# Ass#6 ```COURSE: ENGINEERING PROBABILITY &amp; STATISTICS
ASSIGNMENT #6
Lecturer: Ho Thanh Vu
Student: Phạm Nhật T&acirc;n
-ID :IEIEIU16002
-Class: Friday morning
Question 1:
Let 𝜇 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 𝑙𝑖𝑓𝑒. We want to test whether the new process affects 𝜇 or not
Let 𝐻0 : 𝜇 = 125
𝐻1 : 𝜇 ≠ 125
We have:
𝑋̅ = 131 ℎ𝑜𝑢𝑟𝑠 ; 𝑛 = 39 &gt; 30 ; 𝑠 = 11 ℎ𝑜𝑢𝑟𝑠
𝑋̅ −𝜇
131−125
𝑍𝑡 = 𝑠/ 𝑛0 = 11/√39 = 3.406 =&gt; p-value is 2*(1-0.9997)=0.0006
√
𝐼𝑓 𝛼 = 0.05 ⇒ 𝑍𝛼/2 = &plusmn;1.96 &lt; 3.406 =&gt; 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 =&gt; 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑎𝑓𝑓𝑒𝑐𝑡𝑠 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 𝑙𝑖𝑓𝑒
𝐼𝑓 𝛼 = 0.01 ⇒ 𝑍𝛼/2 = &plusmn;2.57 &lt; 3.406 =&gt; 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 =&gt; 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑎𝑓𝑓𝑒𝑐𝑡𝑠 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 𝑙𝑖𝑓𝑒
Both cases give very significant results
Question 2:
Let 𝜇 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑚𝑎𝑛 𝑖𝑛 𝑉𝑖𝑒𝑡𝑛𝑎𝑚
Let 𝐻0 : 𝜇 ≥ 𝜇0 = 170
𝐻1 : 𝜇 &lt; 𝜇0 = 170
Sample information:
𝑋̅ = 171.283 𝑐𝑚 ; 𝑛 = 20 &lt; 30 ; 𝑠 = 3.829 𝑐𝑚, df=19
𝑋̅ − 𝜇0 171.283 − 170
𝑡𝑡 =
=
= 1.4985
𝑠/√𝑛
3.829/√20
𝐼𝑓 𝛼 = 0.01 ⇒ 𝑡𝛼;19 = −2.539 &lt; 1.4985 =&gt; 𝑑𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0
𝐼𝑓 𝛼 = 0.05 ⇒ 𝑡𝛼;19 = −1.729 &lt; 1.4985 =&gt; 𝑑𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0
Therefore, the average height of men in Vietnam is more than or equal to 170cm
Question 3:
Let 𝜇 = 𝑠𝑢𝑙𝑓𝑢𝑟 𝑐𝑜𝑛𝑡𝑒𝑛𝑡 𝑖𝑛 𝑑𝑖𝑒𝑠𝑒𝑙 𝑓𝑢𝑒𝑙
Let 𝐻0 : 𝜇 ≤ 𝜇0 = 0.15
𝐻1 : 𝜇 &gt; 𝜇0 = 0.15
We have:
𝑋̅ = 0.162 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑠 ; 𝑛 = 40 &gt; 30 ; 𝑠 = 0.040 𝑐𝑚; 𝛼 = 0.05
𝑋̅ − 𝜇0 0.162 − 0.15
𝑍𝑡 =
=
= 1.8974
𝑠/√𝑛
0.040/√40
𝐼𝑓 𝛼 = 0.05 ⇒ 𝑍𝛼 = 1.65 &lt; 1.8974 =&gt; 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 =&gt; 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
𝐿𝑒𝑡 𝜇𝐷 = 𝜇𝐴 − 𝜇𝐵 . We want to test if this drug changes blood pressure so:
Let 𝐻0 : 𝜇𝐷 = 0
𝐻1 : 𝜇𝐷 ≠ 0
̅ = 5.563 ; 𝑛 = 16 &lt; 30 ; 𝑠𝐷 = 6.603; 𝛼 = 0.05 ; 𝑑𝑓 = 15
Sample information: 𝑋̅ = 𝐷
𝑋̅ − 0
5.563 − 0
𝑡𝑡 =
=
= 3.37
𝑠𝐷 /√𝑛 6.603/√16
𝐼𝑓 𝛼 = 0.05 ⇒ 𝑡𝛼;15 = &plusmn;2.131 &lt; 3.37 =&gt; 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 ⇒ this drug actually changes blood pressure
2
Question 5:
Let 𝐻0 : 𝑝 ≥ 0.45
𝐻1 : 𝑝 &lt; 0.45
49
𝑝̂ =
= 0.392
125
𝑝̂ − 𝑝0
0.392 − 0.45
𝑍𝑡 =
=
= −1.31
𝑝
∗
(1
−
𝑝
)
0.45
∗
0.55
0
0
√
√
𝑛
125
𝐼𝑓 𝛼 = 0.05 ⇒ 𝑍𝛼 = −1.65 &lt; −1.31 =&gt; 𝐷𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0
Therefore, the program doesn’t reduce the proportion of executives who show signs of the crisis
Question 6:
Let 𝐻0 : 𝑝 ≥ 0.95
𝐻1 : 𝑝 &lt; 0.95
1380
𝑝̂ =
= 0.92
1500
𝑝̂ − 𝑝0
0.92 − 0.95
𝑍𝑡 =
=
= −1.686
𝑝
∗
(1
−
𝑝
)
0.95
∗
0.05
0
0
√
√
𝑛
1500
𝐶ℎ𝑜𝑜𝑠𝑒 𝛼 = 0.05 ⇒ 𝑍𝛼 = −1.65 &gt; −1.686 =&gt; 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0
Therefore lower interest rates for mortgages during the following period actually reduced the percentage of
households living in rental units.
Question 7:
𝑛 = 25; 𝑠 2 = 175; 𝛼 = 0.05
Assume the population is normally distributed
Let 𝐻0 : 𝜎 2 ≤ 156
𝐻1 : 𝜎 2 &gt; 156
(𝑛 − 1)𝑠 2 24 ∗ 175
𝜒 𝑡=
=
= 26.923
𝜎2
156
𝐹𝑜𝑟 𝛼 = 0.05 ⇒ 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 𝑖𝑠 𝜒 2 (24,0.05) = 36.42 &gt; 26.923 ⟹ 𝐷𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0
Therefore, we can’t prove that the variance is above the required level, corrective action should not be taken.
2
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