Inference (CI / Tests) forComparing 2 Proportions

To compare proportions from distinct populations

p1 vs. p2

Conditions

1. Appropriate randomization.Without this, there’s rarely any way to proceed making sensible conclusions; and when there are ways, they are very complex (and often still produced results for which in general conclusions may be easily disputed)

To compare proportions from distinct populations

p1 vs. p2

Conditions

2. For the particular method we will use

Large populations (20 x larger than sample size).

Large enough samples so that all cells of a data summary table are at least 5 (ie, 5 units).

Without either of these, there are methods that still give conclusions similar in scope to our method.

To compare proportions from distinct populations

p1 vs. p2

There are numerous ways to compare two quantities. Here are two (in red – what we will do):

Assess the differenceWhat is the value of p1 – p2 ? (CI)

Is p1 – p2 = 0 or not? (Test)

Assess the ratio:

What is the value of p1 / p2 ? (CI)

Is p1 / p2 = 1 or not? (Test)

Sample proportions (possibly error margins)

Confidence interval for the value of p1 – p2

Point estimate of difference:

Error margin:

If you know the error margin for each, use the Pythagorean Theorem.

11 andˆ Ep 22 andˆ Ep

21 ˆˆ pp

22

21 EEEDIFF

Sample proportions (possibly error margins)

Confidence interval for the value of p1 – p2

Point estimate of difference:

Error margin:

If you don’t know the error margin for each

11 andˆ Ep 22 andˆ Ep

21 ˆˆ pp

2

22

1

112

ˆ1ˆˆ1ˆ

n

pp

n

ppZEDIFF

Hypothesis test about the difference p1 – p2

H0: p1 = p2 (equivalent to) H0: p1 – p2 = 0

Test statistic:

is the “pooled proportion” (H0 true)

SE

ppZ 21 ˆˆ

21

21

111

11

nnpp

n

pp

n

ppSE

p

The P-value depends on the direction(s) specified in the alternative hypothesis.

HA: p1 > p2 (equivalent to) H0: p1 – p2 > 0

P-value = probability right of Z

HA: p1 < p2 (equivalent to) H0: p1 – p2 < 0

P-value = probability left of Z

HA: p1 p2 (equivalent to) H0: p1 – p2 0

P-value = probability outside of ±Z

= twice probability outside of Z

Pregnant women with AIDS were involved in a study. Each woman was randomly allocated to either treatment with AZT or a Placebo. Researchers want to know if there is any difference in the proportions of babies born HIV+ for (all) women who might be treated with AZT and (all) women if not treated with AZT.

pAZT = proportion of all babies born to mothers with AIDS using AZT that are HIV+.

pPBO = proportion of all babies born to mothers with AIDS using a placebo that are HIV+.

H0: pAZT = pPBO H1: pAZT pPBO

Of 164 babies of women getting AZT, 13 were HIV+.

Of 160 babies of women getting the Placebo, 40 were HIV+.

    Treatment      AZT Placebo Total

OutcomeHIV+HIV-

  Total

Of 164 babies of women getting AZT, 13 were HIV+.

Of 160 babies of women getting the Placebo, 40 were HIV+.

    Treatment      AZT Placebo Total

OutcomeHIV+HIV-      

  Total 164

Of 164 babies of women getting AZT, 13 were HIV+.

Of 160 babies of women getting the Placebo, 40 were HIV+.

    Treatment      AZT Placebo Total

OutcomeHIV+ 13HIV-  

  Total 164

Of 164 babies of women getting AZT, 13 were HIV+.

Of 160 babies of women getting the Placebo, 40 were HIV+.

    Treatment      AZT Placebo Total

OutcomeHIV+ 13 40HIV-    

  Total 164 160

Of 164 babies of women getting AZT, 13 were HIV+.

Of 160 babies of women getting the Placebo, 40 were HIV+.

    Treatment      AZT Placebo Total

OutcomeHIV+ 13 40 53HIV-      

  Total 164 160 324

Of 164 babies of women getting AZT, 13 were HIV+.

Of 160 babies of women getting the Placebo, 40 were HIV+.

    Treatment      AZT Placebo Total

OutcomeHIV+ 13 40 53HIV- 151 120 271

  Total 164 160 324

Condition Check:

Appropriate randomization

Large populations

All cells at least 5

    Treatment      AZT Placebo Total

OutcomeHIV+ 13 40 53HIV- 151 120 271

  Total 164 160 324

Of 164 babies of women getting AZT, 13 were HIV+.

Of 160 babies of women getting the Placebo, 40 were HIV+.

    Treatment      AZT Placebo Total

OutcomeHIV+ 13 40 53HIV- 151 120 271

  Total 164 160 324

0793.016413ˆ AZT p

2500.016040ˆ PBO p

Pooled over both conditions:

Of 324 babies, 53 were HIV+.

    Treatment      AZT Placebo Total

OutcomeHIV+ 13 40 53HIV- 151 120 271

  Total 164 160 324

1636.032453 p

Result of hypothesis test:

Z = -4.15 P-value = 0.0000323 1 / 31000

Reject H0. The (population) proportions differ. And in fact the mothers treated with AZT had babies with better outcomes.

There is sufficient evidence in the sample data to conclude that the proportion of HIV+ babies born to mothers treated with AZT is less than that for mothers treated with a Placebo.

Result of hypothesis test:

Z = -4.15 P-value = 0.0000323 1 / 31000

Suppose AZT and Placebo had the same efficacy (effectiveness):

In only 1 in 31000 studies would the difference in HIV rates between the groups be as large as observed here.

In only 1 in 31000 randomizations would the difference in HIV rates between the groups be as large as observed here.

Result of hypothesis test:

Z = -4.15 P-value = 0.0000323 1 / 31000

95% CI for the difference in proportions

Point Estimate: 0.0793 – 0.2500 = –0.1707

Error Margin: 0.0795

Bounds: –0.1707 ± 0.0795

–0.2502 < pAZT – pPBO < –0.0912

OR (better?)

0.0912 < pPBO – pAZT < 0.2502

95% CI for the difference in proportions

– 0.2502 < pAZT – pPBO < – 0.0912

add pPBO to all “3” sides

pPBO – 0.2502 < pAZT < pPBO – 0.0912

We are 95% confident that the proportion of babies born HIV+ is between 0.0912 and 0.2505 lower for mothers treated with AZT.

Result of hypothesis test:

Z = -4.15 P-value = 0.0000323 1 / 31000

95% CI for the difference in proportions

Point Estimate: 0.1707 Error Margin: 0.0795

0.0912 < pPBO – pAZT < 0.2502

We are 95% confident that the percent of babies born HIV+ is between 9.12% and 25.02% lower for mothers treated with AZT.

We can be quite sure that it is the AZT that caused the observed reduction in HIV+ rates.

The randomization alone is very unlikely to have been responsible for such a difference. (The P-value quantifies this likelihood.)

The randomization also insures that whatever other variables might be important are similarly distributed among the women in the AZT and Placebo groups.

This is the advantage of an experimental study over an observational one. It allows us to infer causation.

95% CI for the difference in proportions

Point Estimate: 0.1707 Error Margin: 0.0795

0.0912 < pPBO – pAZT < 0.2502

Number Needed to Treat

Estimated NNT =

= = 5.8582

Difference ofEst Pt

1ˆˆ

1

21

pp

1707.0

1

58582 pregnant women with AIDS

If not treated with AZT:

0.2502 58582 = 14657 HIV+ babies

If treated with AZT:

0.0795 58582 = 4657 HIV+ babies

Treating 58582 womenwith AZT results in a reduction of 10000 HIV+ babies

Treating 5.8582 womenresults in 1 fewer HIV+ baby

NNT:

The number of units that must be switched from one level of the explanatory variable (one “group” or “treatment”) to the other in order to result in a change of 1 unit from one level of the response variable to the other.

How many units must be “treated” in order to produce 1 additional favorable outcome.

95% CI for the difference in proportions

Point Estimate: 0.1707 Error Margin: 0.0795

0.0912 < pPBO – pAZT < 0.2502

Number Needed to Treat

Estimated NNT = 5.8582

We estimate that for every 5.8582 women treated with AZT (instead of nothing/placebo) there is one fewer HIV+ baby.

95% CI for the difference in proportions

Point Estimate: 0.1707 Error Margin: 0.0795

0.0912 < pPBO – pAZT < 0.2502

Number Needed to Treat

Estimated NNT = 5.8582

A CI for NNT can be formed by taking reciprocals of the CI for the difference:

1/0.0912 = 10.965 1/0.2502 = 3.9968

4 < NNT (population) < 11

Motorcycle Helmet Color

Percent of accidents resulting in harm…

…among those wearing dark helmets:

190/681 = 0.279 27.9%

…among those wearing light helmets:

116/493 = 0.235 23.5%

Point Estimate of Difference: 4.4% = 0.044

(The P-value also happens to be 0.044!)

Estimated NNT = 1/0.044 = 22.7