Two sample: Binary Response ST551 Lecture 21

Two sample: Binary Response

Setting: two independent samples

$Y_1,\dots ,Y_n$ i.i.d from Bernoulli$\left(p_Y\right)$
$X_1,\dots ,X_m$ i.i.d from Bernoulli$\left(p_X\right)$

Parameter: Difference in population proportions $p_Y-p_X$

$p_Y=E\left(Y_i\right)=P(Y_i=1)$
$p_X=E\left(X_i\right)=P(X_i=1)$

As a contingency table

Represent resulting data in a 2 x 2 contingency table:

Two sample: Binary Response - Alternate view

Setting: two independent samples

$$\begin{array}{c}(Y_1,G_1),(Y_2,G_2),\dots ,(Y_n,G_n),(Y_{n+1},G_{n+1}),\dots ,(Y_{n+m},G_{n+m})\end{array}$$

where $G$ is a binary grouping variable which indicates which population the observation came from: $$G_i=\{\begin{array}{cc}0,& \text{observation from}Y\\ 1,& \text{observation from}X\end{array}$$

As a contingency table - Alternate view

Represent resulting data in a 2 x 2 contingency table:

Two views are equivalent

If we are interested in the response variable given the group.

• I sample 40 OSU graduate students and 20 OSU undergraduate students:

  • $Y_i$ = graduate student, did you vote in 2016? $i=1,\dots ,40$
  • $X_i$ = undergraduate student did you vote in 2016? $i=1,\dots ,20$

• I sample 60 OSU students and record:

  • $Y_i$ = did you vote in 2016?, $i=1,\dots ,60$
  • $G_i$ = student’s level (0 = graduate, 1 = undergraduate), $i=1,\dots ,60$

Inference focuses on:

Comparing $P(Y_i=1)$ and $P(X_i=1)$ - first view
Comparing $P(Y_i=1|G_i=0)$ and $P(Y_i=1|G_i=1)$ - second view

Ways to compare two proportions

$Y_1,\dots ,Y_n$ i.i.d from Bernoulli$\left(p_Y\right)$
$X_1,\dots ,X_m$ i.i.d from Bernoulli$\left(p_X\right)$

Typical null hypothesis: $H_0:p_Y=p_X$)

Difference in population proportions: $p_Y-p_X$

• $H_0:p_Y-p_X=0$

Relative risk: $p_Y/p_X$

• $H_0:p_Y/p_X=1$

Odds ratio: $\frac{p_Y}{1-p_Y}/\frac{p_X}{1-p_X}$

• $H_0:\frac{p_Y}{1-p_Y}/\frac{p_X}{1-p_X}=1$

Example

(From class_data, assuming you are like some random sample from a larger population)

$G_i$: Do you prefer cats or dogs? $Y_i$: Did you east breakfast this morning?

##        ate_breakfast
## cat_dog no yes
##    cats  6   9
##    dogs  6  14

Your turn: Fill in the table margins

Estimates

Probability of eating breakfast, given you prefer cats: $$p_Y=P(Y_i=1|G_i=0)=\frac{P(Y_i=1\&G_i=0)}{P(G_i=0)}$$ Estimate $${\hat{p}}_Y=\frac{b/N}{R_1/N}=\frac{9}{15}=0.6$$

Probability of eating breakfast, given you prefer dogs: $$p_X=P(Y_i=1|G_i=1)=\frac{P(Y_i=1\&G_i=1)}{P(G_i=1)}$$

Estimate $${\hat{p}}_X=\frac{d/N}{R_2/N}=\frac{14}{20}=0.7$$

Estimates

Difference in proportions $${\hat{p}}_Y-{\hat{p}}_X=0.6-0.7=-0.1$$

Relative Risk $$\frac{{\hat{p}}_Y}{{\hat{p}}_X}=\frac{0.6}{0.7}=0.86$$

Odds Ratio $$\frac{{\hat{p}}_Y}{1-{\hat{p}}_Y}/\frac{{\hat{p}}_X}{1-{\hat{p}}_X}=\frac{0.6}{1-0.6}/\frac{0.7}{1-0.7}=0.64=\frac{bc}{ad}$$

Two sample Z-test of proportions

(Comes from considering proportion as mean and looking at two sample Z-test)

Null hypothesis: $H_0:p_Y=p_X$

$$Z=\frac{{\hat{p}}_Y-{\hat{p}}_X}{\sqrt{{\hat{p}}_c(1-{\hat{p}}_c)\left(\frac{1}{n}+\frac{1}{m}\right)}}$$

where $p_c=\frac{(n{p}_Y+m{p}_X)}{n+m}=\frac{b+d}{N}$

When null is true $Z$ has a N(0, 1) distribution.

Confidence interval for difference in proportions

$(1-\alpha )100\%$ CI: $${\hat{p}}_Y-{\hat{p}}_X\pm z_{1-\alpha /2}\sqrt{\frac{{\hat{p}}_Y(1-{\hat{p}}_Y)}{n}+\frac{{\hat{p}}_X(1-{\hat{p}}_X)}{m}}$$

Like in one sample case, binomial test and CI may not agree because they use different estimates of the variance of the difference in sample proportions.

Your Turn

$p_c=\frac{(n{p}_Y+m{p}_X)}{n+m}=\frac{b+d}{N}$

What is $p_c$ for our table?

##        ate_breakfast
## cat_dog no yes Sum
##    cats  6   9  15
##    dogs  6  14  20
##    Sum  12  23  35

Example: Z-stat

$$Z=\frac{{\hat{p}}_Y-{\hat{p}}_X}{\sqrt{{\hat{p}}_c(1-{\hat{p}}_c)\left(\frac{1}{n}+\frac{1}{m}\right)}}=\frac{-0.1}{\sqrt{0.66(1-0.66)(\frac{1}{15}+\frac{1}{20})}}=-0.62$$

Compare to $z_{1-\alpha /2}=1.96$

p-value (for two sided alternative) =0.54

95% confidence interval: $$\begin{array}{c}{\hat{p}}_Y-{\hat{p}}_X\pm z_{1-\alpha /2}\sqrt{\frac{{\hat{p}}_Y(1-{\hat{p}}_Y)}{n}+\frac{{\hat{p}}_X(1-{\hat{p}}_X)}{m}}\\ =-0.1\pm \sqrt{\frac{0.6(1-0.6)}{15}+\frac{0.7(1-0.7)}{20}}\\ =(-0.26,0.06)\end{array}$$

Pearson’s Chi-squared Test

$H_0:p_Y-p_X=0$

$$X=\sum _{j,k=1,2}\frac{(O_{jk}-E_{jk}{)}^2}{E_{jk}}$$

$O_{jk}=n_{jk}$

$E_{jk}=\frac{R_j{C}_k}{N}$

If null is true, $X$ has ${\chi }_1^2$ distribution

Example: Chi-squared test

##        ate_breakfast
## cat_dog no yes Sum
##    cats  6   9  15
##    dogs  6  14  20
##    Sum  12  23  35

##        no   yes
## cats 5.14  9.86
## dogs 6.86 13.14

E.g $\frac{15\times 12}{35}=5.14$

Example: Chi-squared test

$$\begin{array}{cc}X& =\frac{(6-5.14{)}^2}{5.14}+\frac{(9-9.86{)}^2}{9.86}+\frac{(6-6.86{)}^2}{6.86}+\frac{(14-13.14{)}^2}{13.14}\\ & =0.38\end{array}$$

Compare to ${\chi }_1^2(1-\alpha )=3.84$

p-value: =0.54

Summary

Pearson’s Chi-squared test for homogeneity of proportions across groups is equivalent (i.e. results in the same p-value) to the Z-test for proportions (when there are two groups).

$X=Z^2$