Chapter 22. Comparing Two Proportions. Bin Zou STAT 141 University of Alberta Winter / 15

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1 Chapter 22 Comparing Two Proportions Bin Zou STAT 141 University of Alberta Winter / 15

2 Introduction In Ch.19 and Ch.20, we studied confidence interval and test for proportions, respectively. There is only one population in those questions. What we want to study is the actual percentage of the proportion(s). For instance, we are interested about the pass rate p of all students registered in STAT 141. But some may want to know whether girls do better than boys or not. Let p 1 and p 2 denote the pass rate of female students and male students, respectively. Then, question becomes p 1 p 2 > 0? Similarly, we can construct confidence intervals for the difference of proportions, p 1 p 2. Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

3 Standard Deviation Proportions observed from independent samples are independent. Recall Var(aX + by ) = a 2 Var(X) + b 2 Var(Y ) provided X andy are independent. Hence, Var(p 1 p 2 ) = p 1q 1 n 1 + p 2q 2 n 2, and p1 q 1 S.D.(p 1 p 2 ) = + p 2q 2. n 1 n 2 Here, q 1 = 1 p 1 and q 2 = 1 p 2, while n 1 and n 2 are the sample size of two samples. Since p 1 and p 2 are usually known, instead we probably only know ˆp 1 and ˆp 2 (sample proportions). ˆp 1 ˆq 1 S.E.( ˆp 1 ˆp 2 ) = + ˆp 2 ˆq 2. n 1 n 2 Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

4 Assumptions and Conditions 1 Randomization condition: samples are obtained independently and randomly from populations. 2 10% condition: sample should not exceed 10% of the population. 3 Independent groups condition: two groups (populations) must be independent. 4 Sample size condition: n 1 and n 2 should be big enough. 5 Success/Failure condition: we must observe at least 10 successes and at least 10 failures in each group. Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

5 The Sampling Distribution of ˆp 1 ˆp 2 Provided all the assumptions and conditions are met, the sampling distribution of ˆp 1 ˆp 2 is a normal distribution with mean standard deviation µ = p 1 p 2, S.D.( ˆp 1 ˆp 2 ) = p1 q 1 n 1 + p 2q 2 n 2. Since p 1 and p 2 are usually unknown, we estimate S.D. by standard error (S.E.), ˆp 1 ˆq 1 S.E.( ˆp 1 ˆp 2 ) = + ˆp 2 ˆq 2. n 1 n 2 Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

6 Confidence Interval of p 1 p 2 The confidence interval for the difference of two proportions is given by where ( ˆp 1 ˆp 2 ) ± Z S.E.( ˆp 1 ˆp 2 ), S.E.( ˆp 1 ˆp 2 ) = ˆp 1 ˆq 1 n 1 + ˆp 2 ˆq 2 n 2. The critical value Z is determined by the confidence level C. If we take Z to be positive (which is usual), then Prob.(Z < Z ) = C + 1 C 2 ; Prob.(Z < Z ) = 1 C 2. Why? Recall CI is always the middle interval, hence given confidence level C, each tail accounts for 1 C 2. Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

7 Example 22.1 A survey of randomly chosen adults found that 30 of the 64 women and 49 of the 74 men follow regular exercise programs. Construct a 95% confidence interval for the difference in the proportions of women and men who have regular exercise programs. (Q17 on PF1) A) (-0.356, ) B) (-0.387, 0.631) C) (-0.387, 0.662) D) (0.306, 0.631) E) (0.276, 0.662) Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

8 Example 22.2 A marketing survey involves product recognition in Ontario and British Columbia. Suppose the proportion of Ontario residents who recognized a product is p 1 and the proportion of British Columbia residents who recognized the product is p 2. The survey found a 98% confidence interval for p 1 p 2 is (-0.023, ). (Q18 on PF1) We are 98% confident that the proportion of British Columbia residents who recognized the product is between 1.9% and 2.3% higher than the proportion of Ontario residents who recognized the product. Assume we want to test whether there is any difference between the proportion of British Columbia residents who recognized the product and the proportion of Ontario residents who recognized the product. What is your conclusion given α = 2%? Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

9 Test for p 1 p 2 Null Hypothesis H 0 : p 1 p 2 = 0. You should assign a value for the parameter, p 1 p 2. Although it is possible to assign a non-zero value for p 1 p 2, that is beyond the scope of this course. Our objective is investigate whether two groups are homogenous (that is, whether two groups have the same proportion). Alternative Hypothesis H a : 1 p 1 p 2 > 0; (one right tail P-value) 2 p 1 p 2 < 0; (one left tail P-value) 3 p 1 p 2 0. (two tails on both sides P-value) Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

10 Pooled Proportion In any test, we start by assume the null hypothesis H 0 is true, which, in this case, is p 1 p 2 = 0 (or p 1 = p 2 ). Under this assumption, we think two groups are homogeneous (a naive understanding would be two groups are the same). Thus, the two samples should be pooled into one, and we need to calculate the proportion for the pooled sample. Assume, there are k 1 successes in the first sample with size n 1 and k 2 successes in the second sample of size n 2. Alternatively, if you know n 1 and ˆp 1, then k 1 = n 1 ˆp 1. Then the pooled proportion ˆp pooled = k 1 + k 2 n 1 + n 2 = n 1 ˆp 1 + n 2 ˆp 2 n 1 + n 2. Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

11 Pooled Proportion (Cont.) We then put this pooled value into the formula, substituting it for both sample proportions in the standard error formula: ( ) S.E.(ˆp pooled ) = ˆp pooled ˆq pooled + 1n1 1n2, where ˆq pooled = 1 ˆp pooled. Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

12 Test Statistic Since the sampling distribution of ˆp 1 ˆp 2 is ( N p 1 p 2,S.D.(p 1 p 2 ) = p1 q 1 + p ) 2q 2, n 1 n 2 then ( ˆp 1 ˆp 2 ) (p 1 p 2 ) S.D.(p 1 p 2 ) Z = N(0,1). Under the assumption of H 0, we have p 1 p 2 = 0. When p 1 and p 2 are unknown, we use S.E.(ˆp pooled ) to replace S.D.(p 1 p 2 ). Finally, we can obtain the test statistic for p 1 p 2 Z = ( ˆp 1 ˆp 2 ) 0 S.E.(ˆp pooled ). Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

13 Making Conclusions Once we have calculated the test statistic z, we can easily obtain the P-value (based on H a ). For instance, if H a : p 1 p 2 > 0, then P-value=P(Z > z). Remark: we use Z to denote a random variable that follows a standard normal distribution, and z for the value of Z. If P-value< α, we reject H 0. If P-value α, we fail to reject H 0. Of course, we can also make conclusions based on the critical value criterion (review Ch.21). Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

14 Example 22.3 A poll reported that 30% of 60 Canadians between the ages of 25 and 29 had started saving money for retirement. Of the 40 Canadians surveyed between the ages of 21 and 24, 25% had started saving for retirement. Carry out an appropriate hypothesis test and see whether there is any difference between the proportions of Canadians between the ages of 25 and 29 and the ages of 21 and 24 who had started saving for retirement. (Q19 of PF1) Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15

15 Example 22.4 In a random sample of 360 women, 65% watched CBC News. In a random sample of 220 men, 60% watched CBC News. Test the claim that the proportion of women who watched CBC News is higher than the proportion of men who watched CBC News. Use a significance level of Bin Zou (bzou@ualberta.ca) STAT 141 University of Alberta Winter / 15