Inferences About Two Proportions

Transcription

1 Inferences About Two Proportions Quantitative Methods II Plan for Today Sampling two populations Confidence intervals for differences of two proportions Testing the difference of proportions Examples 1

2 Sampling two populations Consider an example: a poll of 1250 voters in Alberta revealed that 676 of them would vote Conservative and a poll of 980 voters in British Columbia showed that 487 out of those would vote Conservative. Are the proportions of all Conservative voters in Alberta and B.C. significantly different? Can we establish a confidence interval for the difference in proportions of Conservative voters in these two provinces? Sampling two populations Suppose that we have two independent samples from two different populations. First sample of size n 1 has x 1 successes. Second sample of size n 2 has x 2 successes. From the first sample we compute the sample proportion for the first population: p 1 = x 1 Τn 1 and q 1 = 1 p 1 Similarly, we compute: p 2 = x 2 Τn 2 and q 2 = 1 p 2 2

3 Confidence intervals for p 1 p 2 Suppose that the population proportions are equal to p 1 and p 2 respectively. We would like to figure out an interval estimate for the difference of two population proportions p 1 p 2 based on our knowledge of two samples. The level of confidence (1 α) is given, so we can use Table of standard normal distribution to find the confidence coefficient z( ατ2). Requirements In order to be able to use the normal distribution, the sizes of samples should be sufficiently large. Without any other assumptions, we need the following four inequalities to be satisfied: n 1 p 1 > 5 n 1 q 1 > 5 n 2 p 2 > 5 n 2 q 2 > 5 3

4 Confidence intervals for p 1 p 2 Next, we compute the error bound for the difference in two proportions: EDP = z( ατ2) p 1 q 1 + p 2 q 2 n 1 n 2 The point estimate is the difference between sample proportions, p 1 p 2, and the confidence interval is from p 1 p 2 EDP to p 1 p 2 + EDP It also can be expressed in percentages. Example: voters We compute p 1 = 676 = and 1250 p 2 = 487 = , q = , q 2 = We will construct a 95% confidence interval for the difference in population proportions p 1 p 2. First, z ατ2 = 1.96 and p 1 p 2 = Next, EDP = 1.96 p 1 q 1 + p q =

5 Example: voters Then we compute p 1 p 2 EDP = , and p 1 p 2 + EDP = Thus, the 95% confidence interval for the difference in population proportions p 1 p 2 (i.e. the difference in proportions of Conservative voters in Alberta and British Columbia) is from 0.21% to 8.57% Example: smoking workers A random sample of 145 office workers has found that 35 of them are smokers. Whereas a random sample of 190 factory workers revealed that 59 are smokers. Construct a 94% confidence interval for the difference in the proportions of smokers among the office and factory workers. Do not forget to check the requirements! 5

6 Example: smoking workers Compute: p 1 = , q 1 = p 2 = , q 2 = p 1 p 2 = , z ατ2 = 1.88 p EDP= z( ατ2) 1 q 1 + p 2 q 2 n 1 n 2 = p 1 p 2 EDP = p 1 p 2 + EDP = Thus, the 94% confidence interval for the difference in population proportions p 1 p 2 is from 16.10% to 2.28% Hypotheses testing for proportions The null hypotheses will always be that there is no difference in two population proportions. And the alternative hypothesis depends on the situation: Left-tailed test Right-tailed test Two-tailed test H 0 : p 1 p 2 = 0 H 0 : p 1 p 2 = 0 H 0 : p 1 p 2 = 0 H A : p 1 p 2 < 0 H A : p 1 p 2 > 0 H A : p 1 p 2 0 6

7 Hypotheses testing for proportions We need to compute the pooled probability: p = x 1+x 2 n 1 +n 2 and q = 1 p The pooled standard deviation: σ p = The test statistic: p q + n 1 z = p 1 p 2 σ p p q n 2 Example: voters Let us test at a 5% level of significance whether the proportions of Conservative voters are different in Alberta and B.C. We state the hypotheses: H 0 : p 1 p 2 = 0 H A : p 1 p 2 0 This is a two-tailed test. Compute: p = , q = and σ p = p q + n 1 p q n 2 =

8 Example: voters The test statistics: z = = From this point, we can either use the classical or the p-value approach. The classical approach. The critical values ±z( ατ2) = ±1.96 Draw the curve, indicate the regions. Thus, the test statistic is in the region of rejection. Decision: reject H0. Example: voters Now we will use the p-value approach. Recall that the test statistics: z = 2.06 p-value = = (We must multiply by two, because this is a two-tailed test.) Compare: p-value = < 0.05 = α Decision: reject H0. 8

9 Example: smoking workers Let us test at a 2% level of significance whether the proportion of smokers among factory workers is higher than among office workers. We state the hypotheses: H 0 : p 1 p 2 = 0 H A : p 1 p 2 < 0 This is a left-tailed test. Compute: p = , q = and σ p = p q + n 1 p q n 2 = Example: smoking workers The test statistics: z = = From this point, we can either use the classical or the p-value approach. The classical approach. The critical values z α = Draw the curve, indicate the regions. Thus, the test statistic is in the region of acceptance. Decision: fail to reject H0. 9

10 Example: smoking workers Now we will use the p-value approach. Recall that the test statistics: z = 1.40 p-value = = Compare: p-value = > 0.02 = α Decision: fail to reject H0. Practice example: students and children A poll was conducted among college students, where 150 females a 120 males were asked whether they want to have children in the future. Among those, 116 females and 101 males responded yes. (a) Construct a 98% confidence interval for the difference in proportions among female and male students who want to have children. (b) Test at a 5% level of significance whether this proportion is higher for male than female students. 10