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Samples from Normal Populations with Known Variances If the population variances are known to be σ12 andσ22 , then the two-sided confidence interval for the difference of the population means µ1 − µ2 with confidence level 1 − α is ! r r σ12 σ22 σ12 σ22 X̄ − Ȳ + zα/2 + , + X̄ − Ȳ − zα/2 m n m n Samples from Normal Populations with Known Variances In case of known population variances, the procedures for hypothesis testing for the difference of the population means µ1 − µ2 is similar to the one sample test for the population mean: Null hypothesis H0 : µ1 − µ2 = ∆0 Test statistic value z= Alternative Hypothesis Ha : µ1 − µ2 > ∆0 Ha : µ1 − µ2 < ∆0 Ha : µ1 − µ2 6= ∆0 (X̄ − Ȳ ) − ∆0 q σ12 σ22 m + n Rejection Region for Level α Test z ≥ zα (upper-tailed) z ≤ −zα (lower-tailed) z ≥ zα/2 or z ≤ −zα/2 (two-tailed) Large Size Samples When the sample size is large, both X̄ and Ȳ are approximately normally distributed, and Z= (X̄ − Ȳ ) − (µ1 − µ2 ) q S12 S22 m + n is approximately a standard normal rv. Large Size Samples In case both m and n are large (m, n > 30), the procedure for constructing confidence interval and testing hypotheses for the difference of two population means are similar to the one sample case. The two-sided confidence interval for the difference of the population means µ1 − µ2 with confidence level 1 − α is s s 2 2 2 2 S1 S S1 S X̄ − Ȳ − z + 2, X̄ − Ȳ + zα/2 + 2 α/2 m n m n Large Size Samples In case both m and n are large (m, n > 30), the procedures for hypothesis testing for the difference of the population means µ1 − µ2 is : Null hypothesis H0 : µ1 − µ2 = ∆0 Test statistic value z= Alternative Hypothesis Ha : µ1 − µ2 > ∆0 Ha : µ1 − µ2 < ∆0 Ha : µ1 − µ2 6= ∆0 (X̄ − Ȳ ) − ∆0 q S12 S22 m + n Rejection Region for Level α Test z ≥ zα (upper-tailed) z ≤ −zα (lower-tailed) z ≥ zα/2 or z ≤ −zα/2 (two-tailed) Two-Sample t Test and C.I. Theorem When the population distributions are both normal, the standardized variable T = (X̄ − Ȳ ) − (µX − µY ) q 2 S2 S Y X m + n has approximately a t distribution with df ν estimated from the data by ν= s2 X m (s 2 /m)2 X m−1 + + s2 Y n 2 (s 2 /n)2 Y n−1 (round ν down to the nearest integer.) Two-Sample t Test and C.I. The two-sample t confidence interval for µX − µY with confidence level 100(1 − α)% is given by s s sX2 sX2 sY2 sY2 (x̄ − ȳ ) − t + , (x̄ − ȳ ) + tα/2,ν + α/2,ν m n m n A one-sided confidence bound can be obtained by replacing tα/2,ν with tα,ν . Two-Sample t Test and C.I. The two-sample t test for testing H0 : µX − µY = ∆0 is as follows: (x̄ − ȳ ) − ∆0 Test statistic value: t = q 2 s s2 X Y m + n Alternative Hypothesis Ha : µ X − µ Y > ∆ 0 Ha : µ X − µ Y < ∆ 0 Ha : µX − µY 6= ∆0 Rejection Region for Approximate Level α Test t ≥ tα,ν (upper-tailed) t ≤ tα,ν (lower-tailed) t ≥ tα/2,ν or t ≤ tα/2,ν (two-tailed) Analysis of Paired Data Assumptions: The data consists of n independently selected pairs (X1 , Y1 ), (X2 , Y2 ), . . . , (Xn , Yn ), with E [Xi ] = µ1 and E [Yi ] = µ2 . Analysis of Paired Data Assumptions: The data consists of n independently selected pairs (X1 , Y1 ), (X2 , Y2 ), . . . , (Xn , Yn ), with E [Xi ] = µ1 and E [Yi ] = µ2 . Let D1 = X1 − Y1 , D2 = X2 − Y2 , . . . , Dn = Xn − Yn , so the Di s are the differences within pairs. Analysis of Paired Data Assumptions: The data consists of n independently selected pairs (X1 , Y1 ), (X2 , Y2 ), . . . , (Xn , Yn ), with E [Xi ] = µ1 and E [Yi ] = µ2 . Let D1 = X1 − Y1 , D2 = X2 − Y2 , . . . , Dn = Xn − Yn , so the Di s are the differences within pairs. Then the Di s are assumed to be normally distributed with mean value µD and variance σD2 (this is usually a consequence of the Xi s and Yi s themselves being normally distributed.) Paired t Test The paired t test for testing H0 : µD = ∆0 (where D = X − Y is the difference between the first and second observations within a pair, and µD = µ1 − µ2 ) is as follows: Test statistic value: t = d̄ − ∆0 √ sD / n (where d̄ and sD are the sample mean and standard deviation, respectively, of the di s) Alternative Hypothesis Ha : µ D > ∆ 0 H a : µD < ∆ 0 Ha : µD 6= ∆0 Rejection Region for Approximate Level α Test t ≥ tα,n−1 (upper-tailed) t ≤ tα,n−1 (lower-tailed) t ≥ tα/2,n−1 or t ≤ tα/2,n−1 (two-tailed) Confidence Interval for µD The Paired t CI for µD with confidence level 100(1 − α)% is sD sD √ √ d̄ − tα/2,n−1 · , d̄ + tα/2,n−1 · n n A one-sided confidence bound results from retaining the relevant sign and replacing tα/2,n−1 by tα,n−1 . Analysis of Paired Data Example: (Problem 40) Lactation promotes a temporary loss of bone mass to provide adequate amounts of calcium for milk production. The paper “Bone Mass Is Recovered from Lactation to Postweaning in Adolescent Mothers with Low Calcium Intakes” (Amer. J. Clinical Nutr., 2004: 1322-1326) gave the following data on total body bone mineral content (TBBMC) (g) for a sample both during lactation (L) and in the postweaning period (P). L P 1 1928 2126 2 2549 2885 3 2825 2895 4 1924 1942 5 1628 1750 6 2175 2184 7 2114 2164 8 2621 2626 9 1843 2006 10 2541 2627 Analysis of Paired Data Example: (Problem 40) Lactation promotes a temporary loss of bone mass to provide adequate amounts of calcium for milk production. The paper “Bone Mass Is Recovered from Lactation to Postweaning in Adolescent Mothers with Low Calcium Intakes” (Amer. J. Clinical Nutr., 2004: 1322-1326) gave the following data on total body bone mineral content (TBBMC) (g) for a sample both during lactation (L) and in the postweaning period (P). L P 1 1928 2126 2 2549 2885 3 2825 2895 4 1924 1942 5 1628 1750 6 2175 2184 7 2114 2164 8 2621 2626 9 1843 2006 Does the data suggest that true average total body bone mineral content during postweaning exceeds that during lactation by more than 25g? 10 2541 2627 Analysis of Paired Data The differences for the sample are: -198 -336 -70 -18 -122 -9 -50 -5 -163 with mean d̄ = −105.7 and standard deviation sD = 103.8 -86 Analysis of Paired Data The differences for the sample are: -198 -336 -70 -18 -122 -9 -50 -5 -163 with mean d̄ = −105.7 and standard deviation sD = 103.8 The Quantile-Quantile for this sample differences is -86 Difference Between Population Proportions Proposition Let X ∼ Bin(m, p1 ) and Y ∼ Bin(n, p2 ) with X and Y independent. Then E [p̂1 − p̂2 ] = p1 − p2 so p̂1 − p̂2 is an unbiased estimator of p1 − p2 . Here p̂1 = p̂2 = Yn . Furthermore, p1 q1 p2 q2 V [p̂1 − p̂2 ] = + m n where qi = 1 − pi . X m and Difference Between Population Proportions Large-Sample Test Procedure Null hypothesis: H0 : p1 − p2 = 0 Test statistic value (large samples): p̂1 − p̂2 z=q p̂q̂ m1 + n1 where p̂ = X +Y m+n = m m+n p̂1 Alternative Hypothesis Ha : p 1 − p 2 > 0 Ha : p 1 − p 2 < 0 Ha : p1 − p2 6= 0 + n m+n p̂2 . Rejection Region for Approximate Level α Test z ≥ zα (upper-tailed) z ≤ zα (lower-tailed) z ≥ zα/2 or z ≤ zα/2 (two-tailed) Difference Between Population Proportions Large-Sample Confidence Interval for p1 − p2 The 100(1 − α)% confidence interval for p1 − p2 is given by ! r r p̂1 q̂1 p̂2 q̂2 p̂1 q̂1 p̂2 q̂2 (p̂1 − p̂2 ) − zα/2 + , (p̂1 − p̂2 ) + zα/2 + m n m n Difference Between Population Proportions Example: (Problem 50) Do teachers find their work rewarding and satisfying? The article “Work-Related Attitudes” (Psychological Reports, 1991: 443-450) reports the results of a survey of 395 elementary school teachers and 266 high school teachers. Of the elementary school teachers, 224 said they were very satisfied with their jobs, whereas, 126 of the high school teachers were very satisfied with their work. Is there any difference between the proportion of all elementary school teachers who are satisfied and all high school teachers who are satisfied?