ch18comparisonoftwonormalpopulationmeans.studentview

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Unformatted text preview: Chapter 18: Chapter 18: Comparison of Two Normal Population Means On many occasions, we need to compare the means, µ2 µ 1 and of two normal populations. For example, we might like to compare the mean education levels of Hong Kong people and Singapore people, or to compare the mean travel times of going from A to B by bus and by boat, etc. A simplified situation (to allow for generalization into a more realistic situation later) is given below: N ( µ2 , σ 2 ) N (µ ,σ2 ) 1 Let and be two independent Let and be two independent σ2 populations, with the same variance, but with possibly different and . µ1 µ2 µ1 − µ 2 We are interested in the difference, . µ1 − µ 2 To make inferences on , we take a sample x1 , x2 ,........., xn , from the first population, and a sample, y1 , y2 ,........., yn , from the second population. X ~ N (µ , ) 1 Then, by N.S.T. and n1 σ2 Y ~ N ( µ2 , σ2 n2 ) By the theorem of Normal Assembly (see chapter By the theorem of Normal Assembly (see chapter x−y 9(E)), the difference, , has a normal µ −µ distribution with mean , but with variance equal to the sum of the individual variances, namely, 1 2 X −Y ~ N ( µ − µ2 , 1 σ2 n1 + σ2 n2 1 1 =σ ( + )) n1 n2 2 σ Here, let us suppose that is known, first. µ −µ (i) Confidence interval for can be constructed as in Chapter 15(E): 1 2 CI ( µ − µ2 ) 0.90 = ( x − y ) ±1.645σ 1 1 1 + .........(1) n1 n2 H 0 : µ1 = µ 2 H A : µ1 > µ 2 To test versus is the same as to To test versus is the same as to test versus . H 0 : µ1 − µ 2 = 0 H A : µ1 − µ 2 > 0 The test statistic is, as in Chapter 16(B), Z T .S = ( x − y ) −( µ − µ2 ) 1 .............( 2) 1 1 σ + n1 n2 HA With the decision rule: “ is accepted at the αsignificance level if zT .S > zα ” B. Pooled Estimate of σ B. Pooled Estimate of The above two formulae, (1) and (2), are not practical as σ² is hardly known by the experimenter. To overcome this difficulty, we need to estimate σ² from the samples. Now from Chapter 14(C) , this σ² can be estimated from two sources, the first and second samples. From the first sample, we have that This is called the pooled estimate for This is called the s2 p σ² s2 = p ( xi − x ) + ∑ ( yi − y ) 2 ∑ 2 n1 + n2 − 2 2 (n1 − 1) s12 + (n2 − 1) s2 = n1 + n2 − 2 and 2 (n1 − 1) s12 + (n2 − 1) s2 sp = , d . f = n1 + n2 − 2 n1 + n2 − 2 is called the pooled estimate of σ C. The case of Unknown, Common C. The case of Unknown, Common Population Variance (i) Confidence interval for µ 1 − µ2 In (1), we simply replace σby s p and the zcoefficient 1.645, by t-coefficient t (n +n Then Then 1 2 −2) 0.05 CI ( µ − µ2 ) 0.90 = ( x − y ) ±t ( n1 + n2 −2) 0.05 ⋅ s p 1 1 1 + .........( 4) n1 n2 Example 1: Example 1 A flock of seven chickens, fed with a high­protein diet, each weighing 342, 428, 385, 454, 400, 417 and 423 g. Another flock of six similar chickens, fed with a low­ protein diet, each weighing 230, 283, 340, 408, 373 and 361 g. Assuming normal populations with equal variances, µ1 − µ 2 construct a 90% confidence interval for the difference, between the two population means. Solution for Example 1: Solution for Example 1 Group 1: x = 407, s12 = 1294, n1 = 7 Group 2: y = 332.5, s22 = 4233.1, n2 = 6 6 ×1294 + 5 × 4233.1 ∴sp = = 51.2831, d . f = 7 + 6 − 2 = 11 7+6−2 From Table 7, t (11) 0.05 =1.796 1 1 + n1 n2 ∴ ( µ − µ2 ) 0.90 = ( x − y ) ±t ( n1 + n2 −2) 0.05 ⋅ s p CI 1 = ( 407 −332.5) 1.796 ×51.2831 = 74.5 ±51.24 = ( 23.26,125.74) 11 + 76 (ii) Testing versus H (ii) Testing versus H :µ = µ 0 1 2 A : µ1 > µ 2 In (2), we replace σ by s p , and z by t, so that the test so statistic is statistic ( x − y ) − (µ − µ ) t= 1 2 sp 11 + n1 n2 , d . f = n1 + n2 − 2.............(6) And, the decision rule becomes: “Accept H A ,at α And, significance level if t > t (n + n −2)α ” 1 2 Example 2: Example 2 Using the data in Example 1, test H :µ = µ versus H : µ > µ 0 1 2 A 1 2 Solution: From Example 1, we have x = 407, s 2 y = 332.5, s2 = 4233.1, n2 = 6 2 1 = 1294, n1 = 7 s p = 51.2831, d . f = 11 t= ( x − y ) −( µ − µ2 ) ( 407 −332.5) −0 1 = = 2.6112 1 1 11 sp + 51.2831 + n1 n2 76 0.025 From Table 7, an appropriate C.V. is t (11) =2.201 HA As t =2.6112 > 2.201, we accept at 0.025 S.L. Example 3: Example 3 Two groups of students were taught the same course by two different methods, A and B. At the end of the course, 5 students from group A and 7 students from group B were randomly selected to take an extensive test. The scores they achieved have these summary statistics: GroupA : x = 60, n1 = 5, S xx = 650 GroupB : y = 72, n2 = 7, S yy = 450 Are we justified to say that the two methods have different mean scores? Why? Solution: Solution µ1 µ2 Let and denote the mean scores for methods A and B respectively. H 0 : µ1 = µ 2 We propose to test versus H A : µ1 ≠ µ 2 H0 Under , the pooled s.d. is sp = 650 + 450 = 10.488, d . f = 5 + 7 − 2 = 10 5+7−2 The test statistic is, by (6), t= ( x − y ) −( µ − µ2 ) (60 −72) −0 1 = = −1.954 1 1 11 sp + 10.488 + n1 n2 57 Table 7 gives, for a two­tailed t­test, the appropriate Table 7 gives, for a C.V’s as C.V = ±t (10) = ±2.228 0.05 0.025 As ­1.954 > ­2.228, we accept H 0 Conclusion: The 2 methods are not different. The above t­test is called the 2­sample t–test. ...
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