Homework 8 solutionsHomework 9 solutions

Week 9 practice problems - solutions Exercise 1

Exercise 2

Exercise 2

Exercise 3

Exercise 4Exercise 5

Exercise 1

Exercise 5

Exercise 2

X � Y ⇠ N(µ1 � µ2,

q�219 +

�22

12 ). Therefore we can write:

Z =X � Y � (µ1 � µ2)q

�219 +

�22

12

.

And since �21 = 3�2

2 we get:

Z =X � Y � (µ1 � µ2)p

�22(

39 + 1

12 ).

Now we need to define a �2 random variable. Because X and Y are independent we have:

(9 � 1)S21

�21

+(12 � 1)S2

2

�22

⇠ �212+9�2 ⇠ �2

19.

Using again �21 = 3�2

2 we get:

13 8S

21 + 11S2

2

�22

⇠ �219.

Now we can define a variable that has a t distribution as follows:

t =

X�Y �(µ1�µ2)p�22( 39+ 1

12)

r138S2

1+11S2

2�22

19

⇠ t19

Finally we get:

t =X � Y � (µ1 � µ2)p

13 8S

21 + 11S2

2

p19p

39 + 1

12

=X � Y � (µ1 � µ2)p

13 8S

21 + 11S2

2

q228

5.

We can use the above t19 random variable to construct a 95% confidence interval for µ1 � µ2. We want:

P (�t↵2

;19 X � Y � (µ1 � µ2)p

13 8S

21 + 11S2

2

q228

5 t↵

2;19) = 1 � ↵.

After some manipulation we find that µ1 � µ2 will fall in the following interval with 95% confidence:

X � Y ± t↵2

;19

q8

3S21 + 11S2

2

q5

228.

Exercise 6