Chapter6

# Chapter6 - 205 CHAPTER 6 Section 6.1 1 a We use the sample...

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Unformatted text preview: 205 CHAPTER 6 Section 6.1 1. a. We use the sample mean, x to estimate the population mean m . 1407 . 8 27 80 . 219 ˆ = = Σ = = n x x i m b. We use the sample median, 7 . 7 ~ = x (the middle observation when arranged in ascending order). c. We use the sample standard deviation, ( 29 660 . 1 26 94 . 1860 27 8 . 219 2 2 =- = = s s d. With “success” = observation greater than 10, x = # of successes = 4, and 1481 . ˆ 27 4 = = = n x p e. We use the sample (std dev)/(mean), or 2039 . 1407 . 8 660 . 1 = = x s 2. a. With X = # of T’s in the sample, the estimator is , 10 ; ˆ = = x p n X so 50 . , 20 10 ˆ = = p . b. Here, X = # in sample without TI graphing calculator, and x = 16, so 80 . 20 16 ˆ = = p Chapter 6: Point Estimation 206 3. a. We use the sample mean, 3481 . 1 = x b. 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. c. 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 s m . d. 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 e. The estimated standard error of 0846 . 16 3385 . ˆ = = = = n s n x s 4. a. ( 29 ( 29 ( 29 2 1 m m- =- =- 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 s s s s + = + = + =- ( 29 ; 2 2 2 1 2 1 n n Y X V Y X s s s + =- =- 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 = + = + = + =- s s Y V X V Y X V 5. N = 5,000 T = 1,761,300 6 . 374 = y 6 . 340 = x . 34 = d 000 , 703 , 1 ) 6 . 340 )( 000 , 5 ( ˆ 1 = = = x N q 300 , 591 , 1 ) . 34 )( 000 , 5 ( 300 , 761 , 1 ˆ 2 =- =- = d N T q 281 . 438 , 601 , 1 6 . 374 6 . 340 300 , 761 , 1 ˆ 3 = = = y x T q Chapter 6: Point Estimation 207 6. a. Let ) ln( i i x y = for I = 1, .., 31. It is easily verified that the sample mean and sample sd of the s y i ' are 102 . 5 = y and . 4961 . = y s Using the sample mean and sample sd to estimate m and s , respectively, gives 102 . 5 ˆ = m and 4961 . ˆ = s (whence 2461 . ˆ 2 2 = = y s s ). b. + ≡ 2 exp ) ( 2 s m X E . It is natural to estimate E(X) by using m ˆ and 2 ˆ s in place of m and 2 s in this expression: 87 . 185 ) 225 . 5 exp( 2 2461 . 102 . 5 exp ) ˆ ( = = + = X E 7. a. 6 . 120 10 1206 ˆ = = = = ∑ n x x i m b. 000 , 10 ˆ = t 000 , 206 , 1 ˆ = m c. 8 of 10 houses in the sample used at least 100 therms (the “successes”), so . 80 ....
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Chapter6 - 205 CHAPTER 6 Section 6.1 1 a We use the sample...

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