AP Statistics Curriculum 2007 Estim Proportion
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: <math>z_{\alpha\over 2}\times SE_{\tilde{p}} = z_{\alpha\over 2}\sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2}.</math> | : <math>z_{\alpha\over 2}\times SE_{\tilde{p}} = z_{\alpha\over 2}\sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2}.</math> | ||
- | This equation has one unknown parameter (n), which we can solve for if we are given an upper limit for the margin of error | + | This equation has one unknown parameter (n), which we can solve for if we are given an upper limit for the margin of error (remember that the critical value <math>z_{\alpha\over 2} \approx 2</math>): |
- | : <math>z_{\alpha\over 2}\times SE_{\tilde{p}} \geq 2\sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2} \longrightarrow n \geq {\tilde{p}(1-\tilde{p})\over { | + | : <math>z_{\alpha\over 2}\times SE_{\tilde{p}} \geq z_{\alpha \over 2}\sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2} \longrightarrow n \geq {\tilde{p}(1-\tilde{p})\over {z_{\alpha \over 2}SE_{\tilde{p}}^2} } -z_{\alpha \over 2}^2.</math> |
===Examples=== | ===Examples=== |
Revision as of 01:17, 11 March 2011
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 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 slightly and obtain the corrected-sample-proportion :
where is the normal critical value we saw earlier.
The standard error of also needs a slight modification
Confidence intervals for proportions
The confidence intervals for the sample proportion and the corrected-sample-proportion are given by
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% (α = 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 :
- ;
- ;
And the corresponding confidence intervals are given by
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 product of the critical value (t or z) and the standard-error of the sample-proportion:
This equation has one unknown parameter (n), which we can solve for if we are given an upper limit for the margin of error (remember that the critical value ):
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 ?
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 1^{st} child is a girl are more likely to have a girl, as a second child, compared to mothers with boys as 1^{st} 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 | |
2 (Previous child was boy) | n_{2} = 5978 | 2776 |
- Test Statistics: and Z_{o} = 5.4996.
- . 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 , or when using the corrected sample proportion the CI is . 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.
- This SOCR Analysis Activity illustrates how to use the SOCR Analyses to compute the p-values and answer the hypothesis testing challenge.
References
Problems
- SOCR Home page: http://www.socr.ucla.edu
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