# Ch10 - Chapter 10 Statistical Inference about Means and...

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10 - 1 Chapter 10 Statistical Inference about Means and Proportions with Two Populations Learning Objectives 1. Be able to develop interval estimates and conduct hypothesis tests about the difference between two population means when 1 σ and 2 are known. 2. Know the properties of the sampling distribution of 12 x x . 3. Be able to use the t distribution to conduct statistical inferences about the difference between two population means when 1 and 2 are unknown. 4. Learn how to analyze the difference between two population means when the samples are independent and when the samples are matched. 5. Be able to develop interval estimates and conduct hypothesis tests about the difference between two population proportions. 6. Know the properties of the sampling distribution of pp .

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Chapter 10 10 - 2 Solutions: 1. a. xx 12 = 13.6 - 11.6 = 2 b. /2 .05 zz α == 1.645 22 1.645 nn σσ −± + (2.2) (3) 21 . 6 4 5 50 35 ±+ 2 ± .98 (1.02 to 2.98) c. .025 1.96 (2.2) (3) . 9 6 50 35 2 ± 1.17 (.83 to 3.17) 2. a. () 0 (25.2 22.8) 0 2.03 (5.2) 6 40 50 xx D z −− = + + b. p -value = .5000 - .4788 = .0212 c. p -value .05, reject H 0 . 3. a. 0 2 2 (104 106) 0 1.53 (8.4) (7.6) 80 70 z = + + b. p -value = 2(.5000 - .4370) = .1260 c. p -value > .05, do not reject H 0 . 4. a. x x = 2.04 - 1.72 = .32 b. .025 z + (.10) (.08) 1.96 1.96(.0208) .04 40 35 += = c. .32 ± .04 (.28 to .36)
Statistical Inference about Means and Proportions with Two Populations 10 - 3 5. a. 12 x x = 14.9 - 10.3 = 4.6 years b. 22 /2 (5.2) (3.8) 1.96 1.3 100 85 z nn α σσ += + = c. 4.6 ± 1.3 (3.3 to 5.9) 6. 1 µ = Mean loan amount for 2002 2 = Mean loan amount for 2001 H 0 : 0 µµ −≤ H a : 0 −> () 0 2 2 (175 165) 0 2.17 55 50 270 250 xx D z −− == = + + p -value = .5000 - .4850 = .0150 p -value .05; reject H 0 . The mean loan amount has increased between 2001 and 2002. 7. a. H 0 : 0 −= H a : 0 −≠ b. 0 2 2 (40 35) 0 2.41 91 0 36 49 z = + + p -value = 2(.5000 - .4920) = .0160 p -value .05; reject H 0 . There is a difference between the population mean ages at the two stores. 8. a. 2 2 0 (69.95 69.56) 0 1.08 2.5 2.5 112 84 xx z = + + b. p -value = 2(.5000 - .3599) = .2802 c. p -value > .05; do not reject H 0 . Cannot conclude that there is a difference between the population mean scores for the two golfers.

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Chapter 10 10 - 4 9. a. 12 x x = 22.5 - 20.1 = 2.4 b. 2 2 22 222 2 2 1122 2.5 4.8 20 30 45.8 1 2.5 1 4.8 11 19 20 29 30 ss nn df nnnn  + +   == = + + −− Use df = 45. c. t .025 = 2.014 .025 2.5 4.8 2.014 2.1 20 30 t += + = d. 2.4 ± 2.1 (.3 to 4.5) 10. a. () 2 2 0 (13.6 10.1) 0 2.18 5.2 8.5 35 40 xx t = + + b. 2 2 2 2 5.2 8.5 35 40 65.7 15 . 2 18 . 5 34 35 39 40 df + + =   + +   Use df = 65 c. Using t table, area in tail is between .01 and .025 two-tail p -value is between .02 and .05. Actual p -value = .0329 d. p -value .05, reject H 0 . 11. a. 1 54 9 6 x 2 42 7 6 x b. 2 1 1 1 2.28 1 i s n Σ− 2 2 2 2 1.79 1 i s n c. x x = 9 - 7 = 2
Statistical Inference about Means and Proportions with Two Populations 10 - 5 d. 2 2 22 12 222 2 2 1122 2.28 1.79 66 9.5 . 2 8 11 . 7 9 56 ss nn df nnnn   + +    == =    + +    −− Use df = 9, t .05 = 1.833 2.28 1.79 1.833 xx −± + 2 ± 2.17 (-.17 to 4.17) 12. a. x x = 22.5 - 18.6 = 3.9 b. 2 2 2 2 8.4 7.4 50 40 87.1 1 8.4 1 7.4 49 50 39 40 df + + =   + +   Use df = 87, t .025 = 1.988 8.4 7.4 3.9 1.988 50 40 ±+ 3.9 ± 3.3 (.6 to 7.2) 13. Computer used to obtain the following: Miami Los Angeles n 1 = 50 n 2 = 50 1 x = 6.34 2 x = 6.72 s 1 = 2.16 s 2 = 2.37 x x = 6.34 - 6.72 = -.38 Los Angeles is slightly higher rated.

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## This note was uploaded on 03/01/2012 for the course ECON 371 taught by Professor Staff during the Spring '08 term at UVA.

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Ch10 - Chapter 10 Statistical Inference about Means and...

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