1. Section 8.3 - A Confidence Interval for the Difference of Two Proportions Objectives: 1.To find the mean and standard error of the sampling distribution for the difference of two proportions. 2.To construct and interpret a confidence interval for the difference of two proportions.

2. Introduction A recent poll of 29,700 U.S. households found that 63% owned a pet. The percentage in 1994 was 56%. What was the change in the percentage of U.S. households that own a pet? Section 8.3 - A Confidence Interval for the Difference of Two Proportions

3. Introduction A recent poll of 29,700 U.S. households found that 63% owned a pet. The percentage in 1994 was 56%. What was the change in the percentage of U.S. households that own a pet? The obvious answer, that the percentage increased by 7 percentage points, is only an estimate because 7% is the difference of two sample percentages. These sample percentages, 56% and 63%, are probably not equal to the population percentages. We would like to find a confidence interval and margin of error to go with the difference of 7%. Section 8.3 - A Confidence Interval for the Difference of Two Proportions

4. The Formula for the Confidence Interval Section 8.3 - A Confidence Interval for the Difference of Two Proportions

5. The Formula for the Confidence Interval Section 8.3 - A Confidence Interval for the Difference of Two Proportions

6. The Formula for the Confidence Interval Section 8.3 - A Confidence Interval for the Difference of Two Proportions

7. Confidence Interval for the Difference of Two Proportions Section 8.3 - A Confidence Interval for the Difference of Two Proportions

8. Confidence Interval for the Difference of Two Proportions Section 8.3 - A Confidence Interval for the Difference of Two Proportions

9. E50. The USC Annenberg School Center for the Digital Future found that 66.9% of Americans used the Internet in 2000 and 78.6% used the Internet in 2005. Assume that the samples were independently and randomly selected and that the sample size was 2000 in both years. a.Check the conditions for constructing a confidence interval for the difference of two proportions. b.Find a 99% confidence interval for the difference of proportions. c.Interpret the resulting interval in the context of the problem. d.Is 0 in the confidence interval? What does your answer imply? Section 8.3 - A Confidence Interval for the Difference of Two Proportions

10. E50a. The USC Annenberg School Center for the Digital Future found that 66.9% of Americans used the Internet in 2000 and 78.6% used the Internet in 2005. Assume that the samples were independently and randomly selected and that the sample size was 2000 in both years. Section 8.3 - A Confidence Interval for the Difference of Two Proportions

11. E50b. The USC Annenberg School Center for the Digital Future found that 66.9% of Americans used the Internet in 2000 and 78.6% used the Internet in 2005. Assume that the samples were independently and randomly selected and that the sample size was 2000 in both years. Section 8.3 - A Confidence Interval for the Difference of Two Proportions

12. E50c. The USC Annenberg School Center for the Digital Future found that 66.9% of Americans used the Internet in 2000 and 78.6% used the Internet in 2005. Assume that the samples were independently and randomly selected and that the sample size was 2000 in both years. We are 99% confident that the difference between the proportion of Americans who used the Internet in 2005 and the proportion who used the Internet in 2000 is between 8.1% and 15.3%. We are 99% confident that the interval from 0.081 to 0.153 contains the difference in the proportions of Americans who used the Internet in 2005 and 2000. Section 8.3 - A Confidence Interval for the Difference of Two Proportions

13. E50d. The USC Annenberg School Center for the Digital Future found that 66.9% of Americans used the Internet in 2000 and 78.6% used the Internet in 2005. Assume that the samples were independently and randomly selected and that the sample size was 2000 in both years. No, 0 is not in the interval [0.081, 0.153]. The proportions in 2000 and 2005 were different. We are 99% confident that the proportion of Americans who used the Internet increased from 2000 to 2005. Section 8.3 - A Confidence Interval for the Difference of Two Proportions

14. Objectives: 1.To use simulation to construct an approximate sampling distribution for the difference of two proportions 2.To review the sampling distribution for the difference of two proportions when p 1 = p 2. 3.To use a test of significance to decide whether to reject a claim that two samples were drawn from two binomial populations that have the same proportion of successes. Section 8.4 - A Significance Test for the Difference of Two Proportions

15. Introduction In Section 8.3, we extended our knowledge of confidence intervals and sampling distributions and learned how to compute a confidence interval for the difference of two proportions. We will now extend our knowledge of tests of significance to differences of two proportions. We want to be able to determine if an observed difference can reasonably be attributed to chance, or if the observed difference is large enough to be able to rule out chance as a likely explanation. Section 8.4 - A Significance Test for the Difference of Two Proportions

16. The Test Statistic Section 8.4 - A Significance Test for the Difference of Two Proportions

17. The Test Statistic Section 8.4 - A Significance Test for the Difference of Two Proportions

18. The Test Statistic Section 8.4 - A Significance Test for the Difference of Two Proportions

19. The Test Statistic Section 8.4 - A Significance Test for the Difference of Two Proportions

20. The USC Annenberg School Center for the Digital Future found that 66.9% of Americans used the Internet in 2000 and 78.6% used the Internet in 2005. Assume that the samples were independently and randomly selected and that the sample size was 2000 in both years. Test the claim that the proportion of Americans using the Internet increased between 2000 and 2005. Use a significance level of  = 0.01. Section 8.4 - A Significance Test for the Difference of Two Proportions

21. Check conditions for inference. Section 8.3 - A Significance Test for the Difference of Two Proportions

22. Write a null and alternative hypothesis. Section 8.4 - A Significance Test for the Difference of Two Proportions

23. Compute the test statistic. Section 8.4 - A Significance Test for the Difference of Two Proportions

24. Compute the P-value. Section 8.4 - A Significance Test for the Difference of Two Proportions

25. Determine the critical value. Section 8.4 - A Significance Test for the Difference of Two Proportions

26. Write a conclusion. Critical value method: Because the test statistic z = 8.31 falls to the right of the critical value z* = 2.33, we reject the null hypothesis that p 2005 - p 2000 = 0 at the  = 0.01 level. P-value method: Because the P-value of 0.0001 is less than the significance level 0.01, we reject the null hypothesis that p 2005 - p 2000 = 0 at the  = 0.01 level. Section 8.4 - A Significance Test for the Difference of Two Proportions

27. Write a conclusion. We conclude that there is sufficient sample evidence to support the claim that the proportion of Americans who use the Internet has increased between 2000 to 2005. Section 8.4 - A Significance Test for the Difference of Two Proportions