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UNC-Chapel Hill STOR 151 - General Method- Difference of Means

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General Method: Difference of Means1. Calculate ¯x1, ¯x2, SE1, SE2.2. Combined SE =qSE21+ SE22.ASSUMES INDEPENDENT SAMPLES.3. Calculate df: either Welch-Satterthwaite formula or simplerdf = min(n1, n2) − 1.4. For a hypothesis test, t =¯x1−¯x2SE; convert to a P-value usingtable of t statistics5. For a confidence interval, calculate critical value t∗corre-sponding to desired confidence level (e.g. in our example wehad df = 13, confidence level 95%, led to t∗= 2.16). Thenthe confidence interval is¯x2− ¯x1± t∗× SE.1Example (Question 10.28, page 493)Following are the numbers of newspapers read by a sample ofwomen and of menWomen: 5,3,6,3,7,1,1,3,0,4,7,2,2,7,3,0,5,0,4,4,5,14,3,1,2,1,7,2,5,3,7Men: 0,3,7,4,3,2,1,12,1,6,2,2,7,7,5,3,14,3,7,6,5,5,2,3,5,5,2,3,3(a) Construct and interpret a plot comparing responses of malesand females(b) Construct and interpret a 95% confidence intervals compar-ing populations means(c) Show the five steps of a significance test comparing popula-tions means(d) State and check the assumptions20 2 4 6 8 10 12 14Newspapers ReadMENWOMENBox plot for number of newspapers read by women and by men.3(a) See boxplots; male and female distributions look very similar(b) With suffix 1 indicating women, 2 for men: ¯x1= 3.774, ¯x2=4.414, s1= 2.929, s2= 3.100, n1= 31, n2= 29, SE =rs21n1+s22n2= 0.78. We have df = 57.1 according to Welch-Satterthwaite formula, df = 28 by simpler formula. Based ondf = 28, the critical value of t for a 95% confidence intervalis t∗= 2.048, so the 95% confidence interval for µ2− µ1is4.414 − 3.774 ± 2.048 × 0.78 = (−0.957, 2.237).(c) The t statistic for a hypothesis test is¯x2−¯x1SE= 0.82; this iswell below the critical value for a test at significance level.05 (the critical value is t∗= 2.048, as in part 2) so we DONOT REJECT the null hypothesis that the means for menand women are equal.4(d) The assumptions require independence of the two samples(probably more or less correct); randomness of the two sam-ples (depends on how the samples were obtained); and ap-proximately normal distributions for the samples themselves(probably true to a reasonable approximation).5Paired Comparison TestsConsider the following dataset, based on the midterm and finalexam scores of a recent course of mine (not STOR 151!):Student 1 2 3 4 5 6 7 8 9 10Midterm 86 100 90 90 94 73 76 76 95 87Final 84 95 77 83 70 76 54 81 90 84Difference 2 5 13 7 24 –3 22 –5 5 3Mean midterm score=86.7; Mean final score=79.4; Difference7.3Is this significant evidence of a difference?6We could apply the same test as previously, which would lead to¯x1= 86.7,¯x2= 79.4,s1= 9.06,s2= 11.37,SE =s9.06210+11.37210= 4.60,t =86.7 − 79.44.60= 1.59,df = 9.The two-sided P-value is 0.15 — this is greater than 0.05, there-fore the result is not significant.However, there is an error in this calculation....7The two samples are not independent since they represent scoresfrom the same students.Instead, we apply a matched pairs test:1. Calculate the difference in scores for each student.2. Carry out a single-sample test for the null hypothesis thatthe mean difference is 0.8DetailsThe 10 differences 2, 5, 13, 7, 24, –3, 22, –5, 5, 3 have a mean¯x = 7.3 and a standard deviation s = 9.67.The standard error is9.67√10= 3.06.The t statistic is7.33.06= 2.39.The corresponding two-sided P-value is 0.04.Therefore, the result is statistically significant9Message of this example:The standard two-sample test is valid only when the two sam-ples are independent (along with other assumptions: quantitativevariables, random sampling, approximately normal distributions)When the observations are directly paired (e.g. two exam scoresfrom the same student, two medical results from the same pa-tient, etc.) it is possible to apply a paired comparison test in-stead.The main difference is a different method of computing the stan-dard error. This has implications for both hypothesis testing andconfidence interval calculations.10Example: Problem 10.47, page 511/512.In a study to determine whether exercise reduced blood pressure,a sample of three patients was tested, with the following results:Subject Before After1 150 1302 165 1403 135 120(a) Explain why the three “before” observations and the three“after” observations are dependent samples.(b) Find the sample mean of the before scores, sample mean ofthe after scores, and the sample mean of d = before − after.(c) Find a 95% confidence interval for the mean difference, andinterpret it.11(a) They are from the same subjects; it is a “matched pairs”design.(b)150+165+1353= 150;130+140+1203= 130; difference 20.(c) The three differences 20, 25, 15 have mean 20 and standarddeviationr52+0+522= 5. The standard error is5√3= 2.887.Based on the t table with df = 2, the critical value of t∗for a 95% confidence interval is 4.303. Therefore, a 95%confidence interval for the difference is 20 ±4.303 ×2.887 =(7.58, 32.42). This does not contain 0, so despite the verysmall sample size, the difference is statistically


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UNC-Chapel Hill STOR 151 - General Method- Difference of Means

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