Université d’Ottawa/University of Ottawa
Faculté des sciences sociales /Faculty of Social Sciences
Science Économique/Economics
ECO 3150B
Midterm 2
Lecture 5 : THE NEXT TWO QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
The life of a new type of light bulb is uniformly distributed between 1,200 and 1,600 hours.
1. The probability is 70% that a randomly-selected light bulb will last at least how long.
ANSWER:
70% of the distance from 1200 to 1600 is 280. Then, t =1200 + 280 = 1480.
2. What is the probability that a randomly-selected light bulb burns out in less than 1,500
hours?
ANSWER:
P(X<1500) = (1500-1200)/(1600-1200)=300/400=0.75
THE NEXT TWO QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
A random variable X is normally distributed with mean of 50 and variance of 50, and a random
variable Y is normally distributed with mean of 100 and variance of 200.
3. Given the random variables X and Y have a correlation coefficient equal to 0.50, find the
mean and variance of the random variable W = 4X+3Y.
ANSWER:
X  50,  X2  50, Y  100,  Y2  200, Corr( X ,Y )  0.50
W  4 X  3Y  (4)(50) + (3)(100) = 500
W2  (4)2 X2  (3)2 Y2  2(4)(3)Corr( X , Y ) X  Y
= (16)(50) +(9)(200) + (24)(0.50)(7.071)(14.142) = 3,799.977
4. Given the random variables X and Y have a correlation coefficient equal to 0.50, find the
mean and variance of the random variable W = 4X - 3Y.
ANSWER:
X  50,  X2  50, Y  100,  Y2  200, Corr( X ,Y )  0.50
W  4 X  3Y  (4)(50) - (3)(100) = -100
W2  (4)2 X2  (3)2 Y2  2(4)(3)Corr( X , Y ) X  Y
= (16)(50) +(9)(200) - (24)(0.50)(7.071)(14.142) = 1,400.023
Lecture 6 : THE NEXT TWO QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Monthly rates of return on the shares of a particular common stock are independent of one
another and normally distributed with a standard deviation of 1.8. A sample of 15 months is
taken.
5. Find the probability that the sample standard deviation is less than 2.6.
ANSWER:
P(s  2.6)  P(s 2  6.76)  P[(n  1)s 2 /  2  (14)(6.76) /(1.8)2 ]  P( 142  29.21)
 1  .01  0.99
6.
Find the probability that the sample standard deviation is more than 1.2.
ANSWER:
P(s  1.2)  P(s 2  1.44)  P[(n  1)s 2 /  2  (14)(1.44) /(1.8)2 ]  P( 142  6.22)
 between 0.95 and 0.975.
Lecture 7 : THE NEXT THREE QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Suppose that x1 and x2 are random samples of observations from a population with mean 
and variance  2 . Consider the following three point estimators, X, Y, Z, of  : X  ( x1  x2 ) / 2 ,
Y  ( x1  3 x2 ) / 4 , and Z  ( x1  2 x2 ) / 3 .
7. Show that all three estimators X, Y, and Z are unbiased.
ANSWER:
E ( X )  [ E ( X1 )  E ( X 2 )]/ 2  (    ) / 2  
E (Y )  [ E ( X1 )  3E ( X 2 )]/ 4  (   3 ) / 4  
E (Z )  [ E ( X1 )  2 E ( X 2 )]/ 3  (   2 ) / 3  
Since E ( X )  E (Y )  E (Z )   , then all three estimators are unbiased.
8.
Which of the estimators X,Y and Z is the most efficient?
ANSWER:
Var ( X )  [Var ( X 1 )  Var ( X 2 )] / 4  ( 2   2 ) / 4   2 / 2
Var (Y )  [Var ( X 1 )  9Var ( X 2 )] /16  ( 2  9 2 ) /16  5 2 / 8
Var ( Z )  [Var ( X 1 )  4Var ( X 2 ) / 9  ( 2  4 2 ) / 9  5 2 / 9
Hence, X is the most efficient estimator since Var ( X )  Var ( Z )  Var (Y ) .
Note that X  ( X1  X 2 ) / 2  X , then, Var ( X )  Var ( X )   2 / n   2 / 2 .
Lecture 8 : THE NEXT TWO QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Independent random sampling from two normally distributed populations gives the following
results: nx = 64, x = 441,  x = 20, ny = 36, y = 361, and  y = 25.
9. Find the margin of error for a 90% confidence interval for the difference in the means of the
two populations.
ANSWER:
  0.10  z / 2  z0.05  1.645
Margin of error: ME  z / 2 . ( x2 / nx )  ( y2 / ny )  1.645 (400/ 64)  (625/ 36)
= (1.645)(4.859) = 7.993
10.
Find the 95% confidence interval for the difference in the means of the two populations.
Based on the confidence interval in the second question, is there evidence that the population
means are different?
ANSWER:
( x  y )  ME  80  7.993 or 72.007 <  X  Y < 87.993
This confidence interval does not include zero, indicating strong evidence that the
population means are different.