AP Statistics Curriculum 2007 Estim Proportion
Contents
General Advance-Placement (AP) Statistics Curriculum - Estimating a Population Proportion
Estimating a Population Proportion
When the sample size is large, 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 \(\tilde{p}\): \[\hat{p}={y\over n} \longrightarrow \tilde{p}={y+0.5z_{\alpha \over 2}^2 \over n+z_{\alpha \over 2}^2},\] where \(z_{\alpha \over 2}\) is the normal critical value we saw earlier.
The standard error of \(\hat{p}\) also needs a slight modification \[SE_{\hat{p}} = \sqrt{\hat{p}(1-\hat{p})\over n} \longrightarrow SE_{\tilde{p}} = \sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2}.\]
Confidence intervals for proportions
The confidence intervals for the sample proportion \(\hat{p}\) and the corrected-sample-proportion \(\tilde{p}\) are given by \[\hat{p}\pm z_{\alpha\over 2} SE_{\hat{p}}\]
\[\tilde{p}\pm z_{\alpha\over 2} SE_{\tilde{p}}\]
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. Calculate a 95% (\(\alpha=0.05\)) confidence interval for the true (unknown) proportion of subjects with early heart disease that have a heart attack while taking aspirin daily. Note that \(z_{\alpha \over 2} = z_{0.025}=1.96\):
\[\hat{p} = {17\over 500}=0.034\] ; \(\tilde{p} = {17+0.5z_{0.025}^2\over 500+z_{0.025}^2}== {17+1.92\over 500+3.84}=0.038\)
\[SE_{\hat{p}}= \sqrt{0.034(1-0.034)\over 500}=0.0036\]; \(SE_{\tilde{p}}= \sqrt{0.038(1-0.038)\over 500+3.84}=0.0085\)
And the corresponding confidence intervals are given by \[\hat{p}\pm 1.96 SE_{\hat{p}}=[0.026944, 0.041056]\]
\[\tilde{p}\pm 1.96 SE_{\tilde{p}}=[0.0213, 0.0547]\]
Sample-size Estimation
For a given margin of error we can derive the minimum sample-size that guarantees an interval estimate within the given margin of error. The margin of error is the standard-error of the sample-proportion:
\[SE_{\tilde{p}} = \sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2}.\]
This equation has one unknown parameter (n), which we can solve for if we are given an upper limit for the margin of error.
\[SE_{\tilde{p}} \geq \sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2} \longrightarrow n \geq {\tilde{p}(1-\tilde{p})\over {SE_{\tilde{p}}^2} } -z_{\alpha \over 2}^2.\]
Examples
Sample-Size Estimation
How many subjects are needed if the heart-researchers want \(SE < 0.005\) for a 95% CI, and have a guess based on previous research that \(\tilde{p}= 0.04\)? \[n \geq {0.04(1-0.04)\over 0.005^2} - 1.96^2=1533.16 \approx 1534.\]
Siblings Genders
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 \(\hat{p}_1-\hat{p}_2 \pm 1.96 SE(\hat{p}_1 - \hat{p}_2)\), or when using the corrected sample proportion the CI is \(\tilde{p}_1-\tilde{p}_2 \pm 1.96 SE(\tilde{p}_1 - \tilde{p}_2)\). Clearly, this is a practically neglegable effect and no reasonable person would make important prospective family decisions based on the gender of their (first) child.
- This SOCR Analysis Activity illustrates how to use the SOCR Analyses to compute the p-values and answer the hypothesis testing challenge.
References
- SOCR Home page: http://www.socr.ucla.edu
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