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soultions

# soultions - 1.Because we assume normality the mean = median...

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1.Because we assume normality, the mean = median, so we also use the sample mean 3481 . 1 = x . We could also easily use the sample median. We use the 90 th percentile of the sample: ( 29 ( 29 7814 . 1 3385 . 28 . 1 3481 . 1 28 . 1 ˆ ) 28 . 1 ( ˆ = + = + = + s x σ μ . Since we can assume normality, ( 29 ( 29 6736 . 45 . 3385 . 3481 . 1 5 . 1 5 . 1 5 . 1 = < = - < = - < < Z P Z P s x Z P X P The estimated standard error of 0846 . 16 3385 . ˆ = = = = n s n x σ a. ( 29 ( 29 ( 29 2 1 μ μ - = - = - Y E X E Y X E ; 434 . 575 . 8 141 . 8 = - = - y x b. ( 29 ( 29 ( 29 2 2 2 1 2 1 2 2 n n Y V X V Y X V Y X σ σ σ σ + = + = + = - ( 29 ; 2 2 2 1 2 1 n n Y X V Y X σ σ σ + = - = - The estimate would be 5687 . 20 104 . 2 27 66 . 1 2 2 2 2 2 1 2 1 = + = + = - n s n s s Y X . c. 7890 . 104 . 2 660 . 1 2 1 = = s s d. ( 29 ( 29 ( 29 1824 . 7 104 . 2 66 . 1 2 2 2 2 2 1 = + = + = + = - σ σ Y V X V Y X V 3.Let θ = the total audited value. Three potential estimators of θ are X N = 1 ˆ θ , D N T - = 2 ˆ θ , and Y X T = 3 ˆ θ . From the data, y = 374.6, x = 340.6, and d = 34.0. Knowing N = 5,000 and T = 1,761,300, the three corresponding estimates are 000 , 703 , 1 ) 6 . 340 )( 000 , 5 ( ˆ 1 = = θ , 300 , 591 , 1 ) 0 . 34 )( 000 , 5 ( 300 , 761 , 1 ˆ 2 = - = θ , and 281 . 438 , 601 , 1 6 . 374 6 . 340 300 , 761 , 1 ˆ 3 = = θ .// 4.The interval (114.4, 115.6) has the 90% confidence level. The higher confidence level will produce a wider interval. (next 5) 82% confidence 09 . 18 . 82 . 1 2 = = = - α α α , so 34 . 1 09 . 2 = = z z α and the interval is ( 29 ( 29 7 . 58 , 9 . 57 100 3 34 . 1 3 . 58 = ± . 5. ( 29 62 . 239 1 3 58 . 2 2 2 = = n so n = 240. 6. If n z L σ α 2 2 = and we increase the sample size by a factor of 4, the new length is 2 2 1 2 4 2 2 2 L n z n z L = = = σ σ α α . Thus halving the length requires n to be increased fourfold. If n n 25 = , then 5 L L = , so the length is decreased by a factor of 5// 7. 50 43 . 164 96 . 1 16 . 654 025 . ± = ± n s z x = (608.58, 699.74). We are 95% confident that the true average CO 2 level in this population of homes with gas cooking appliances is between 608.58ppm and 699.74ppm ( 29 ( 29 ( 29 ( 29 72 . 13 50 175 96 . 1 2 175 96 . 1 2 50 = = = = n n w n = (13.72) 2 = 188.24 ↑ 189 8. 5646 . 356 201 ˆ = = p ; We calculate a 95% confidence interval for the proportion of all dies that pass the probe: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 615 ,. 513 . 01079 . 1 0518 . 5700 . 356 96 . 1 1 356 4 96 . 1 356 4354 . 5646 .

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soultions - 1.Because we assume normality the mean = median...

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