Two sample: Binary Response ST551 Lecture 21

Two sample: Binary Response

Setting: two independent samples

\(Y_1, \ldots, Y_n\) i.i.d from Bernoulli\((p_Y)\)
\(X_1, \ldots, X_m\) i.i.d from Bernoulli\((p_X)\)

Parameter: Difference in population proportions \(p_Y - p_X\)

\(p_Y = E(Y_i) =P(Y_i = 1)\)
\(p_X = E(X_i) =P(X_i = 1)\)

As a contingency table

Represent resulting data in a 2 x 2 contingency table:

0 1 Total
\(Y_i\) a b n = a+b
\(X_i\) c d m = c+d
Total a+c b+d m + n

Two sample: Binary Response - Alternate view

Setting: two independent samples

\[ \begin{aligned} (Y_1, G_1), (Y_2, G_2), \ldots,(Y_n, G_n), (Y_{n+1}, G_{n+1}), \ldots, (Y_{n+m}, G_{n+m}) \end{aligned} \]

where \(G\) is a binary grouping variable which indicates which population the observation came from: \[ G_i = \begin{cases} 0, & \text{observation from } Y \\ 1, & \text{observation from } X \end{cases} \]

As a contingency table - Alternate view

Represent resulting data in a 2 x 2 contingency table:

\(Y_i = 0\) \(Y_i = 1\) Total
\(G_i = 0\) \(n_{11} = a\) \(n_{12} = b\) n = a+b = \(R_1\)
\(G_i = 1\) \(n_{21} = c\) \(n_{22} = d\) m = c+d = \(R_2\)
Total a+c = \(C_1\) b+d=\(C_2\) a + b + c + d = N

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? \(i = 1, \ldots, 40\)
    • \(X_i\) = undergraduate student did you vote in 2016? \(i = 1, \ldots, 20\)

  • I sample 60 OSU students and record:

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

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

\(Y_1, \ldots, Y_n\) i.i.d from Bernoulli\((p_Y)\)
\(X_1, \ldots, X_m\) i.i.d from Bernoulli\((p_X)\)

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 \(p_c = \frac{( np_Y + mp_X)}{n + m} = \frac{b + d}{N}\)

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

\(p_c = \frac{( np_Y + mp_X)}{n + m} = \frac{b + d}{N}\)

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: \[ \begin{aligned} \hat{p}_Y - \hat{p}_X \pm z_{1-\alpha/2} \sqrt{\frac{\hat{p}_Y(1 - \hat{p}_Y)}{n} + \frac{\hat{p}_X(1 - \hat{p}_X)}{m}} \\ = -0.1 \pm \sqrt{\frac{0.6(1 - 0.6)}{15} + \frac{0.7(1 - 0.7)}{20}} \\ = (-0.26, 0.06) \end{aligned} \]

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 \(\chi^2_{1}\) 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

\[ \begin{aligned} X &= \frac{(6 - 5.14)^2}{5.14} + \frac{(9 - 9.86)^2}{9.86} + \frac{(6 - 6.86)^2}{6.86} + \frac{(14 - 13.14)^2}{13.14} \\ &= 0.38 \end{aligned} \]

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

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).

\(X = Z^2\)