Sections7

2 Solution. ¯ x - s √ n z α/ 2 , ¯ x + s √ n z α/ 2 = [135 - 25 √ 365 × 1 . 96 , 135 + 25 √ 365 × 1 . 96] = [132 . 44 , 137 . 56] . 5. A new drug given to 49 patients resulting in lowering their systolic blood pressure on average by 20 units with sample standard deviation of 10 units. Let μ be the mean drop in the blood pressure achived by the drug. Compute large sample lower 95% confidence bound for μ. Solution. μ ≥ ¯ x - s √ n × z α = 20 - 10 7 × 1 . 65 = 17 . 64 . 6. 62 out of 100 Maryland students admitted that they will not do homework assignment if it is ungraded. (a) Find the score confidence interval with confidence level 95% for the proportion of students who will not do homework if it is ungraded. (b) Compare this interval with large sample confidence interval using estimated variance. Solution. (a) The score CI is   ˆ p + z 2 α/ 2 2 n - z α/ 2 q ˆ p (1 - ˆ p ) n + z α/ 2 4 n 2 1 + z 2 α/ 2 /n , ˆ p + z 2 α/ 2 2 n + z α/ 2 q ˆ p (1 - ˆ p ) n + z α/ 2 4 n 2 1 + z 2 α/ 2 /n   =   0 . 62 + 1 . 96 2 200 - 1 . 96 q 0 . 68 × 0 . 32 100 + 1 . 96 2 40000 1 + 1 . 96 2 / 100 , 0 . 62 + 1 . 96 2 200 = 1 . 96 q 0 . 68 × 0 . 32 100 + 1 . 96 2 40000 1 + 1 . 96 2 / 100   = [0 . 522 , 0 . 709] (b) The large sample CI is " ˆ p - z α/ 2 r ˆ p (1 - ˆ p ) n , ˆ p + z α/ 2 r ˆ p (1 - ˆ p ) n # = " 0 . 62 - 1 . 96 r 0 . 62 × 0 . 38 100 , 0 . 62 + 1 . 96 r 0 . 62 × 0 . 38 100 # = [0 . 525 , 0 . 715] . 7. Calls to a technical support center of a certain company form Poisson process. During 168 hours the center received 210 calls. (a) Find the score confidence interval with confidence level 95% for the intensity of calls. (b) Compare this interval with large sample confidence interval using estimated variance. Solution. (a) We have ¯ x ≈ N ( λ, λ n ) . We score CI has form | ¯ x - λ | p λ/n = z α/ 2 . Thus ( λ - ¯ x ) 2 = z 2 α/ 2 λ n ⇔ λ 2 - 2¯ xλ + ¯ x 2 = z 2 α/ 2 n λ ⇔ λ 2 - 2 ¯ x + z 2 α/ 2 2 n ! λ + ¯ x 2 = 0 .