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10 - Comparing Two Population Means ( 1 vs. 2 ) General Idea: Example of Independent vs. Dependent Samples - 89 10.1 - Comparing Two Population Means Using Dependent Samples When using dependent samples each observation from population 1 has a one-to-one correspondence with an observation from population 2. One of the most common cases where this arises is when we measure the response on the same subjects before and after treatment. This is commonly called a “pre-test/post-test” situation. However, sometimes we have pairs of subjects in the two populations meaningfully matched on some prespecified criteria. For example, we might match individuals who are the same race, gender, socio-economic status, height, weight, etc... to control for the influence these characteristics might have on the response of interest. When this is done we say that we are “controlling for the effects of race, gender, etc...”. By using matched-pairs of subjects we are in effect removing the effect of potential confounding factors, thus giving us a clearer picture of the difference between the two populations being studied. DATA FORMAT Matched Pair X 1i 1 2 3 ... n X 2i X 11 X 21 X 12 X 22 X 13 X 23 ... ... X 1n X 2 n d i X 1i X 2i d1 d2 d3 ... dn The hypotheses are H o : d 0 H a : d 0 or H a : d 0 or H a : d 0 For the sample paired differences ( d i ' s ) find the sample mean (d ) and standard deviation ( s d ) . We actually can hypothesize any size difference for the mean of the paired differences that we want. For example if wanted to show a certain diet resulted in at least a 10 lb. decrease in weight then we could test if the paired differences: d = Initial weight – After diet weight had mean greater than 10 ( H a : d 10 lbs. ) Test Statistic for a Paired t-test d d t ~ t-distribution with df = n - 1 sd n Note: d = the hypothesized value for the mean paired difference. Confidence Interval for d s where t comes from the appropriate quantile of t-distribution df = n – 1. d t d n This interval has a specified % chance of covering the true mean paired difference. 90 Example 1: Effect of Captopril on Blood Pressure In order to estimate the effect of the drug Captopril on blood pressure (both systolic and diastolic) the drug is administered to a random sample n = 15 subjects. Each subjects blood pressure was recorded before taking the drug and then 30 minutes after taking the drug. The data are shown below. Syspre – initial systolic blood pressure Syspost – systolic blood pressure 30 minutes after taking the drug Diapre – initial diastolic blood pressure Diapost – diastolic blood pressure 30 minutes after taking the drug Research Questions: Is there evidence to suggest that Captopril results in a systolic blood pressure decrease of at least 10 mmHg on average in patients 30 minutes after taking it? Is there evidence to suggest that Captopril results in a diastolic blood pressure decrease of at least 5 mmHg on average in patients 30 minutes after taking it? For each blood pressure we need to consider paired differences of the form d i BPpre i BPpost i . For paired differences defined this way, positive values correspond to a reduction in their blood pressure ½ hour after taking Captopril. To answer research questions above we need to conduct the following hypothesis tests: H o : syspre syspost 10 mmHg and H o : diaprediapost 5 mmHg H a : syspre syspost 10 mmHg H a : diaprediapost 5 mmHg Below are the relevant statistical summaries of the paired differences for both blood pressure measurements. 91 The t-statistics for both tests are given below: Systolic BP Diastolic BP We can use the t-Probability Calculator in JMP to find the associated p-values or better yet use JMP to conduct the entire t-test. Systolic Blood Pressure Diastolic Blood Pressure Both tests result in rejection of the null hypotheses. This we have sufficient evidence to suggest that taking Captopril will result in mean decrease in systolic blood pressure exceeding 10 mmHg (p = _______) and a mean decrease in diastolic blood pressure exceeding 5 mmHg (p = _______). Furthermore we estimate that the mean change in systolic blood pressure will be somewhere between _______ mmHg and ______ mmHg, and that the mean change in diastolic blood pressure could be as large as ______ mmHg. 92 Example 2 - Middle Ear Effusion in Breast-Fed and Bottle-Fed Infants (Hw 3) A common symptom of otitus media in young children in the prolonged presence of fluid in the middle ear, known a middle-ear effusion. The presence of fluid may result in termporary hearing loss and interfere with normal learning skills in the first two years of life. One hypothesis is that babies who are breast-fed for at least 1 month build up some immunity against the effects of the disease and have less prolonged effusion than do bottle-fed babies. A small study of 24 pairs of babies is set up, where the babies are matched on a one-to-one basis according to age, sex, socioeconomic status, and type of medications taken. One member of the matched pair is a breast-fed baby, and other member is a bottle fed baby. The outcome variable is the duration of middle-ear effusion after the first episode of otitus media. The results are shown below. Pair Duration of effusion in Duration of effusion in Paired Difference Number breast-fed baby bottle-fed baby d = Bottle - Breast 1 20 18 -22 2 11 35 24 3 3 7 4 4 24 182 156 5 7 6 -1 6 28 33 5 7 58 223 165 8 7 7 0 9 39 57 18 10 17 76 59 11 17 186 169 12 12 29 17 13 52 39 -13 14 14 15 1 15 12 21 9 16 30 28 -2 17 7 8 1 18 15 27 12 19 65 77 12 20 10 12 2 21 7 8 1 22 19 16 -3 23 34 28 -6 24 25 20 -5 Do these data provide evidence that breast-fed babies have shorter durations of effusion when compared to bottle-fed babies that are the same age, sex, socioeconomic status, and on the same medications? 93 94 10.2 – Comparing Two Population Means Using Independent Samples Basic Form of Hypotheses: H o : 1 2 H a : 1 2 H a : 1 2 H a : 1 2 or or or or (1 2 ) 0 (1 2 ) 0 (upper-tail test) (1 2 ) 0 (lower-tail test) (1 2 ) 0 (two-tailed test, use CI for (1 2 ) to make decision) Example 1: Normal Body Temperature and Gender These data were taken from the results of a study presented in Mackowiak, P. A., Wasserman, S. S., and Levine, M. M. (1992), "A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich", Journal of the American Medical Association, 268, 1578-1580. The data consists of body temperatures and heart rates for random samples of 65 healthy men and women. Research Questions: 1) Is there evidence that 98.6 Degrees F is not the normal human body temperature? 2) Is there evidence that the normal mean human body temperature differs across gender? Question 1: Is 98.6o F the normal mean body temperature? 95 Research Question 2: Is there evidence to suggest that the mean normal human body temperature differs across gender? Assumptions for Independent Samples Test 1. The two groups must be independent of each other. 2. The two populations should be normally distributed. 3. Decide whether or not we wish to assume the population variances are equal. (see two cases below) Assessing Normality of the Two Sampled Populations To assess normality we select Normal Quantile Plot from the Oneway Analysis pulldown menu as shown below. Normality appears to be satisfied here. Two Cases 96 Case 1 - Equal Populations Variances/Standard Deviations ( 1 2 = 2 common variance to both populations) 2 2 Rule of Thumb for Checking Variance Equality If the larger sample variance is more than twice the smaller sample variance do not assume the variances are equal. Formally Checking the Equality of the Population Variances To test the equality of the population variances select Unequal Variances from the Oneway Analysis pull-down menu. The test is: Ho : F M Ha : F M JMP gives four different tests for examining the equality of population variances. To use the results of these tests simply examine the resulting p-values. If any/all are less than .10 or .05 then worry about the assumption of equal variances and use the unequal variance t-Test instead of the pooled t-Test. p-values for testing equality of population variances 97 Confidence Interval for Difference in Population Means ( 1 2 ) ( X 1 X 2 ) t SE ( X 1 X 2 ) where the standard error of the difference when we assume equal population variances is given by the following: 1 2 1 SE ( X 1 X 2 ) s p n1 n 2 where (n 1) s1 (n2 1) s 2 1 n1 n 2 2 2 sp 2 2 if n1 n 2 s 2p s12 s 22 if n1 n2 2 s p is called the “pooled estimate of the common variance ( 2 ) ”. The degrees of 2 freedom for the t-distribution is df n1 n2 2 . CI for ( F M ) 98 Hypothesis Testing ( 1 vs. 2 ) The general null hypothesis says that the two population means are equal, or equivalently there difference is zero. The alternative or research hypothesis can be any one of the three usual choices (upper-tail, lower-tail, or two-tailed). For the two-tailed case we can perform the test by using a confidence interval for the difference in the population means discussed above. H o : 1 2 or equivalently ( 1 2 ) 0 H a: 1 2 or equivalently ( 1 2 ) 0 (upper - tail) or H a : 1 2 or equivalently ( 1 2 ) 0 (two - tailed, USE CI! ) etc.... Test Statistic ( X X 2 ) ( 1 2 ) t 1 ~ t-distribu tion with df n1 n2 2 SE ( X 1 X 2 ) where the SE ( X 1 X 2 ) is as defined in the confidence interval section above. Computing the Test Statistic for Body Temperature Example Summary Statistics x F 98.39 x M 98.10 s F .743 s M .699 n F 65 n M 65 99 Performing the Test to Compare the Population Means in JMP To perform the two-sample t-test for independent samples: assuming equal population variances select the Means/Anova/Pooled t option from Oneway-Analysis pull-down menu. assuming unequal population variances select t-Test from the Oneway-Analysis pull-down menu. Because we have no evidence against the equality of the population variances assumption we will use a pooled t-Test to compare the population means. Several new boxes of output will appear below the graph once the appropriate option has been selected, some of which we will not concern ourselves with. The relevant box for us will be labeled t Test as shown below for the mean body temperature comparison. Because we have concluded that the equality of variance assumption is reasonable for these data we can refer to the output for the t-Test assuming equal variances. 100 Example 2: Effect of Cadmium Oxide on Hemoglobin Levels in Dogs An experiment was conducted to determine the examine the potential effect cadmium oxide might have on the hemoglobin levels of dogs. It is thought that cadmium oxide exposure would lead to decreased hemoglobin levels. 10 dogs were randomly assigned to the control group and 15 were randomly assigned to the cadmium oxide exposure group. Research Question: Is there evidence to suggest the cadmium oxide exposure lowers the hemoglobin level found in dogs? To answer the question of interest we need tools for comparing the population mean hemoglobin level for dogs not exposed to cadmium oxide vs. that for dogs that have had cadmium oxide exposure, i.e. how does control compare to exp osed . Assumptions for Independent Samples Test 1. The two groups must be independent of each other. 2. The observation from each group should be normally distributed. 3. Decide whether or not we wish to assume the population variances are equal. 101 Cadmium Exposure and Hemoglobin Levels: 95% CI for C E Hypothesis Test 102 Case 2 ~ Unequal Populations Variances/Standard Deviations ( 1 2 ) Assumptions: For this case we make the following assumptions 1. The samples from the two populations were drawn independently. 2. The population variances/standard deviations are NOT equal. (This can be formally tested or use rule o’thumb presented above.) 3. The populations are both normally distributed. This assumption can be relaxed when the samples from both populations are “large”. Confidence Interval for ( 1 2 ) ( X 1 X 2 ) t SE ( X 1 X 2 ) where 2 SE ( X 1 X 2 ) 2 s1 s 2 n1 n2 and df s1 2 s 2 2 n n 2 1 2 rounded down to the nearest integer 2 2 s1 2 s2 2 n n 1 2 n1 1 n2 1 The t-quantiles are the same as those we have seen previously. Hypothesis Testing Test Statistic t ( X 1 X 2 ) ( 1 2 ) ~ t-distribution with df = (see formula above) SE ( X 1 X 2 ) where the SE ( X 1 X 2 ) is as defined above. 103 Example: Cell Radii of Malignant vs. Benign Breast Tumors In your previous work with these data you noticed that the radii of malignant breast tumor cells were generally larger than the radii of benign breast tumor cells. Assuming the researchers initially hypothesized that cancerous breast tumor cells have larger radii than non-cancerous cells, conduct a test to see if this is supported by these data. The cell radii of the malignant tumors certainly appear to be larger than the cell radii of the benign tumors. The summary statistics support this with sample means/medians of rough 17 and 12 units respectively. The 95% CI’s for the mean cell radius for the two tumor groups do not overlap, which further supports a significant difference in the cell radii exists. FYI: Formally Testing the Equality of Population Variances Ho : 2 1 2 2 H a : 12 22 or equivalently Ho :1 2 In JMP H a : 1 2 Test Statistic s12 s 22 F max 2 , 2 which has an F-distribution with s 2 s1 numerator df = n1 1 and denominator df = n2 1 if s12 s 22 and are reversed if s12 s 22 . If F is large then one variance is several times larger than the other and we should reject the null in favor of the alternative. There is separate F-table for each level of significance. If our test statistic value exceeds the value in the table for appropriate level of significance and degrees of freedom we reject the null hypothesis. BETTER TO JUST USE JMP!!! 104 Because we conclude that the population variances are unequal we should use the nonpooled version to the two-sample t-test. No one does this by hand, so we will use JMP. Conclusion: 105 10.3 - Comparing Two Population Proportions Using Independent Samples ( p1 vs. p2 ) Confidence Interval for ( p1 p2 ) ( pˆ 1 pˆ 2 ) z SE( pˆ 1 pˆ 2 ) (provided n1 & n2 are “large”) where, SE ( pˆ 1 pˆ 2 ) pˆ 1 (1 pˆ 1 ) pˆ 2 (1 pˆ 2 ) n1 n2 “Large” sample sizes Both samples should be larger than 25 and both samples should have more than 5 “successes” and more than 5 “failures” and Confidence Level 95 % ( .05) 90 % ( .10 ) 99 % ( .01 ) z 1.96 1.645 2.576 Hypothesis Testing Basic Hypotheses Test Statistic z ( pˆ 1 pˆ 2 ) 0 ~ standard normal dist. provided n1 , n2 are “large” (see above) SE ( pˆ 1 pˆ 2 ) where the SE ( X 1 X 2 ) is as defined in the confidence interval section above SE ( pˆ 1 pˆ 2 ) 1 1 pq n1 n2 where p # of successes in combined sample n1 n2 n1 pˆ 1 n2 pˆ 2 n1 n2 q 1 p 106 Example: In a study conducted to investigate the non-clinical factors associated with the method of surgical treatment received for early-stage breast cancer, some patients underwent a modified radical mastectomy while others had a partial mastectomy accompanied by radiation therapy. We are interested in determining whether the age of the patient affects the type of treatment she receives. In particular, we want to know whether the proportions of women under 55 are identical in the two treatment groups. In a sample of n = 658 women who underwent a partial mastectomy and subsequent radiation therapy contains 292 women under 55, which is a sample percentage of 44.4%. In another independently drawn sample of n = 1580 women who received a modified radical mastectomy 397 women were under 55, which is a sample percentage of 25.1%. Conduct a test comparing the proportion of women each group under the age of 55 and construct a 95% confidence interval for the difference in these two proportions. 107 Fisher’s Exact Test for Comparing Two Proportions (in JMP) Enter these data as you would for setting up a 2 X 2 contingency table. In JMP, select Analyze > Fit Y by X and place Surgery in the X box and Age in the Y. The following output from JMP is obtained. The results of Fisher's Exact Test are always included in the JMP output whenever we are working with a 2 X 2contingency table. The three p-values given are for testing the following: (1) Left, p-value = 1.000 is for testing if the proportion of women under age 55 is larger for the modified radical mastectomy group than the partial mastectomy group. This is clearly not supported as the p-value >> .05. Obviously we would not conclude this when only 25% of women in the mod. rad. Group were under age 55 compared to 44.4% in the partial mastectomy group. (2) Right, p-value < .0001 is for testing if the proportion of women under age 55 is larger for the partial mastectomy group than the modified radical mastectomy group. The fact this p-value is highly significant suggests that the proportion of women under age 55 in partial mastectomy group is indeed greater than the proportion under 55 in the modified radical mastectomy group. This was the research hypothesis. (3) 2-Tail, p-value < .0001 is for testing if the proportion of women under age 55 differs between the two surgery groups. The fact this p-values is highly significant suggests that these proportion of women under age 55 is not the same for both surgery groups. 108 Example 2: Low Birth Weight and Smoking These data come from a study looking at the effects of smoking during pregnancy on birth weight. Amongst the 381 non-smokers in the study, 13 had babies with low birth weight, while amongst the 299 mothers who smoked during pregnancy, 28 had babies with low birth weight. Is there evidence to suggest that the proportion of babies born with low birth weight is greater for mothers who smoked during pregnancy? Normal Birth Weight Low Birth Weight Nonsmoker 368 96.59% 13 3.41% 381 Smoker 271 90.64% 28 9.36% 299 Column Totals 639 41 Smoking Status Row Totals 680 Hypothesis Test: 1) 2) 3) 4) 5) 109 Construct and interpret a 95% CI for ( p smo ker pnonsmo ker ) Fisher’s Exact Test Results from JMP Conclusion: 110