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COURSE: ENGINEERING PROBABILITY & STATISTICS ASSIGNMENT #6 Lecturer: Ho Thanh Vu Student: Phạm Nhật Tân -ID :IEIEIU16002 -Class: Friday morning Question 1: Let = . We want to test whether the new process affects or not Let 0 : = 125 1 : ≠ 125 We have: ̅ = 131 ℎ ; = 39 > 30 ; = 11 ℎ ̅ − 131−125 = / 0 = 11/√39 = 3.406 => p-value is 2*(1-0.9997)=0.0006 √ = 0.05 ⇒ /2 = ±1.96 < 3.406 => 0 => ℎ = 0.01 ⇒ /2 = ±2.57 < 3.406 => 0 => ℎ Both cases give very significant results Question 2: Let = ℎℎ Let 0 : ≥ 0 = 170 1 : < 0 = 170 Sample information: ̅ = 171.283 ; = 20 < 30 ; = 3.829 , df=19 ̅ − 0 171.283 − 170 = = = 1.4985 /√ 3.829/√20 = 0.01 ⇒ ;19 = −2.539 < 1.4985 => 0 = 0.05 ⇒ ;19 = −1.729 < 1.4985 => 0 Therefore, the average height of men in Vietnam is more than or equal to 170cm Question 3: Let = Let 0 : ≤ 0 = 0.15 1 : > 0 = 0.15 We have: ̅ = 0.162 ; = 40 > 30 ; = 0.040 ; = 0.05 ̅ − 0 0.162 − 0.15 = = = 1.8974 /√ 0.040/√40 = 0.05 ⇒ = 1.65 < 1.8974 => 0 => the sulfur content in the diesel fuel is actually larger than 0.15 percent. Question 4: Patient 1 2 3 4 5 6 7 8 9 10 11 Before 134 122 132 130 128 140 118 127 125 142 137 After Difference 140 6 130 135 126 134 8 3 -4 6 138 124 -2 6 126 132 -1 7 144 137 2 0 12 13 14 15 16 136 130 139 128 128 144 137 8 7 143 146 4 18 149 21 = − . We want to test if this drug changes blood pressure so: Let 0 : = 0 1 : ≠ 0 ̅ = 5.563 ; = 16 < 30 ; = 6.603; = 0.05 ; = 15 Sample information: ̅ = ̅ − 0 5.563 − 0 = = = 3.37 /√ 6.603/√16 = 0.05 ⇒ ;15 = ±2.131 < 3.37 => 0 ⇒ this drug actually changes blood pressure 2 Question 5: Let 0 : ≥ 0.45 1 : < 0.45 49 ̂ = = 0.392 125 ̂ − 0 0.392 − 0.45 = = = −1.31 ∗ (1 − ) 0.45 ∗ 0.55 0 0 √ √ 125 = 0.05 ⇒ = −1.65 < −1.31 => 0 Therefore, the program doesn’t reduce the proportion of executives who show signs of the crisis Question 6: Let 0 : ≥ 0.95 1 : < 0.95 1380 ̂ = = 0.92 1500 ̂ − 0 0.92 − 0.95 = = = −1.686 ∗ (1 − ) 0.95 ∗ 0.05 0 0 √ √ 1500 ℎ = 0.05 ⇒ = −1.65 > −1.686 => 0 Therefore lower interest rates for mortgages during the following period actually reduced the percentage of households living in rental units. Question 7: = 25; 2 = 175; = 0.05 Assume the population is normally distributed Let 0 : 2 ≤ 156 1 : 2 > 156 ( − 1) 2 24 ∗ 175 = = = 26.923 2 156 = 0.05 ⇒ 2 (24,0.05) = 36.42 > 26.923 ⟹ 0 Therefore, we can’t prove that the variance is above the required level, corrective action should not be taken. 2