1. Confidence Intervals for µ 1 - µ 2 and p 1 - p 2 1

2. We are interested in: Confidence intervals for the difference between two means. Confidence intervals for the difference between two proportions. 2 Inference about Two Populations

3. Two random samples are drawn from the two populations of interest. Because we compare two population means, we use the statistic. 3 Confidence Intervals for the Difference between Two Population Means µ 1 - µ 2 : Independent Samples

4. 4 Population 1Population 2 Parameters: µ 1 and  1 2 Parameters: µ 2 and  2 2 (values are unknown) (values are unknown) Sample size: n 1 Sample size: n 2 Statistics: x 1 and s 1 2 Statistics: x 2 and s 2 2 Estimate µ 1  µ 2 with x 1  x 2

5. Confidence Interval for    –   5 Note: when the values of  1 2 and  2 2 are unknown, the sample variances s 1 2 and s 2 2 computed from the data can be used.

6. Do people who eat high-fiber cereal for breakfast consume, on average, fewer calories for lunch than people who do not eat high-fiber cereal for breakfast? A sample of 150 people was randomly drawn. Each person was identified as a consumer or a non-consumer of high- fiber cereal. For each person the number of calories consumed at lunch was recorded. 6 Example: confidence interval for    –  

7. 7 Solution: The parameter to be tested is the difference between two means. The claim to be tested is: The mean caloric intake of consumers (  1 ) is less than that of non-consumers (  2 ). Use s 1 2 = 4,103 for  1 2 and s 2 2 = 10,670 for  2 2

8. The confidence interval estimator for the difference between two means is 8 Example: confidence interval for    –  

9. The 95% CI is (-56.59, -1.83). We are 95% confident that the interval (-56.59, -1.83) contains the true but unknown difference    –   Since the interval is entirely negative (that is, does not contain 0), there is evidence from the data that µ 1 is less than µ 2. We estimate that non- consumers of high-fiber breakfast consume on average between 1.83 and 56.59 more calories for lunch. 9 Interpretation

10. Does smoking damage the lungs of children exposed to parental smoking? Forced vital capacity (FVC) is the volume (in milliliters) of air that an individual can exhale in 6 seconds. FVC was obtained for a sample of children not exposed to parental smoking and a group of children exposed to parental smoking. We want to know whether parental smoking decreases children’s lung capacity as measured by the FVC test. Is the mean FVC lower in the population of children exposed to parental smoking? Parental smokingFVCsn Yes75.59.330 No88.215.130

11. Parental smokingFVCsn Yes75.59.330 No88.215.130 We are 95% confident that lung capacity in children of smoking parents is between 19.05 and 6.35 milliliters LESS than in children without a smoking parent. 95% confidence interval for (µ 1 − µ 2 ):  1 = mean FVC of children with a smoking parent;  2 = mean FVC of children without a smoking parent

12. The data below show the sugar content (as a percentage of weight) of 10 brands of cereal randomly selected from a supermarket shelf that is at a child’s eye level and 8 brands selected from the top shelf. 12 Bunny Rabbits and Pirates on the Box Eye level 40.35545.743. 3 50.345.953.54344.244 Top202. 2 7.54.422.216.614.510 Create and interpret a 95% confidence interval for the difference  1 –  2 in mean sugar content, where  1 is the mean sugar content of cereal at a child’s eye level and  2 is the mean sugar content of cereal on the top shelf.

13. Eye level 40.35545.743.350.345.953.54344.244 Top202.27.54.422.216.614.510 13

14. Interpretation We are 95% confident that the interval (28.46, 40.22) contains the true but unknown value of  1 –  2. Note that the interval is entirely positive (does not contain 0); therefore, it appears that the mean amount of sugar  1 in cereal on the shelf at a child’s eye level is larger than the mean amount  2 on the top shelf. 14

15. Do left-handed people have a shorter life-expectancy than right-handed people?  Some psychologists believe that the stress of being left- handed in a right-handed world leads to earlier deaths among left-handers.  Several studies have compared the life expectancies of left- handers and right-handers.  One such study resulted in the data shown in the table. We will use the data to construct a confidence interval for the difference in mean life expectancies for left- handers and right-handers. Is the mean life expectancy of left-handers less than the mean life expectancy of right-handers? HandednessMean age at deathsn Left66.825.399 Right75.215.1888 left-handed presidents star left-handed quarterback Steve Young

16. We are 95% confident that the mean life expectancy for left- handers is between 3.32 and 13.48 years LESS than the mean life expectancy for right-handers. 95% confidence interval for (µ 1 − µ 2 ):  1 = mean life expectancy of left-handers;  2 = mean life expectancy of right-handers HandednessMean age at deathsn Left66.825.399 Right75.215.1888 The “Bambino”,left-handed hitter Babe Ruth, baseball’s all-time best hitter

17. Example An ergonomic chair can be assembled using two different sets of operations (Method A and Method B) The operations manager would like to know whether the assembly time under the two methods differ. 17 Example: confidence interval for    –  

18. Example Two samples are randomly and independently selected A sample of 25 workers assembled the chair using method A. A sample of 25 workers assembled the chair using method B. The assembly times were recorded Do the assembly times of the two methods differs? 18 Example: confidence interval for    –  

19. 19 Assembly times in Minutes Solution The parameter of interest is the difference between two population means. The claim to be tested is whether a difference between the two methods exists. Use s 1 2 =.848 for  1 2 and s 2 2 = 1.303 for  2 2

20. 20 Example: confidence interval for    –   A 95% confidence interval for  1 -  2 is calculated as follows: We are 95% confident that the interval (-0.3029, 0.8469) contains the true but unknown  1 -  2 Notice: “Zero” is included in the confidence interval

21. In this section we deal with two populations whose data are qualitative. For qualitative data we compare the population proportions of the occurrence of a certain event. Examples Comparing the effectiveness of new drug versus older one Comparing market share before and after advertising campaign Comparing defective rates between two machines 21 Confidence Intervals for the difference p 1 – p 2 between two population proportions

22. Parameter When the data are qualitative, we can only count the occurrences of a certain event in the two populations, and calculate proportions. The parameter we want to estimate is p 1 – p 2. Statistic An estimator of p 1 – p 2 is (the difference between the sample proportions). 22 Parameter and Statistic

23. Two random samples are drawn from two populations. The number of successes in each sample is recorded. The sample proportions are computed. 23 Point Estimator: Sample 1 Sample size n 1 Number of successes x 1 Sample proportion Sample 1 Sample size n 1 Number of successes x 1 Sample proportion Sample 2 Sample size n 2 Number of successes x 2 Sample proportion Sample 2 Sample size n 2 Number of successes x 2 Sample proportion x n 1 1 ˆ  p 1

24. Confidence Interval for p 1  p 2 24

25. Estimating the cost of life saved Two drugs are used to treat heart attack victims: Streptokinase (available since 1959, costs $460) t-PA (genetically engineered, costs $2900). The maker of t-PA claims that its drug outperforms Streptokinase. An experiment was conducted in 15 countries. 20,500 patients were given t-PA 20,500 patients were given Streptokinase The number of deaths by heart attacks was recorded. 25 Example: confidence interval for p 1 – p 2

26. Solution The problem objective: Compare the outcomes of two treatments. The data are qualitative (a patient lived or died) The parameter to be estimated is p 1 – p 2. p 1 = death rate with Streptokinase p 2 = death rate with t-PA 26 Example: confidence interval for p 1 – p 2 (cont.)

27. Experiment results A total of 1497 patients treated with Streptokinase died. A total of 1292 patients treated with t-PA died. Estimate the difference in the death rates when using Streptokinase and when using t-PA. 27 Example: confidence interval for p 1 – p 2 (cont.)

28. Compute: Manually Sample proportions: The 95% confidence interval estimate is 28 Example: confidence interval for p 1 – p 2 (cont.)

29. Interpretation The interval (.0051,.0149) for p 1 – p 2 does not contain 0; it is entirely positive, which indicates that p 1, the death rate for streptokinase, is greater than p 2, the death rate for t-PA. We estimate that the death rate for streptokinase is between.51% and 1.49% higher than the death rate for t-PA. 29 Example: confidence interval for p 1 – p 2 (cont.)

30. Example: 95% confidence interval for p 1 – p 2 30 The age at which a woman gives birth to her first child may be an important factor in the risk of later developing breast cancer. An international study conducted by WHO selected women with at least one birth and recorded if they had breast cancer or not and whether they had their first child before their 30 th birthday or after. CancerSample Size Age at First Birth > 30 683322021.2% Age at First Birth <= 30 149810,24514.6% The parameter to be estimated is p1 – p2. p1 = cancer rate when age at 1 st birth >30 p2 = cancer rate when age at 1 st birth <=30 We estimate that the cancer rate when age at first birth > 30 is between.05 and.082 higher than when age <= 30.