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ch23comparisonoftwonormalpopulationmeans.studentview

# Ondsamples fromthefirstsamplewehavethat

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Unformatted text preview: n 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 ∴ ( 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 1 1 + n1 n2 (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 sp 2 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 1 2 y = 332.5, s2 = 4233.1, n2 = 6 t= = 1294, n1 = 7 s p = 51.2831, d . f = 11 ( x − y ) −( µ − µ2 ) ( 407 −332.5) −0 1 = = 2.6112 1 1 11 sp + 51.2831 + n1 n2 76 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. 0.025 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.V0.05 = ±t (10) 0.025 = ±2.228 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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