chapter11 - Chapter 11 Inferences on Two Samples 11.1...

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561 Chapter 11 Inferences on Two Samples 11.1 Inferences about Two Means: Dependent Samples 1. independent 2. dependent 3. Since the researcher claims the mean of population 1, 1 μ , is less than the mean of population 2, 2 , in matched pair data, the difference 12 should be negative. Thus, define 1 :0 d H < with ii i dX Y = . 4. To test a claim regarding the differences of two means with dependent sampling, (1) the sample must be obtained using simple random sampling; (2) the sample must be matched pairs; and (3) the differences must either be normally distributed with no outliers or the sample size must be large (i.e., 30 n ). 5. Since the members of the two samples are married to each other, the sampling is dependent. 6. Because the 100 subjects are randomly allocated to one of two groups, the sampling is independent. 7. Because the 80 students are randomly allocated to one of two groups, the sampling is independent. 8. Because the samples are obtained by giving different treatments to the same subjects, the sampling is dependent. 9. Because the two sets of twins are chosen at random, the sampling is independent. 10. Because the 30 subplots are randomly allocated to one of two groups, the sampling is independent. 11. (a) 7.6 7.6 7.4 5.7 8.3 6.6 5.6 8.1 6.6 10.7 9.4 7.8 9.0 8.5 0.5 1.0 3.3 3.7 0.5 2.4 2.9 i i i X Y dXY =− O b s e r v a t i o n 1 234 567 (b) Using technology, 1.614 d ≈− and 1.915 d s . (c) The hypotheses are 0 d H = versus 1 d H < . The level of significance is 0.05 α = . The test statistic is 0 1.614 2.230 / 1.915/ 7 d d t sn == . Classical approach : Since this is a left-tailed test with 6 degrees of freedom, the critical value is 0.05 1.943 t −= . Since the test statistic 0 2.230 t ≈ − is less than the critical value 0.05 1.943 t (i.e., since the test statistic fall within the critical region), we reject 0 H .
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Chapter 11 Inferences on Two Samples 562 P -value approach : The P -value for this left-tailed test is the area under the t -distribution with 6 degrees of freedom to the left of the test statistic 0 2.230 t = − , which by symmetry is equal to the area to the right of 0 2.230 t = . From the t -distribution table in the row corresponding to 6 degrees of freedom, 2.230 falls between 1.943 and 2.447 whose right- tail areas are 0.05 and 0.025, respectively. So, 0.025 < P -value < 0.05. (Using technology, we find P -value = 0.0336.) Because the P -value is less than the level of significance 0.05 α = , we reject 0 H . Conclusion : There is sufficient evidence at the 0.05 = level of significance to support the claim that 0 d μ < . (d) For 0.05 = and df = 6, /2 0.025 2.447 tt == . Then: Lower bound: 0.025 1.915 1.614 2.447 3.385 7 d s dt n =− ≈− ; Upper bound: 0.025 1.915 1.614 2.447 0.157 7 d s n +⋅ = + . We can be 95% confident that the mean difference is between 3.385 and 0.157. 12. (a) 19.4 18.3 22.1 20.7 19.2 11.8 20.1 18.6 19.8 16.8 21.1 22.0 21.5 18.7 15.0 23.9 0.4 1.5 1.0 1.3 2.3 6.9 5.1 5.3 i i ii i X Y dX Y Observation 1 2 3 4 5 6 7 8 (b) Using technology, 1.075 d and 3.833 d s .
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This note was uploaded on 11/21/2011 for the course NGN 111 taught by Professor Ahmad during the Spring '11 term at American Dubai.

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chapter11 - Chapter 11 Inferences on Two Samples 11.1...

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