CH 10 Solutions

# CH 10 Solutions - Chapter 10 Statistical Infer ence About...

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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 xx . 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. x x 1 2 = 13.6 - 11.6 = 2 b. / 2 .05 zz α = = 1.645 22 12 1.645 xx nn σσ −± + (2.2) (3) 2 1.645 50 35 ±+ 2 ± .98 (1.02 to 2.98) c. / 2 .025 = = 1.96 (2.2) (3) 2 1.96 50 35 2 ± 1.17 (.83 to 3.17) 2. a. ( ) 0 2 2 (25.2 22.8) 0 2.03 (5.2) 6 40 50 xx D z −− = = = + + b. p -value = 1.0000 - .9788 = .0212 c. -value .05, reject H 0 . 3. a. ( ) 0 2 2 (104 106) 0 1.53 (8.4) (7.6) 80 70 z = = = − + + b. -value = 2(.0630) = .1260 c. -value > .05, do not reject H 0 . 4. a. = 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 xx = 135.67 – 68.64 = 67.03 b. 22 /2 (35) (20) 2.576 17.08 40 30 z nn α σσ += + c. 67.03 ± 17.08 (49.95 to 84.11) We estimate that men spend \$67.03 more than women on Valentine’s Day with a margin of error of \$17.08. 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 = 1.0000 - .9850 = .0150 -value .05; reject H 0 . The mean loan amount has increased between 2001 and 2002. 7. a. 1 = Population mean 2002 2 = Population mean 2003 H 0 : 0 H a : 0 b. With time in minutes, = 172 - 166 = 6 minutes c. ( ) 0 2 2 (172 166) 0 2.61 12 12 60 50 z = = = + + -value = 1.0000 - .9955 = .0045 -value .05; reject H 0 . The population mean duration of games in 2003 is less than the population mean in 2002.

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Chapter 10 10 - 4 d. 22 12 1 2 .025 xxz nn σσ −± + 12 12 (172 166) 1.96 60 50 + 6 ± 4.5 (1.5 to 10.5) e. Percentage reduction: 6/172 = 3.5%. Management should be encouraged by the fact that steps taken in 2003 reduced the population mean duration of baseball games. However, the statistical analysis shows that the reduction in the mean duration is only 3.5%. The interval estimate shows the reduction in the population mean is 1.5 minutes (.9%) to 10.5 minutes (6.1%). Additional data collected by the end of the 2003 season would provide a more precise estimate. In any case, most likely the issue will continue in future years. It is expected that major league baseball would prefer that additional steps be taken to further reduce the mean duration of games. 8. a. ( ) 2 2 0 (69.95 69.56) 0 1.08 2.5 2.5 112 84 xx z −− = = = + + b. p -value = 2(1.0000 - .8599) = .2802 c. -value > .05; do not reject H 0 . Cannot conclude that there is a difference between the population mean scores for the two golfers. 9. a. = 22.5 - 20.1 = 2.4 b. 2 2 2 2 2 2 1122 2.5 4.8 20 30 45.8 1 2.5 1 4.8 11 19 20 29 30 ss df nnnn  + +   = = = + + Use df = 45.
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CH 10 Solutions - Chapter 10 Statistical Infer ence About...

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