Chapter9

# Chapter9 - CHAPTER 9 Section 9.1 1 a E(X Y = E X E(Y = 4.1...

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261 CHAPTER 9 Section 9.1 1. a. ( 29 ( 29 ( 29 4 . 5 . 4 1 . 4 - = - = - = - Y E X E Y X E , irrespective of sample sizes. b. ( 29 ( 29 ( 29 ( 29 ( 29 0724 . 100 0 . 2 100 8 . 1 2 2 2 2 2 1 = + = + = + = - n m Y V X V Y X V s s , and the s.d. of 2691 . 0724 . = = - Y X . c. A normal curve with mean and s.d. as given in a and b (because m = n = 100, the CLT implies that both X and Y have approximately normal distributions, so Y X - does also). The shape is not necessarily that of a normal curve when m = n = 10, because the CLT cannot be invoked. So if the two lifetime population distributions are not normal, the distribution of Y X - will typically be quite complicated. 2. The test statistic value is n s m s y x z 2 2 2 1 + - = , and H o will be rejected if either 96 . 1 z or 96 . 1 - z . We compute 85 . 4 33 . 433 2100 45 1900 45 2200 400 , 40 500 , 42 2 2 = = + - = z . Since 4.85 > 1.96, reject H o and conclude that the two brands differ with respect to true average tread lives. 3. The test statistic value is ( 29 n s m s y x z 2 2 2 1 5000 + - - = , and H o will be rejected at level .01 if 33 . 2 z . We compute ( 29 76 . 1 93 . 396 700 45 1500 45 2200 5000 800 , 36 500 , 43 2 2 = = + - - = z , which is not > 2.33, so we don’t reject H o and conclude that the true average life for radials does not exceed that for economy brand by more than 500.

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Chapter 9: Inferences Based on Two Samples 262 4. a. From Exercise 2, the C.I. is ( 29 ( 29 ( 29 33 . 849 2100 33 . 433 96 . 1 2100 96 . 1 2 2 2 1 ± = ± = + ± - n s m s y x ( 29 33 . 2949 , 67 . 1250 = . In the context of this problem situation, the interval is moderately wide (a consequence of the standard deviations being large), so the information about 1 m and 2 m is not as precise as might be desirable. b. From Exercise 3, the upper bound is ( 29 95 . 6352 95 . 652 5700 93 . 396 645 . 1 5700 = + = + . 5. a. H a says that the average calorie output for sufferers is more than 1 cal/cm 2 /min below that for nonsufferers. ( 29 ( 29 1414 . 10 16 . 10 04 . 2 2 2 2 2 1 = + = + n m s s , so ( 29 ( 29 90 . 2 1414 . 1 05 . 2 64 . - = - - - = z . At level .01, H o is rejected if 33 . 2 - z ; since – 2.90 < -2.33, reject H o . b. ( 29 0019 . 90 . 2 = - Φ = P c. ( 29 8212 . 92 . 1 1414 . 1 2 . 1 33 . 2 1 = - Φ - = + - - - Φ - = b d. ( 29 ( 29 15 . 65 2 . 28 . 1 33 . 2 2 . 2 2 = - + = = n m , so use 66.
Chapter 9: Inferences Based on Two Samples 263 6. a. H o should be rejected if 33 . 2 z . Since ( 29 33 . 2 53 . 3 32 96 . 1 40 56 . 2 87 . 16 12 . 18 = + - = z , H o should be rejected at level .01. b. ( 29 ( 29 3085 . 50 . 3539 . 0 1 33 . 2 1 = - Φ = - - Φ = b c. ( 29 06 . 37 0529 . 96

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## This note was uploaded on 03/30/2008 for the course STAT 211 taught by Professor Parzen during the Spring '07 term at Texas A&M.

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Chapter9 - CHAPTER 9 Section 9.1 1 a E(X Y = E X E(Y = 4.1...

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