ch23comparisonoftwonormalpopulationmeans.studentview

Ondsamples fromthefirstsamplewehavethat

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

This document was uploaded on 09/28/2013.

Ask a homework question - tutors are online