STA 106 Homework 1 Solutions

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STA 106 Homework 1 Solutions


Book Problems

1.  (a)  The table follows:

A

B

C

D

U

E

A,U

A,E

B,U

B,E

C,U

C,E

D, U D, E

(b)  The table follows:

Y

M

O

C

N

C,Y

N,Y

C,M N,M

C,O N,O

(c)  The table follows:

R

U

N

S

N

S,R

N,R

S,U

N,U

S,N

N,N

(d)  The table follows:

BC

GS

D

H

M

L

BC,H

GS,H

D,H

BC,M

GS,M

D,M

BC,L

GS,L

D,L

2.  (a)  Let a = 3, b = 4. Then, using the properties of a linear combination of random variables;

µU1   = 3 + (4)µY = 19

‡U1   = (4)2 ‡Y(2)  = (1024) = 32

(b)  Let a = -10, b = 2. Then, using the properties of a linear combination of random variables;

µU2   = ≠10 + (2)µY = -2

‡U2   = (2)2 ‡Y(2)  = (256) = 16

(c)  Let a = 0.25, b = -1. Then, using the properties of a linear combination of random variables;

µU3   = 0.25 + (≠1)µY = -3.75

‡U3   = (≠1)2 ‡Y(2)  = (64) = 8

(d)  Let a = 0.75, b = -0.25. Then, using the properties of a linear combination of random variables;

µU4   = 0.75 + (≠0.25)µY = -0.25

‡U4   = (≠0.25)2 ‡Y(2)  = (4) = 2

3.  (a)  E{} = µY  = 20, and ‡2 {} = ‡ 2 /n = 52 /10 = 2.5

(b)  E{q Yi } = nµY  = 10(20) = 200, and ‡2 {q Yi } = n‡2  = (10)52  = 250

(c)  E{a + b} = a + bµ, and ‡2 {a + b} = b ‡ {22 } = b2 2.5


(d)  E{5 ≠ 2 q Yi } = 5 ≠ 2E{q Yi } = 5 ≠ 2(200) = ≠395 , and ‡2 {5 ≠ 2 q Yi } = (≠2)2 ‡ 2 {q Yi } = 4(250) = 1000

4.  (a) 1 ≠ 2 ≥ N (µ1 ≠ µ2 , Ò + ) (b) 1 + 2 ≥ N (µ1 + µ2 , Ò + )

(c) ≠ 3 = 1 + 2 ≠ 3 , and E{1 + 2 ≠ 3 } = µ 1 + µ2 ≠ µ3

‡  {2 1 + 2 ≠ 3 } = ( )2 ‡ 2 {1 } +()2 ‡ 2 {2 } + ‡  {2 1 } = + + .

≠ 3 ≥ N ( µ 1 + µ2 ≠ µ3 , Ò + + )

(d) E{1 + 3 ≠ 22 } = µ 1 + µ3 ≠ 2µ2 , and ‡2 {1 + 3 ≠ 22 } = ‡ {2 1 } + ‡ {2 3 } +4‡ {2 2 } = + +

5.  (a) The null hypothesis is: H0 : µ 1 = µ2 vs HA : µ 1 µ2

(b) The test-statistic is ts = = = = 4.4038.

The degress of freedom are : n1 + n2 ≠ 2 = 266 + 234 ≠ 2 = 498

The number of estimated standard deviations our sample difference is from the hypothesized difference of 0 is 4.4038. (c) At approximate d.f. = 140 (the next lowest value on the table), we find the range of the p-value is: p-value < 2(0.0005),

or p-value < 0.001.

If in reality there were no difference between the true average systolic blood pressure for smokers and non-smokers,

we would see our data or more extreme less than 0.1% of the time.

(d) Since the p-value is less than –, we reject the null and conclude there is a significant difference between the true average systolic blood pressure for smokers and non-smokers.


2022-05-25