Chapter 22: Comparing two proportions. Religious identification Current at 16 Same Different Total Catholic Jewish

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1 Chapter 22: Comparing two proportions Example: The table below shows the cross-classification of a sample of individuals according to their religious identification at age 16 and their current religious identification 1 Religious identification Current at 16 Same Different Total Catholic Jewish Is there statistical evidence supporting the contention that Jews and Catholics differ with respect to their fidelity to their parents religion (assuming that a person s religious identification at age 16 is the same as that of their parents)? To answer the question, two populations are identified: C: individuals that identify themselves as Catholics at age 16, and J: individuals that identify themselves as Jewish at age 16 Let p 1 the proportion of C that identify themselves at the time of the survey as Catholic, and p 2 the proportion of J that identify themselves at the time of the survey as Jewish The objective is to compare p 1 to p 2 using point estimates and a confidence interval The difference between population proportions, p 2 p 1 is the parameter of interest The estimator of p 2 p 1 is the difference in sample proportions p 2 p 1 Since sample sizes are n and n 2 31, the estimate is p 2 p The sampling distribution of p 2 p 1 Confidence intervals and hypothesis tests involving p 2 p 1 are derived from the large-sample distribution of p 2 p 1 Suppose that the randomization, 10% and success/failure conditions are met for p 1 and p 2 Then, p 1 N(p 1, p 1 q 1 /n 1 ) and p 2 N(p 2, p 2 q 2 /n 2 ) 1 The General Social Survey (GSS) has been conducted by the National Opinion Research Center annually since 1972 and biennially beginning in 1994 The GSS is designed as part of a program of social indicator research, replicating questionnaire items and wording in order to facilitate time-trend studies Items on religion in the 1980 GSS include religious preference, church attendance, beliefs about life after death, and attitudes toward organized religion 167

2 These results are extended to p 2 p 1 using the Chapter 16 results regarding the expected value and standard deviation of the sum and difference of two random variables Specifically, E( p 1 p 2 ) p 1 p 2 If the two samples are drawn independently from their respective populations, then the variance and standard deviation of p 2 p 1 are Var( p 1 p 2 ) Var( p 1 ) + Var( p 2 ) p 1q 1 + p 2q 2, n 1 n 2 p1 q 1 and σ( p 1 p 2 ) + p 2q 2 The Central Limit Theorem implies that sampling distribution of p 1 p 2 is approximately normal 2 In summary, p 1 p 2 N (p 1 p 2, σ( p 1 p 2 )) (1) Equivalently, ( p1 q 1 p 1 p 2 N p 1 p 2, + p ) 2q 2 The standard deviation of σ( p 1 p 2 ) is estimated by the standard error p 1 q 1 σ( p 1 p 2 ) + p 2 q 2 Several conditions must be met for the confidence interval and hypothesis test (discussed below) to be accurate: 1 Randomization condition: Both samples are random samples or at least representative of the respective populations of interest 2 Large sample conditions: (a) The success/failure condition: The number of successes and number of failures are at least 10 Specifically, n 1 p 1, n 1 q 1, n 2 p 2, and n 2 q 2 all at least 10 (b) The 10% condition specifies that n 1 and n 2 are no more than 10% of the respective population sizes 3 Independence condition: The two samples are independent of one another Random sampling of each population insures that the independence condition is met 2 Large sample conditions stated above must be met if the approximation is to be accurate 168

3 Example: Checking the conditions for the GSS data proceeds as follows: 1 The randomization condition: General Social Survey uses randomization protocols in data collection, so the sample is effectively a random sample 2 The success/failure condition: n 1 p 1 351, n 1 q 1 67, n 2 p 2 28, and n 2 q 2 3 are not all least 10 The confidence interval and hypothesis test may not be accurate 3 10% condition: The 10% condition is easily met by both population/samples since the populations are very large 4 Independence: The samples were drawn by randomly sampling a population consisting of a number of religious groups so the samples are independent The confidence interval and hypothesis test can be derived from the large sample distribution The details are omitted, however A confidence interval for p 2 p 1 : A 100(1 2α)% confidence interval for p 1 p 2 is p 1 p 2 ± zα p 1 q 1 + p 2 q 2 where z α is a critical value from the standard normal distribution for the chosen level of confidence Example Using the GSS data, p and p Hence, p 1 q 1 + p 2 q For a 95% confidence level, z α 196, and the 95% confidence interval for p 1 p 2 is p 1 p 2 ± z α σ( p 1 p 2 ) 063 ± [ 0464, 1734] Because 0 is bracketed by the interval, it appears that there is insufficient evidence to conclude that the proportions p 1 and p 2 are different Example: In the 1970 s, the majority of oncologists and surgeons believed that total mastectomies were more effective at curing breast cancer than less radical approaches (principally involving targeted excisions of cancerous tumors) Yet the survival rates of patients treated 169

4 by radical mastectomy was dismally poor The National Surgical Adjuvant Breast and Bowel Project (NSABP) conducted a clinical trial to evaluate lumpectomy for the treatment of breast cancer Between April 1976, and January 1984 volunteer patients were randomly assigned to one of three groups: total mastectomy, lumpectomy, and lumpectomy followed by irradiation The events identified in the analyses of disease-free survival were first recurrences of disease, second cancers, and death without recurrence of cancer A portion of the 12-year data are shown below Total Lumpectomy Status mastectomy and irridation Lumpectomy No recurrence Recurrence Total Let p 1 denote the proportion of patients receiving a total mastectomy that had no recurrence in the 12 years after treatment, and p 2 denote the proportion of patients receiving a lumpectomy that had no recurrence in the 12 years after treatment The sample estimates are p 1 439/ (total mastectomy) p 2 482/ (lumpectomy) The sample estimates are different and contradict prevailing medical beliefs in 1975, but do the data imply a difference among the population of all subjects undergoing total mastectomy or lumpectomy? A hypothesis test is needed to answer this question The test described below is the two-proportion z-test A hypothesis test for comparing two proportions: When comparing population proportions, the usual null hypothesis is H 0 : p 1 p 2, or equivalently, H 0 : p 1 p 2 0 The alternative hypothesis is one of the following three: H a : p 1 p 2 > 0 (or H A : p 1 > p 2 ) H a : p 1 p 2 < 0 (or H A : p 1 < p 2 ) H a : p 1 p 2 0 (or H A : p 1 p 2 ) 170

5 Result (1) states the approximate sampling distribution of p 1 p 2 is normal provided that the conditions enumerated at the bottom of page 168 are met Because the test statistic is computed assuming that H 0 : p 1 p 2 is true, the expected value and standard deviation of the estimator p 1 p 2 are slightly (but importantly) different Let p p 1 p 2 the common value of the population proportions E( p 1 p 2 H 0 ) p 1 p 2 p p 0, and σ( p 1 p 2 H 0 ) p1 q 2 + p 2q 2 pq + pq ( 1 pq + 1 ) An estimator of p is needed First, let X 1 denote the number of successes observed in the sample from population 1 and X 2 denote the number of successes in the second sample Then p 1 X 1 /n 1 and p 2 X 2 /n 2 The two samples are combined to estimate p since the samples are from the same population (according to H 0 ) The pooled estimator of p is p pooled X 1 + X 2 n 1 + n 2 The estimated standard deviation of the difference 3 p 1 p 2 is ( ) σ pooled ( p 1 p 2 ) ppooled qpooled + 1n1 1n2 Example: For the breast cancer treatment study, and ppooled qpooled ( 1n1 + 1n2 ) p pooled X 1 + X 2 n 1 + n , (1 4209) 3 Sometimes called the standard error of the difference of sample proportions ( )

6 The test computes the standardized value of p 1 p 2 (which is approximately N(0, 1) in distribution if H 0 is correct), and determines the (approximate) probability of obtaining a value as or more extreme than the observed value (assuming H 0 is correct) The test statistic is Z p 1 p 2 σ pooled ( p 1 p 2 ) p 1 p 2 ppooled qpooled ( 1n1 + 1n2 ) Example: For the breast cancer treatment study, Z ( (1 4209) ) The p-value is a numerical measure of the strength of evidence against H 0 and in favor of H a Specifically, the p-value is the probability of observing Z as or more extreme than observed if H 0 were true Extreme means contradicting H 0 and supporting H a H a determines how the p-value is computed: H a : p 1 p 2 < 0 p-value P (Z z H 0 ) H a : p 1 p 2 > 0 p-value P (Z z H 0 ) H a : p 1 p 2 0 p-value 2P (Z z H 0 ) Example: Returning to the breast cancer treatment example, the hypothesis test of interest compares total mastectomy (treatment 1) and lumpectomy (treatment 2) The parameters p 1 and p 2 represent the 12-year cancer recurrence rate As of 1975, the prevailing belief was that at best, lumpectomy is no more effective than total mastectomy (ie, p 1 p 2 ) 4 To convince the medical community that this may not be true, the hypotheses to be test are H 0 : p 1 p 2 0 H a : p 1 p 2 < 0 Consequently, 4 More accurately, the prevailing belief was that lumpectomy is is not as effective as total mastectomy 172

7 p-value P (Z < 1374) 0852 There is weak evidence (p-value 0852) supporting the hypothesis that 12-year survival rates are smaller for patients receiving total mastectomies than those receiving lumpectomies Ignoring the specifics of individual cases, the study supports the position that lumpectomies are preferable to total mastectomies 5 Before finishing the example, several conditions need to be checked: 1 Independence between samples is met because subjects were randomly assigned to treatment group 2 Randomization: In a large expensive study of this type, I expect that the subjects (volunteers meeting a set of medical conditions) are representative of very large populations of individuals that receive surgical treatment for breast cancer 3 The 10% condition (n 1 and n 2 both must be less than 10% of the respective population sizes) is met for both samples since the populations are very large 4 The success/failure condition is met, because X 1 n 1 p > 10 n 1 X 1 n 1 q > 10 X 2 n 2 p > 10 and n 2 X 2 n 2 q > 10 Remark: The medical establishment in the 1970 s might have objected to H 0, as their position was that total mastectomy leads to equal or larger 12-year survival rates than lumpectomy So their favored hypothesis is p 1 p 2 and a corresponding alternative is p 1 < p 2 (same as above), which states than lumpectomy is better than total mastectomy These hypotheses can be used without modifying the test So, the hypotheses of interest are H 0 : p 1 p 2 0 H a : p 1 p 2 < 0 No modification to the test is needed because evidence against H 0 and in favor of H a is computed under the assumption that p 1 p 2 0 This assumption is consistent with H 0 (as 5 The first sometimes-effective treatment of cancer (radical surgery) because institutionalized through the advocacy of the surgeon William Halstead, whose principal objective was to remove as much tissue around a tumor (including muscles and even bone) as possible without killing the patient While Halstead experienced early success with his approach, the effectiveness of radical surgery appeared to diminish over time as more patients were treated (p 68, S Mukherjee, The Emperor of All Maladies, 2010, Scribner) 173

8 it should be in the hypothesis-testing framework) In addition, any other difference p 1 p 2 that is greater than 0 will lead to a measure of evidence that is even more extreme 6 than the measure obtained from assuming p 1 p 2 0 So, the traditionalist that might complain about a p-value computed under the assumption that p 1 p 2 0 can be silenced by pointing out that assuming p 1 p 2 d < 0 will lead to an even smaller p-value than the observed p-value Once again, the test is formulated to give advantage to the null hypothesis position 6 Contradicting H 0 and supporting H a 174