Comparing Two Population Proportions: Hypothesis Testing and Confidence Intervals - Prof. , Study notes of Probability and Statistics

Download Comparing Two Population Proportions: Hypothesis Testing and Confidence Intervals - Prof. and more Study notes Probability and Statistics in PDF only on Docsity! March 13, 2006, §19: Comparing two population proportions Due March 17th: 18:1-4, 14, 15, 18, 19, 24, 35, 36. 19: 8, 9, 12, 15 Excel assignment. 19: 20a, 21. Review: Hypothesis testing for one-sample proportions. In a certain population, we expect 46% of the population to get a cold in a 3 month period. We give 264 volunteers 1000 mgs of vitamin C per day for 3 months. At the end of the period 119 people (45%) have gotten colds. So the proportion of our sample that have gotten colds is p̂ = .45. Let p be the proportion of people being treated with vitamin C who will get colds. We would like to know if there is evidence that p is smaller than .46. 1. Our null hypothesis is H0 : p = p0(= .46). In other words, that the percent of people who get colds is unaffected by taking vitamin C. Our alternative hypothesis is Ha : p < p0 = .46. 2. Our z-statistic is z = p̂− p0p p0(1− p0)/n = .45− .46p .46(.54)/264 = −.326. 3. Our P -value is P (Z ≤ −.326) = .3722. §19: Comparing two proportions We consider a situation where we wish to compare two proportions. Typically we would like to compare the effect of two different treatments, either in an experiment or in an observation. We assume we have two populations: Population Proportion Sample Size Sample proportion 1 p1 n1 p̂1 2 p2 n2 p̂2 We’d like to use p̂1 − p̂2 to estimate p1 − p2. The critical observations are the following: • if n1 and n2 are large enough, the distribution of p̂1 − p̂2 is approximately normal, • with mean p1 − p2 • and standard deviation s p1(1− p1) n1 + p2(1− p2) n2 . To estimate confidence intervals, we could proceed as usual, using for our error term SE = s p̂1(1− p̂1) n1 + p̂2(1− p̂2) n2 This is the Large Sample standard error. gWe can use this if each population is at least 10 times the sample sizes, and the samples sizes are large enough to contain at least 10 successes and 10 failures each. It turns out that we get more accurate results by using the “Plus 4” confidence estimates (adding 4 imaginary observations - 2 to our first sample, and two to our second sample). So we replace n1 above with n1 + 2, replace n2 with n2 + 2. p̂1 with p1 (where we add two observations, one success and one failure) and similary, p̂2 with p2. So we get SE = s p1(1− p1) n1 + 2 + p2(1− p2) n2 + 2 . Then if z∗ is a critical value for confidence level C, we get p1 − p2 is p1 − p2 ± z∗SE with confidence C. p1 is the proportion of pregnancies preceded by a C-section which result in stillbirths. p2 is the proportion of pregnancies not preceded by a C-section which result in stillbirths. 1. Our null hypothesis is that having a C-section does not affect later possibility of stillbirth. So H0 : p1 = p2. Our alternative hypothesis is Ha : p1 > p2. (That C-sections are associated with higher rates of stillbirths in the subsequent pregnancy.) 2. We calculate a two sample z-statistic as follows: z = p̂1 − p̂2q p̂(1− p̂)( 1n1 + 1 n2 ) = .00383− .00237q .00259(1− .00259)( 117,754 + 1 102,879) = 3.535. 3. We now calculate our P -value. We calculate P (Z ≥ 3.535). This is the probability (if H0 is true) that we would see at least this many more stillbirths in the C-section population. P (Z ≥ 3.535) = .0002 4. Conclusion: If H0 is true, then the chance of results like the ones we see is .0002. This is strong evidence against H0. In words: if C-sections are not associated with higher rates of stillbirth, then the probability of a result like the one we’ve seen is .0002. We consider this strong evidence that C-sections are associated with higher rates of stillbirth. NOTE: this method is approriate if • The populations are at least 10 times as large as the samples. • There are at least 5 failures and at least 5 successes in each sample. Is gun ownership rising? February 1999: Gallup polls 1134 adults finds 408 own a gun. October 2004: Gallup polls 1134 adlts finds 431 own a gun. Is a newspaper correct to run headline “Percentage of gun owners rising?” Let p1 be percentage of population owning a gun in 1999, p2 percentage in 2004. We wish to test the hypothesis that p2 > p1. • H0 : p1 = p2, Ha : p2 > p1. • Test statistic: – pooled proportion is p̂ = 408+4312·1134 = .370. – p̂1 = 408/1134 = .360 – p̂2 = 431/1134 = .380 z = p̂1 − p̂2q p̂(1− p̂)( 11134 + 1 1134) = .36− .38q .37 · .63 · 21134 = −.986