Introductory Statistics. Lecture 26

8.3 A Confidence Interval for the Difference of two Proportions Math 140 Introductory Statistics „ A very common and important situation involves taking two samples independently from two different populations with the goal of estimating the size of the difference between the proportion of successes in one population and the proportion of successes in the other. Lecture 26 Sections 8.3 Example Intuitive Formula „ A recent poll of 29,700 U.S. households „ A confidence interval for the difference of two found that 63% owned a pet. The percentage in 1994 was 56%. [Source: American Pet Products Manufacturers Association, www.appma.org.] „ The two populations are the households in the United States in 1994 and the households now. The question you will investigate is: „ “What was the change in the percentage of U.S. households that own a pet?” proportions, p1 − p2, where p1 is the proportion of successes in the first population and p2 is the proportion of successes in the second population: (pb1 ¡ pb2 ) § z ¤ ¢ standard error of(pb1 ¡ pb2 ) pb11 and pb22 are the proportions of successes in the two samples. „ In our example, the 95% confidence interval for the difference between the proportion of U.S. households that own pets now and the proportion that owned pets in 1994: (0:63 ¡ 0:56) § 1:96 ¢ standard error of(pb1 ¡ pb2 ) 1 Standard Error of the Difference The Formula „ The standard errors of the distributions of pb1 „ The confidence interval for the difference, and pb2 can be estimated respectively as: s s pb1 (1 ¡ pb1 ) and n1 pb2 (1 ¡ pb2 ) n2 „ And if we assume the proportion samples are independent, then the standard error of the difference can be approximated as s pb1 (1 ¡ pb1 ) pb2 (1 ¡ pb2 ) + n1 n2 p1 − p2 , of the proportion of successes in one population and the proportion of successes in the second population is, s pb1 (1 ¡ pb1 ) pb2 (1 ¡ pb2 ) ¤ (pb1 ¡ pb2 ) § z ¢ + n1 n2 „ Where pb11 is the proportion of successes in a random sample of size n1 taken from the first population, and pb2 is the proportion of successes in a random sample of size n2 taken from the second population. (Sample sizes do not need to be equal.) Conditions for use Example: A Difference in Pet Ownership? „ The conditions that must be met in order to „ A recent pet ownership survey found that use this formula are that the two samples are taken randomly and independently from two populations. „ each population is at least 10 times as large as its sample size. b1, n1(1−p b1), n2p b 2, and n2(1−pb2) are all „ n1p „ at least 5. 63% of the 29,700 U.S. households sampled own a pet. A 1994 survey, taken by the same organization, found that 56% of the 6,786 U.S. households sampled owned a pet. Find and interpret a 95% confidence interval for the difference between the proportion of U.S. households that own a pet now and the proportion of U.S. households that owned a pet in 1994. 2 8.4/8.5 Significance Test for the Difference of two Proportions Example: Two AIDS treatments „ Often times we need to decide which is the „ Consider a clinical trial experiment comparing two treatments for greater of two proportions, or whether we can assume they are the same. For example. Is snowboarding or skiing more likely to result in a serious injury? „ Does a new treatment for AIDS result in fewer deaths than an old treatment? „ Is Reggie Jackson’s World Series record so much better than his play during the regular season that the difference can’t reasonably be attributed to chance? „ Notation:Proportions, Sample Proportions, and Sample Sizes. n1 = 69 sample size of patients treated with AZT n2 = 62 sample size of patients treated with AZT+ACV pˆ 1 = pˆ 2 = 41 69 49 62 proportion of patients in the sample treated with AZT that survived proportion of patients in the sample treated with AZT+ACV that survived p1 True proportion of survival if all patients were treated with AZT (unknown) p2 True proportion of survival if all patients were treated with AZT+ACV (unknown) AIDS-related complex (ARC). The investigators want to find out if there is a significant difference on the survival rates of patients who had already developed AIDS. They have two treatments, patients who were given AZT and patients who were given AZT + ACV. Here’s the data Treated with AZT AZT + ACV Total 28 13 41 Yes 41 49 90 Total 69 62 131 No Survived? Assumptions about the difference of two proportions. „ If we have obtain two independent sample proportions p p̂1 and p̂p2 , then the distribution of the difference of the pˆ 1 − pˆ 2 is approximately normal as two proportions p1-p2 long as each proportion satisfies the following three conditions: p̂1 and p̂p2 is a simple random sample from a „ Each sample p „ „ binomial population, and they are independent from each other. or, in case of experiments, subjects were randomly assigned to their treatments. ˆ11,,nn12ppˆ2,2,nn11(1(1–− ppˆ11),),n − pˆ 2 ) All the numbers nn11pp and 2 (1n are at least 5. Each of the two population sizes is at least 10 times the sample size. 3 Checking the conditions Writing the hypothesis „ Check the conditions. „ We assume that subjects were randomly assigned to treatments. „ All of the following are greater than 5 „ The null hypothesis „ H0 : The new therapy (AZ+ACV) is as good as the old therapy (AZT). That is the survival rate if all patients were treated with AZT+ACV would be equal to the survival rate if all patients were treated with AZT. In symbols p 1 = p 2 or p 1 − p 2 = 0 n1pˆ 1 = 69(41/ 69) = 41, n2pˆ 2 = 62(49/ 62) = 49 n1(1− pˆ 1) = 69(1− 41/ 69) = 28 n2 (1− pˆ 2 ) = 62(1− 49/ 62) =13 „ The population size of AIDS patients that could potentially be treated with AZT is clearly greater than 10 times 69 = 690 The population size of AIDS patients that could potentially be treated with AZT+ACV is clearly greater than 10 times 62 = 620 „ The alternate hypothesis „ Ha : The new therapy (AZ+ACV) is better than the old therapy (AZT). That is the survival rate if all patients were treated with AZT+ACV would be greater than the survival rate if all patients were treated with AZT. In symbols p 1 < p 2 or p 1 − p 2 < 0 4