# Difference between revisions of "AP Statistics Curriculum 2007 Infer 2Proportions"

### Testing for equality of Two Proportions

Suppose we have two populations and we are interested in estimating whether the proportions of subjects that have certain characteristic of interest (e.g., fixed gender) in each population are equal. To make this inference we obtain two samples {$$X_1, X_2, X_3, \cdots, X_n$$} and {$$Y_1, Y_2, Y_3, \cdots, Y_k$$}, where each $$X_i$$ and $$Y_i$$ represents whether the ith observation in the sample had the characteristic of interest. That is $X_i = \begin{cases}0,& \texttt{Characteristic-absent},\\ 1,& \texttt{Characteristic-present}.\end{cases}$ and $$Y_i = \begin{cases}0,& \texttt{Characteristic-absent},\\ 1,& \texttt{Characteristic-present}.\end{cases}$$

Since the raw sample proportions of observations having the characteristic of interest are $\hat{p_x}={1 \over n}\sum_{i=1}^n{x_i}$ and $$\hat{p_y}={1 \over k}\sum_{i=1}^k{y_i}$$

The corrected sample proportions (for small samples) are $\tilde{p_x}={\sum_{i=1}^n{x_i}+0.5z_{\alpha \over 2}^2 \over n+z_{\alpha \over 2}^2},$ and $$\tilde{p_y}={\sum_{i=1}^k{y_i}+0.5z_{\alpha \over 2}^2 \over k+z_{\alpha \over 2}^2},$$ where $$z_{\alpha \over 2}$$ is the normal critical value we saw earlier.

By the independence of the samples, the standard error of the difference of the two proportion estimates is:

Raw proportions$SE_{\hat{p_x}-\hat{p_y}} = \sqrt{SE_{\hat{p_x}}^2 + SE_{\hat{p_y}}^2}= \sqrt{ {\hat{p_x}(1-\hat{p_x})\over n} + {\hat{p_y}(1-\hat{p_y})\over k}}.$
Corrected Proportions$SE_{\tilde{p_x}-\tilde{p_y}} = \sqrt{SE_{\tilde{p_x}}^2 + SE_{\tilde{p_y}}^2}= \sqrt{ {\tilde{p_x}(1-\tilde{p_x})\over n+z_{\alpha \over 2}^2} + {\tilde{p_y}(1-\tilde{p_y})\over k+z_{\alpha \over 2}^2}}.$

### Hypothesis Testing the difference of Two Proportions

• Null Hypothesis$H_o: p_x=p_y$, where $$p_x$$ and $$p_x$$ are the sample population proportions of interest.
• Alternative Research Hypotheses:
• One sided (uni-directional)$H_1: p_x > p_y$, or $$H_o: p_x < p_y$$
• Double sided$H_1: p_x \not= p_y$
• Test Statistics$Z_o={\tilde{p_x} - \tilde{p_y} \over SE_{\tilde{p_x}-\tilde{p_y}}}$

### Genders of Siblings Example

Is the gender of a second child influenced by the gender of the first child, in families with >1 child? Research hypothesis needs to be formulated first before collecting/looking/interpreting the data that will be used to address it. Mothers whose 1st child is a girl are more likely to have a girl, as a second child, compared to mothers with boys as 1st child. Data: 20 yrs of birth records of 1 Hospital in Auckland, New Zealand.

 Second Child Male Female Total First Child Male 3,202 2,776 5,978 Female 2,620 2,792 5,412 Total 5,822 5,568 11,390

Let $$p_1$$=true proportion of girls in mothers with girl as first child, $$p_2$$=true proportion of girls in mothers with boy as first child. The parameter of interest is $$p_1- p_2$$.

• Hypotheses$H_o: p_1- p_2=0$ (skeptical reaction). $$H_1: p_1- p_2>0$$ (research hypothesis).
 Second Child Number of births Number of girls Proportion Group 1 (Previous child was girl) $$n_1=5412$$ 2792 $$\hat{p}_1=0.516$$ 2 (Previous child was boy) $$n_2=5978$$ 2776 $$\hat{p}_2=0.464$$
• Test Statistics$Z_o = {Estimate-HypothesizedValue\over SE(Estimate)} = {\hat{p}_1 - \hat{p}_2 - 0 \over SE(\hat{p}_1 - \hat{p}_2)} = {\hat{p}_1 - \hat{p}_2 - 0 \over \sqrt{{\hat{p}_1(1-\hat{p}_1)\over n_1} + {\hat{p}_2(1-\hat{p}_2)\over n_2}}} \sim N(0,1)$ and $$Z_o=5.4996$$.
• $$P_value = P(Z>Z_o)< 1.9\times 10^{-8}$$. This small p-values provides extremely strong evidence to reject the null hypothesis that there are no differences between the proportions of mothers that had a girl as a second child but had either boy or girl as their first child. Hence there is strong statistical evidence implying that genders of siblings are not independent.
• Practical significance: The practical significance of the effect (of the gender of the first child on the gender of the second child, in this case) can only be assessed using confidence intervals. A 95% $$CI (p_1- p_2) =[0.033; 0.070]$$ is computed by $$p_1-p_2 \pm 1.96 SE(p_1 - p_2)$$. Clearly, this is a practically negligible effect and no reasonable person would make important prospective family decisions based on the gender of their (first) child. 