you can choose a one-sided or two-sided test

. tab coldany sex, chi exact

At least |


prior 3 | birth


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

no | 131 100 | 231

yes | 166 140 | 306

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

Total | 297 240 | 537

Pearson chi2(1) = 0.3227 Pr = 0.570

Fisher's exact = 0.599

1-sided Fisher's exact = 0.316



• The chi-square test can be used for more than 2 levels of exposure– The null hypothesis is p1 = p2 = ... = pc

– The alternative hypothesis is is that not all the proportions are the same

• Note that, like ANOVA, a statistically significant result does not tell you which level differed from the others

• Also when you have more than 2 groups, all tests are 2-sided


Chi-square test of independence tab coldany racegrp, chi col

+-------------------+| Key ||-------------------|| frequency || column percentage |+-------------------+

At least | one cold, | prior 3 | racegrp months | White, Ca Asian/PI Other | Total-----------+---------------------------------+---------- no | 132 71 30 | 233 | 42.31 44.65 44.12 | 43.23 -----------+---------------------------------+---------- yes | 180 88 38 | 306 | 57.69 55.35 55.88 | 56.77 -----------+---------------------------------+---------- Total | 312 159 68 | 539 | 100.00 100.00 100.00 | 100.00

Pearson chi2(2) = 0.2614 Pr = 0.877


• Another way to state the null hypothesis for the chi-square test:– Factor A is not associated with Factor B

• The alternative is– Factor A is associated with Factor B

• For more than 2 levels of the outcome variable this would make the most sense

32

. tab auditc_cat racegrp, chi exact col

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

| Key | |-------------------|| frequency || column percentage |+-------------------+

| racegrp auditc_cat | White, Ca Asian/PI Other | Total-------------------+---------------------------------+---------- no alcohol | 30 41 13 | 84 | 9.62 25.79 19.12 | 15.58 -------------------+---------------------------------+---------- low risk | 141 75 25 | 241 | 45.19 47.17 36.76 | 44.71 -------------------+---------------------------------+----------at risk, or higher | 141 43 30 | 214 | 45.19 27.04 44.12 | 39.70 -------------------+---------------------------------+---------- Total | 312 159 68 | 539 | 100.00 100.00 100.00 | 100.00

Pearson chi2(4) = 28.6067 Pr = 0.000 Fisher's exact = 0.000

33

Note that this is a 3x3 table, so the chi-square test has 2x2=4 degrees of freedom

What is the null hypothesis?

Paired data?

• Matched pairs– Matched case-control study– Before and after data

• E.g. Self-reported alcohol consumption before and after being consented for alcohol biomarker specimen collection

34

Self-reported 3 months alcohol consumption

Cohort study (before)

Sub-study (after) Total

Yes 12 25 37

No 50 37 87

Total 62 62 124

35

But there really are only 62 pairs!

Self-reported alcohol consumption in Uganda

McNemar’s test – correct table

36

• Null hypothesis: The groups change their self-reported alcohol consumption equally; there is no association between self-reported alcohol consumption and before versus after measures

• The concordant pairs provide no information

After measure Before measure Total

Alcohol consumption prior 3 months Yes No

Yes 12 13 25

No 0 37 37

Total 12 50 62

• The test statistic for NcNemar’s test is

• r and s represent the discordant cell counts• This statistic has an approximate chi-square

distribution with 1 degree of freedom• The -1 is a continuity correction, not all versions

of the test use this, some use .5• For our example Χ2 = (13-1)2/13 = 11.08• Compare to chi-square distribution, df=1. di chi2tail(1,11.076923).00087409

Reject the null• For small samples (r+s<25), use exact methods

37

)(

]1|[| 22

sr

sr

Matched case-control study command will do the same

. mcci 12 13 0 37

| Controls |

Cases | Exposed Unexposed | Total

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

Exposed | 12 13 | 25

Unexposed | 0 37 | 37

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

Total | 12 50 | 62

McNemar's chi2(1) = 13.00 Prob > chi2 = 0.0003

Exact McNemar significance probability = 0.0002

Proportion with factor

Cases .4032258

Controls .1935484 [95% Conf. Interval]

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

difference .2096774 .0922202 .3271346

ratio 2.083333 1.385374 3.132929

rel. diff. .26 .138419 .381581

odds ratio . 3.04772 . (exact)

38

The odds ratio r/s is not calcuable here because the denominator is 0

Case-control study

39

Controls Total

Cases

Alcohol consumption prior 3 months?

Yes (exposed) No (not exposed)

Yes (exposed) 4 9 13

No (not exposed) 3 11 14

Total 7 20 27

•Cases: Treatment failure: HIV viral load after 6 months of ART >400•Controls: HIV viral load <400Matched on sex, duration on treatment, and treatment regimen class

. mcc lastalc_case lasttime_alc_3mos

| Controls |Cases | Exposed Unexposed | Total-----------------+------------------------+------------ Exposed | 4 9 | 13 Unexposed | 3 11 | 14-----------------+------------------------+------------ Total | 7 20 | 27

McNemar's chi2(1) = 3.00 Prob > chi2 = 0.0833Exact McNemar significance probability = 0.1460

Proportion with factor Cases .4814815 Controls .2592593 [95% Conf. Interval] --------- -------------------- difference .2222222 -.0518969 .4963413 ratio 1.857143 .9114712 3.78397 rel. diff. .3 .0159742 .5840258

odds ratio 3 .7486845 17.228 (exact)

40

Comparison of disease frequencies across groups

• The chi-square test is a test of independence• It does not give us an estimate of how much the two

groups differ, i.e. how much the disease outcome varies by the exposure variable

• We use odds ratios (OR) and relative risks (RR) as measures of ratios of disease outcome

• The odds ratio and the relative risk are just two examples of “measures of association”

41

Comparison of disease frequencies – relative risk

Exposure

Disease + - Total

+ a b a+b

- c d c+d


Risk ratio (or relative risk or relative rate) = P (disease | exposed) / P(disease | unexposed)= Re / Ru = a/(a+c) / b/(b+d)

42

Comparison of disease frequencies – relative risk

Note that you cannot calculate this entity when you have chosen your sample based on disease status

I.e. Case-control study – you have fixed a prior the probability of disease! Relative risk is a NO GO!

Exposure

Disease + - Total

+ a b a+b

- c d c+d


43

Odds

• If an event occurs with probability p, the odds of the event are p/(1-p) to 1

• If an event has probability .5, the odds are 1:1• Conversely, if the odds of an event are a:b, the

probability of a occurring is a/(a+b)– The odds of horse A winning over horse B winning

are 2:1 the probability of horse A winning is .667.

44

Odds ratio

Odds of disease among the exposed persons = P(disease | exposed) / (1-P(disease | exposed))= [ a / (a + c) ] / [ c / (a + c) ] = a/c

Odds of disease among the unexposed persons = P(disease | unexposed) / (1-P(disease | unexposed))

= [ b / (b + d) ] / [ d / (b + d) ] = b/d Odds ratio = a/c / b/d = ad/bc

Exposure

Disease + - Total

+ a b a+b

- c d c+d


45

Odds ratio note

• Note that the odds ratio is also equal to [ P(exposed | disease)/(1-P(exposed |disease) ] / [ P(exposed | no disease)/(1-P(exposed | no disease) ]

• This is needed for case-control studies in which the proportion with disease is fixed (so you can’t calculate the odds of disease)

46

Interpretation of ORs and RRs

• If the OR or RR equal 1, then there is no effect of exposure on disease.

• If the OR or RR >1 then disease is increased in the presence of exposure. (Risk factor)

• If the OR or RR <1 then disease is decreased in the presence of exposure. (Protective factor)

47

Comparison of measures of association

When a disease is rare, i.e. the risk is <10%, the odds ratio approximates the risk ratio

The odds ratio overestimates the risk ratio Why use it? – statistical properties, usefulness in case-

control studies

48

The association of having at least one cold with gender

At least |


prior 3 | birth


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

no | 131 100 | 231

yes | 166 140 | 306

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

Total | 297 240 | 537

What is the (estimated) odds ratio?

49

95% Confidence interval for an odds ratio

• Remember the 95% confidence interval for a mean µLower Confidence Limit: Upper Confidence Limit:

• The odds ratio is not normally distributed (it ranges from 0 to infinity)– But the natural log (ln) of the odds ratio is approximately normal– The estimate of the standard error of the estimated ln OR is

– This is based on a Taylor series approximation

nX /96.1_

nX /96.1_

dcbaORSE

1111)(ln(

50

95% Confidence interval for an odds ratio

• We calculate the 95% confidence interval for the log odds

• Then exponentiate back to obtain the 95% confidence interval for the OR

51

dcba

ORdcba

OR1111

96.1ln,1111

96.1ln

dcbaOR

dcbaOR

ee1111

96.1ln1111

96.1ln

,

Calculating an odds ratio and 95% confidence interval in Stata using tabodds

command

se ln OR1a

1b

1c

1d

Tabodds outcomevar exposurevar , or

. tabodds coldany sex, or

--------------------------------------------------------------------------- sex | Odds Ratio chi2 P>chi2 [95% Conf. Interval]-------------+------------------------------------------------------------- male | 1.000000 . . . . female | 1.104819 0.32 0.5704 0.782925 1.559057---------------------------------------------------------------------------Test of homogeneity (equal odds): chi2(1) = 0.32 Pr>chi2 = 0.5704

Score test for trend of odds: chi2(1) = 0.32 Pr>chi2 = 0.5704

52

Odds ratio for matched pairs

• The odds ratio is r/s• The standard error of ln(OR) is

• So the 95% confidence interval for the estimated OR is

54

rs

srORSE

])(ln[

rs

srOR

rs

srOR

ee96.1ln96.1ln

,

For next time

• Read Pagano and Gauvreau

– Pagano and Gauvreau Chapter 15 (review)– Pagano and Gauvreau Chapter 16

Biostat 200 Lecture 8 1. Hypothesis testing recap Hypothesis testing – Choose a null hypothesis,...

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Transcript of Biostat 200 Lecture 8 1. Hypothesis testing recap Hypothesis testing – Choose a null hypothesis,...