1. June 26, 2008Stat 111 - Lecture 151 Two-Sample Inference for Proportions Statistics 111 - Lecture 15

2. June 26, 2008Stat 111 - Lecture 152 Administrative Notes HW 5 due on July 1 (next Wednesday) Exam is from 10:40-12:10 on July 2 (next Thursday) Will focus on material covered after midterm You should expect a question or two on topics covered before the midterm

3. June 26, 2008Stat 111 - Lecture 153 Count Data and Proportions Last class, we re-introduced count data: X i = 1 with probability p and 0 with probability (1-p) Example: Pennsylvania Primary X i = 1 if you favor Obama, X i = 0 if not What is the proportion p of Obama supporters at Penn? We derived confidence intervals and hypothesis tests for a single population proportion p

4. June 26, 2008Stat 111 - Lecture 154 Two-Sample Inference for Proportions Today, we will look at comparing the proportions between two samples from distinct populations Two tools for inference: Hypothesis test for significant difference between p 1 and p 2 Confidence interval for difference p 1 - p 2 Population 1:p 1 Sample 1: Population 2:p 2 Sample 2:

5. June 26, 2008Stat 111 - Lecture 155 Example: Vitamin C study Study done by Linus Pauling in 1971 Does vitamin C reduce incidence of common cold? 279 people randomly given vitamin C or placebo Is there a significant difference in the proportion of colds between the vitamin C and placebo groups? GroupColdsTotal Vitamin C17139 Placebo31140

6. June 26, 2008Stat 111 - Lecture 156 Hypothesis Test for Two Proportions For two different samples, we want to test whether or not the two proportions are different: H 0 : p 1 = p 2 versus H a : p 1  p 2 The test statistic for testing the difference between two proportions is: is called the pooled standard error and has the following formula: is called the pooled sample proportion

7. June 26, 2008Stat 111 - Lecture 157 Example: Vitamin C study We need the following three sample proportions: = 17/139 =.12 = 31/140 =.22 = 48/279 =.17 Next, we calculate the pooled standard error: = – = = √(.17*.83*(1/139 + 1/140)) =.045 Finally, we calculate our test statistic: z = (.12-.22)/.045 = -2.22 Vitamin C groupY 1 = 17n 1 = 139 Placebo groupY 2 = 31n 2 = 140

8. June 26, 2008Stat 111 - Lecture 158 Hypothesis Test for Two Proportions We use the standard normal distribution to calculate a p-value for our test statistic Since we used a two-sided alternative, our p-value is 2 x P(Z < -2.22) = 2 x 0.0132 = 0.0264 At a  = 0.05 level, we reject the null hypothesis Conclusion: the proportion of colds is significantly different between the Vitamin C and placebo groups Z = -2.22 prob = 0.0132

9. June 26, 2008Stat 111 - Lecture 159 Confidence Interval for Difference We use the two sample proportions to construct a confidence interval for the difference in population proportions p 1 - p 2 between two groups: Interval is centered at the difference of the two sample proportions As usual, the multiple Z * you use depends on the confidence level that is needed eg. for a 95% confidence interval, Z * = 1.96

10. June 26, 2008Stat 111 - Lecture 1510 Example: Vitamin C study Want a C.I. for difference in proportion of colds p 1 - p 2 between Vitamin C and placebo Need sample proportions from before: = 17/139 =.12 = 31/140 =.22 Now, we construct a 95% confidence interval: (.12-.22) +/- √(.12*.88/139 +.22*.78/140) =(-.19,-.01) Vitamin C causes decrease in cold proportions between 1% and 19%

11. June 26, 2008Stat 111 - Lecture 1511 Another Example Has Shaq gotten worse at free throws over his career? Free throws are uncontested shots given to a player when they are fouled…Shaquille O’Neal is notoriously bad at them Two Samples: the first three years of Shaq’s career vs. a later three years of his career Group Free Throws Made Free Throws Attempted Early Years Y 1 = 1353n 1 = 2425 Later Years Y 2 = 1121n 2 = 2132

12. June 26, 2008Stat 111 - Lecture 1512 Another Example: Shaq’s Free Throws We calculate the sample and pooled proportions = 1353/2425=.558 =1121/2132=.526 =2474/4557=.543 Next, we calculate the pooled standard error: = √(.543*.467(1/2425+1/2132))=.015 Finally, we calculate our test statistic: Z = (.558-.526)/.015 = 2.13

13. June 26, 2008Stat 111 - Lecture 1513 Another Example: Shaq’s Free Throws We use the standard normal distribution to calculate a p-value for our test statistic Since we used a two-sided alternative, our p-value is 2 x P(Z > 2.13) = 0.0332 At  = 0.05 level, we reject null hypothesis Conclusion: Shaq’s free throw success is significantly different now than early in his career Z = 2.13 prob = 0.0166

14. June 26, 2008Stat 111 - Lecture 1514 Confidence Interval: Shaq’s FT We want a confidence interval for the difference in Shaq’s free throw proportion: = 1353/2425=.558 =1121/2132=.526 Now, we construct a 95% confidence interval: (.558-.526) +/- 1.96 *√(.558*.442/2425 +.526*.474/2132) (.003,.061) Shaq’s free throw percentage has decreased from anywhere between 0.3% to 6.1%

15. June 26, 2008Stat 111 - Lecture 1515 Is Shaq still bad at Free Throws?