2. Chapter 7 Confidence Intervals for 2 Proportions and 2 Means © 2006 W.H. Freeman and Company

3. Objectives (Chapter 7) Comparing two proportions  Comparing two independent samples  Large-sample CI for two proportions  Test of statistical significance

4. 3 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  Two random samples are drawn from two populations.  The number of successes in each sample is recorded.  The sample proportions are computed. 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 Point Estimator of p 1 – p 2 :

5. Comparing two independent samples We often need to estimate the difference p 1 – p 2 between two unknown population proportions based on independent samples. We can compute the difference between the two sample proportions and compare it to the corresponding, approximately normal sampling distribution for

6. Large-sample CI for two proportions For two independent SRSs of sizes n 1 and n 2 with sample proportion of successes 1 and 2 respectively, an approximate level C confidence interval for p 1 – p 2 is Use this method when C is the area under the standard normal curve between −z* and z*.

7. Cholesterol and heart attacks How much does the cholesterol-lowering drug Gemfibrozil help reduce the risk of heart attack? We compare the incidence of heart attack over a 5-year period for two random samples of middle-aged men taking either the drug or a placebo. So the 90% CI is (0.0414 − 0.0273) ± 1.645*0.00746 = 0.0141 ± 0.0125 = (0.016, 0.0266) We are 90% confident that the interval 1.6% to 2.66% captures the true percentage difference in heart attack rates for middle-aged men when taking a placebo and the cholesterol-lowering drug. Standard error of the difference p 1 − p 2 : H. attackn Drug p 2 5620512.73% Placebo p 1 8420304.14%

8. 7 Example: 95% confidence interval for p 1 – p 2 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.

9. Beware!! Common Mistake !!! A common mistake is to calculate a one-sample confidence interval for p   a one-sample confidence interval for p   and to then conclude that p  and p  are equal if the confidence intervals overlap. This is WRONG because the variability in the sampling distribution for from two independent samples is more complex and must take into account variability coming from both samples. Hence the more complex formula for the standard error.

10. INCORRECT Two single-sample 95% confidence intervals: The confidence interval for the rightie BA and the confidence interval for the leftie BA overlap, suggesting no significant difference between Ryan Howard’s ABILITY to hit right- handed pitchers and his ABILITY to hit left-handed pitchers. Rightie interval: (0.274, 0.366) HitsABphat(BA) Rightie126394.320 Leftie50222.225 Leftie interval: (0.170, 0.280) 0.095.023.167

11. Reason for Contradictory Result 10

12. Two random samples are selected from the two populations of interest. Because we compare two population means, we use the statistic 11 Chapter 7 CI for the Difference m 1 - m 2 Between 2 Population Means: Independent Samples

13. 12 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

14. Sampling distribution model for ? Sometimes used (not always very good) estimate of the degrees of freedom is min(n 1 − 1, n 2 − 1). Shape? Estimate using df 0

15. Confidence Interval for    –   14

16. Example: “Cameron Crazies”. Confidence interval for    –    Do the “Cameron Crazies” at Duke home games help the Blue Devils play better defense?  Below are the points allowed by Duke (men) at home and on the road for the conference games from a recent season. 15 Pts allowed at home 44564454751019181 Pts allowed on road 5856707480676579

17. Example: “Cameron Crazies”. Confidence interval for    –   16 Calculate a 95% CI for  1 -  2 where  1 = mean points per game allowed by Duke at home.  2 = mean points per game allowed by Duke on road n 1 = 8, n 2 = 8; s 1 2 = (21.8) 2 = 475.36 ; s 2 2 = (8.9) 2 = 79.41 n 1 = 8, n 2 = 8; s 1 2 = (21.8) 2 = 475.36 ; s 2 2 = (8.9) 2 = 79.41

18. To use the t-table let’s use df = 9; t 9 * = 2.2622 The confidence interval estimator for the difference between two means is … 17 Example: “Cameron Crazies”. Confidence interval for    –  

19. Interpretation The 95% CI for  1 -  2 is (-19.22, 18.46). Since the interval contains 0, there appears to be no significant difference between  1 = mean points per game allowed by Duke at home.  2 = mean points per game allowed by Duke on road The Cameron Crazies appear to have no affect on the ABILITY of the Duke men to play better defense. 18 How can this be?

20. Example: 95% confidence interval for    –   Example – 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. 19

21. Example: 95% confidence interval for    –   20 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 ).

22. Example: 95% confidence interval for    –   Let’s use df = 122.6; t 122.6 * = 1.9795 The confidence interval estimator for the difference between two means is… 21

23. Interpretation The 95% CI is (-56.862, -1.56). 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 1.56 and 56.862 more calories for lunch. 22

24. Let’s use df = min(43-1, 107-1) = min(42, 106) = 42; t 42 * = 2.0181 The confidence interval estimator for the difference between two means is 23 Previous Example: (cont.) confidence interval for  1 –  2 using min(n 1 –1, n 2 -1) to approximate the df

25. Beware!! Common Mistake !!! A common mistake is to calculate a one-sample confidence interval for    a one-sample confidence interval for    and to then conclude that   and   are equal if the confidence intervals overlap. This is WRONG because the variability in the sampling distribution for from two independent samples is more complex and must take into account variability coming from both samples. Hence the more complex formula for the standard error.

26. INCORRECT Two single-sample 95% confidence intervals: The confidence interval for the male mean and the confidence interval for the female mean overlap, suggesting no significant difference between the true mean for males and the true mean for females. Male interval: (18.68, 20.12) MaleFemale mean19.417.9 st. dev. s2.523.39 n50 Female interval: (16.94, 18.86) 0 1.5.3132.69

27. Reason for Contradictory Result 26

28. 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

29. Parental smokingFVCsn Yes75.59.330 No88.215.130 We are 95% confident that lung capacity is between 19.21 and 6.19 milliliters LESS in children of smoking parents. 95% confidence interval for (µ 1 − µ 2 ), with : df = 48.23  t* = 2.0104:  1 = mean FVC of children with a smoking parent;  2 = mean FVC of children without a smoking parent

30. 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

31. We are 95% confident that the mean life expectancy for left- handers is between 3.27 and 13.53 years LESS than the mean life expectancy for right-handers. 95% confidence interval for (µ 1 − µ 2 ), with : df = 105.92  t* = 1.9826:  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 Babe Ruth, baseball’s all-time best player.