# Difference between revisions of "AP Statistics Curriculum 2007 Hypothesis Proportion"

## General Advance-Placement (AP) Statistics Curriculum - Testing a Claim about Proportion

### Testing a Claim about Proportion

Recall that for large samples, the sampling distribution of the sample proportion $$\hat{p}$$ is approximately Normal, by CLT, as the sample proportion may be presented as a sample average or Bernoulli random variables. When the sample size is small, the normal approximation may be inadequate. To accommodate this, we will modify the sample-proportion $$\hat{p}$$ slightly and obtain the corrected-sample-proportion $$\tilde{p}$$: $\hat{p}={y\over n} \longrightarrow \tilde{y}={y+0.5z_{\alpha \over 2}^2 \over n+z_{\alpha \over 2}^2},$ where $$z_{\alpha \over 2}$$ is the normal critical value we saw earlier.

The standard error of $$\hat{p}$$ also needs a slight modification $SE_{\hat{p}} = \sqrt{\hat{p}(1-\hat{p})\over n} \longrightarrow SE_{\tilde{p}} = \sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2}.$

### Hypothesis Testing about a Sinlge Sample Proportion

• Null Hypothesis$H_o: p=p_o$ (e.g., 0), where p is the population proportion of interest.
• Alternative Research Hypotheses:
• One sided (uni-directional)$H_1: p >p_o$, or $$H_o: p<p_o$$
• Double sided$H_1: p \not= p_o$
• Test Statistics$Z_o={\tilde{p} -p_o \over SE_{\tilde{p}}}$

### Example

Suppose a researcher is interested in studying the effect of aspirin in reducing heart attacks. He randomly recruits 500 subjects with evidence of early heart disease and has them take one aspirin daily for two years. At the end of the two years, he finds that during the study only 17 subjects had a heart attack. Use $$\alpha=0.05$$ to formulate a test a research hypothesis that the proportion of subject on aspirin treatment that have heart attacks within 2 years of treatment is $$p_o=0.04$$.

$\tilde{p} = {17+0.5z_{0.025}^2\over 500+z_{0.025}^2}== {17+1.92\over 500+3.84}=0.038$

$SE_{\tilde{p}}= \sqrt{0.038(1-0.038)\over 500+3.84}=0.0085$

And the corresponding test statistics is $Z_o={\tilde{p} - 0.04 \over SE_{\tilde{p}}}={0.002 \over 0.0085}=0.2353$

The p-value corresponding to this test-statistics is clearly insignificant.

### Genders of Siblings Example

Is the gender of a second child influenced by the gender of the first child, in families with >1 kid? 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 children. 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.