From 2 independent populations each following a

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from 2 independent populations, each following a normal distribution: Population 1: Population 2: We want to test to see if there is a difference in the means at alpha = .01. MUST TEST EQUALITY OF VARIANCES FIRST !!! 15 , 2 . 3 , 3 . 15 = = = n s x 15 , 3 . 3 , 2 . 14 = = = n s x H 0 : σ 2 1 = σ 2 2 H 1 : σ 2 1 does not equal σ 2 2 Test stat: F = 10.24/10.89 = .94 Rejection Region: Reject H 0 if F > F 14,14,.025 =2.48 or if F < F 14,14,.975 = 1 / F 14,14,.025 = 1/2.48 = .403 Decision: Fail to Reject H 0 Use pooled variances test
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11 We want to test to see if there is a difference in the means at alpha = .01. Hypotheses: Test Statistic:
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12 df: P-value: p(t 28 > .9268) + p(t 28 < - .9268) = 2* p(t 28 > .9268) Note: p(t 28 > .9268) is between .1 and .25 .20 < p-value < .50 Decision: Conclusion:
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13 Independent Populations, σ Unknown and Unequal Variances Hypotheses: H 0 : μ 1 2 = ∆ 0 or H 1 : μ 1 2 0 , μ 1 2 < ∆ 0 , μ 1 2 does not equal 0 Test Statistic: 1 ) ( 1 ) ( ) / ( d.f. ) ( ) ( ) ( 2 2 2 2 2 1 2 1 2 1 2 2 2 2 1 2 1 2 2 2 1 2 1 0 2 1 - + - + = = + - - = n n s n n s n s n s n s n s x x t υ #3
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14 An engineer wanted to know whether the strength of 2 different concrete mix designs differed significantly. He poured mixture 67-0-301 into 9 cylinders. He poured mixture 67-0-400 into 10 cylinders. He waited 28 days, then measured the strength of the 19 cylinders (in psi). He wants to test to see if the 400 mixture is stronger than the 301 mixture at alpha = .05. The results are as follows: Strengths for 301 Mixture: 3960, 4090, 3100, 3830, 3200, 3780, 4080, 4040, 1550 Strengths for 400 Mixture: 4070, 4890, 5020, 4330, 4640, 5220, 4190, 3730, 4120, 4620
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MUST TEST EQUALITY OF VARIANCES FIRST !!! H 0 : σ 2 1 = σ 2 2 H 1 : σ 2 1 does not equal σ 2 2 Test stat: F = (839.09 )2 /(473.6630 )2 = 3.138 Rejection Region: Reject H 0 if F > F 9,10,.025 =3.02 or if F < F 9,10,.975 = 1 / F 10,9,.025 = 1/3.14 = .403 Decision: Reject H 0 Use unequal variances test 15
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16 Hypotheses: Test Statistic Calculation: P-value: p(t 12 > 3.09) < .005 Decision: Conclusion: = 3.09 12.4887 Use df = 12
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#4 Paired Data Examples of data on which we should use the paired t-test include the following: each subject took a pre-test and post-test a patient had their cholesterol checked before and after 6 weeks on new medicine two tires are placed on the same vehicle and tested for wear 17
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Why use a Paired T test? The value of the paired t test is best demonstrated in an example. Suppose patient 1 responds to a drug with a 5 mm Hg rise in mean blood pressure from 100 to 105. Patient 2 has a 30 mm Hg rise, from 90 to 120. Likewise for several other subjects. The response to the drug varied widely, but all patients had one thing in common - there was always a rise in blood pressure. Some of that experimental error is avoided by the paired t test, which likely will pick up a significant difference. The independent test, which would be improperly applied in this case, would not be able to reject the null hypothesis.
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