1. Two-sample Proportions Section 12.2.1

2. Starter 12.2.1 One-sample procedures for proportions can also be used in matched pairs experiments. Here is an example: Each of 50 randomly selected subjects tastes two unmarked cups of coffee and says which he/she prefers. One cup in each pair contains instant coffee; the other is fresh-brewed. 31 of the subjects prefer fresh-brewed. 1.Test the claim that a majority of people prefer the taste of fresh-brewed coffee. State hypotheses, check assumptions, find the test statistic and p-value. Is your result significant at the 5% level? 2.Find a 90% confidence interval for the true proportion that prefer fresh-brewed. 3.When you do an experiment like this, in what order should you present the two cups of coffee to the subjects?

3. Today’s Objectives The student will form confidence intervals and perform hypothesis tests on the difference of proportions from samples of two populations California Standards 17.0 Students determine confidence intervals for a simple random sample from a normal distribution of data and determine the sample size required for a desired margin of error. 18.0 Students determine the P- value for a statistic for a simple random sample from a normal distribution.

4. Two-sample Proportions To compare the proportions of two populations, we need to define notation Population Proportion Number of Successes Sample Size Sample Proportion 1p1p1 x1x1 n1n1 2p2p2 x2x2 n2n2

5. We are interested in the difference (p 1 -p 2 ) between the two population proportions If the population proportions are the same, then p 1 = p 2, so (p 1 -p 2 ) = 0 The C.I. estimates the difference (p 1 -p 2 ) The hypothesis test looks at: H o : p 1 = p 2 versus one of these alternatives: H a : p 1 ≠ p 2 H a : p 1 < p 2 H a : p 1 > p 2

6. Assumptions for the: Two-Population Z procedure for proportions In both samples, we have a valid SRS Population at least ten times sample size All counts of success and failure are at least 5 –So including both populations there are still 10

7. Assumptions So Far… Procedure One population t for means Two population t for means One population z for proportions Two population z for proportions Assumptions SRS, Normal Dist SRS, Normal, Independent SRS, large pop, 10 succ/fail SRS, large pop, 5 s/f each

8. How do we estimate p 1 – p 2 ? For large sample size (n>30), the variable (p 1 – p 2 ) is –normally distributed –with mean of 0 –and standard deviation equal to Standard Error For confidence intervals, we use the usual approach –Estimate ± z*SE For hypothesis tests, we find the z statistic and its associated p-value

9. Formulas for SE and z Here is the formula for Standard Error: Here is the formula for the z statistic: where p-hat is the “pooled” proportion found by dividing total successes by total sample size.

10. Do the actual calculations on TI For Hypothesis Tests –Stat : Tests : 2-PropZTest –Enter successes (x), sample size (n), H a –Calculate and write a conclusion For Confidence Intervals –Stat : Tests : 2-PropZInt –Enter successes (x), sample size (n), C-Level –Calculate and write a conclusion

11. Example: Confidence Interval A study asked whether poor children who attended pre-school required less social services as adolescents than those who had not attended pre- school. The study found that 49 of 61 who had not attended pre-school later needed social services. Call them group 1. 38 of 62 who had attended pre-school needed social services. Call them group 2 Use the calculator to form a 95% confidence interval for the difference between the two population proportions. Write your conclusions in a sentence

12. Example Solution Stat:Tests:2-PropZInt Enter x 1 = 49, n 1 = 61, x 2 = 38, n 2 = 62 Choose C-Level =.95 Calculate to find (.033,.347) Conclusion: We are 95% confident that the percent needing social services was between 3.3% and 34.7% lower among those who attended pre-school

13. Example: Hypothesis Test High levels of cholesterol are associated with heart attacks. An experiment was conducted to see whether a new drug reduces cholesterol level and therefore heart attacks. In the treatment group, 2051 men took the drug. 56 had heart attacks. Call them group 1. In the control group, 2030 men took a placebo. 84 of them had heart attacks. Call them group 2. So about 4.1% of the men in the placebo group had heart attacks while only 2.7% in the treatment group had heart attacks. Is this apparent reduction statistically significant? Perform a test and write your answer.

14. Example Solution H o : p 1 = p 2 H a : p 1 < p 2 Stat:Tests:2-PropZTest Enter x 1 = 56, n 1 = 2051, x 2 = 84, n 2 = 2030 Choose p 1 < p 2 Calculate to find z = -2.47 and p =.007 Conclusion: There is strong evidence (p=.007) to support the claim that the drug reduces heart attacks.

15. Today’s Objectives The student will form confidence intervals and perform hypothesis tests on the difference of proportions from samples of two populations California Standards 17.0 Students determine confidence intervals for a simple random sample from a normal distribution of data and determine the sample size required for a desired margin of error. 18.0 Students determine the P- value for a statistic for a simple random sample from a normal distribution.

16. Homework Read pages 678 – 689 Do problems 21 - 24