9-6. Testing the difference between proportions /20

Transcription

1 9-6 Testing the difference between proportions 1

2 Homework Discussion Question p514 Ex 9-6 p514 2, 3, 4, 7, 9, 11 (use both the critical value and p-value for all problems. 2

3 Objective Perform hypothesis tests for two proportions using the z statistic. 3

4 Hypothesis Testing Proportions Compares the proportions of two groups (samples). Is the proportion of glioma higher among regular cell-phone users than non-users? Do women suffer higher rates of depression? Remember, the proportion of a sample is represented by p. For the current scenarios, we are comparing p 1 and p 2. 4

5 z test Comparing a sample proportion against a population value. Test Statistic = Observed Value - Expected Value standard error Z = Sample proportion - Population proportion adjusted standard deviation Z = p! - p pq n n = sample size p! = x n 5

6 Once Again - It is the Difference As we did when comparing means of two samples, we will find the difference between the proportions of two samples and compare that to the expected difference between the population proportions, typically 0. The hypotheses are then stated in terms of the difference in the population proportions, p 1 and p 2. H 0 : p 1 = p 2 (p 1 - p 2 = 0) and H 1 : p 1 p 2 (p 1 - p 2 0) H 0 : p 1 p 2 (p 1 - p 2 0) and H 1 : p 1 > p 2 (p 1 - p 2 > 0) H 0 : p 1 p 2 (p 1 - p 2 0) and H 1 : p 1 < p 2 (p 1 - p 2 < 0) 6

7 p H 0 assumes p 1 = p 2. We do not know the value of p. We do know two estimates of p, p 1 and p 2. Combining (pooling) these two sample proportions gives a better estimate of p. That allows us to use a pooled (weighted) value: p pooled = p = p! 1 + n 2 p! 2 + n 2 q = 1 - p 7

8 p A little arithmetic to simplify notation.! x p = n p = p! 1 + n 2 p! 2 + n 2 = x 1 + n 2 x 2 n 2 + n 2 = x 1 + x 2 + n 2 p = x 1 + x 2 + n 2 q = 1 p This value, p, is used when we calculate the value of the test statistic. 8

9 z test Comparing two sample proportions. Test Statistic = Observed Value - Expected Value standard error Z = Difference in sample proportions - Difference in population proportions pooled standard deviation Z = (! p - p! ) p p ( ) p = x 1 + x 2 + n 2 q = 1 p pq + pq n 2 9

10 Conditions As before the samples must be independent, and np and nq are greater than or equal to 10 (or 5 depending on who s talking). 10

11 Confidence Interval As with the all previous hypothesis testing, the z statistic can be used to determine a confidence interval. We find the interval within which we expect to find the proportion difference. ( p! p! ) z 1 2 α 2 p! q!! p q! 2 2 < p p < ( p! p! ) + z n n α p! 1 q! 1 + p! 2 q! 2 n 2 11

12 Hypothesis Testing 1. Formulate hypotheses: H 0, H 1 2. Determine which test statistic that will assess the evidence against the null hypothesis. (Z or t) 3. Find the critical value of the test statistic (z c or t c ), the calculated value of the test statistic (t or z) and the probability of the calculated test statistic(p-value). 4. Decision: Reject or fail to reject H Conclusion: Answer the question posed in context. 12

13 Example A study was conducted to compare the proportion of smokers under 20 to the proportion of middle-aged smokers. A sample of 200 people under 20 (Y) had 56 smokers. A sample of 175 between the ages of 35 and 50 (M) showed 55 smokers. Is there a lower proportion in the younger group? Test at the.01 level. x y = 56, n y = 200, p y =.28 x m = 55, n m = 175, p m =.3143 x y = young smokers x m = middle-aged smokers 13

14 Conditions x y = 56, n y = 200, p y =.28 x m = 55, n m = 175, p m =.3143 n M p! M = 55 n Y p! Y = 56 q! Y = 1.28 =.72 n Y q! Y = 200 i.72 = 144 q! M = =.6857 n M q! M = 175 i.686 = 120 All values are greater tha0, so we can use the 2-proportion z test. 14

15 Hypotheses x y = 56, n y = 200, p y =.28 x m = 55, n m = 175, p m =.3143 One-tailed. H 0 : p Y p M and H 1 : p Y < p M Or H 0 : p Y p M 0 and H 1 :: p Y p M < 0 Test Statistic and Critical Value Because we are testing proportions, we know σ, thus we use a two proportion z-test. α =.01 z* = invnorm(.01, 0, 1) =

16 Test Statistic and p-value TI-84 x y = 56, n y = 200, p y =.28 x m = 55, n m = 175, p m =.3143 STAT TESTS 6:2-PropZTest x1:56 n1:200 x2:55 n2:175 p-value 2-PropZTest p1<p2 z= p= p1: p2 <p2 >p2 p1: Calculate p2: % p pooled p: n1: 200 n2: 175 1% -3σ -2σ -1σ 0 1σ 2σ 3σ 16

17 Test Statistic and p-value x y = 56, n y = 200, p y =.28 x m = 55, n m = 175, p m =.3143 p = x 1 + x 2 + n 2 = =.2960 q = 1 p = =.7040 Z = (! p - p! ) p p ( ) = ( ) ( 0) =.7259 pq + pq n 2 (.296)(.704) p(p Y - p M ) = p(z ) = Normcdf(-9, , 0, 1) =

18 Decision x y = 56, n y = 200, p y =.28 x m = 55, n m = 175, p m = > -2.33,.2340 >.01 we fail to reject the null. Conclusion There is not sufficient evidence to suggest there is a lower proportion of smokers in the younger group. 18

19 Confidence Interval x y = 56, n y = 200, p y =.28 x m = 55, n m = 175, p m =.3143 TI-84 STAT TESTS B:2-PropZInterval x1: 56 n1: 200 x2: 55 n2: 175 C-Level:.98 Calculate 2-PropZInt (.1444,.0758) p1:.28 p2: n1: 200 n2: 175 We are 98% confident the true difference in proportion of smokers over 20 and under 20 is between -.14 and.08. Since 0 is in the interval there is not sufficient evidence to conclude there is a difference in proportion of smokers in the two groups. 19

20 Confidence Interval x y = 56, n y = 200, p y =.28 x m = 55, n m = 175, p m =.3143 ( p! p! ) + z 1 2 α 2 p! q!! p q! 2 2 < p p < ( p! p! ) + z n n α p! 1 q! 1 + p! 2 q! 2 n 2 ( ) (.28)(.72) (.3143)(.6857) 175 < p 1 p 2 (.28)(.72) < ( ) (.3143)(.6857) < p 1 p 2 < < p 1 p 2 <