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UNIVERSITY OF VICTORIA
Midterm
July 29, 2015
solutions
NAME: _____________________________
STUDENT NUMBER: V00______________
Course Name & No.
Statistical Inference
Economics 246
Section(s)
A01
CRN:
31188
Instructor:
Betty Johnson
Duration:
1hour 50 minutes
This exam has a total of __ pages including this cover page.
Students must count the number of pages and report any discrepancy immediately to the
Invigilator.
This exam is to be answered: In Booklets provided
Marking Scheme:
Part I:
Q1:
Q2:
Q3:
25 marks
6 marks
6 marks
Q4:
Q5:
Q6:
5 marks
5 marks
5 marks
Q7:
8 marks
Part II:
Part III:
Part IV:
Q8:
6 marks
Q9:
10 marks
Bonus Question: 5 marks
Materials allowed: Non-programmable calculator
Econ 246 Summer 2015 CRN#:31188
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Part I: Multiple choice and True/False
1.
In a recent survey of college professors, it was found that the average amount of money
spent on entertainment each week was normally distributed with a mean of $95.25 and a
standard deviation of $27.32. What is the probability that the average spending of a
sample of 25 randomly-selected professors will exceed $102.50?
A) 0.0918
B) 0.1064
C) 0.3936
D) 0.4082
ANSWER:
2.
In a recent survey of high school students, it was found that the average amount of money
spent on entertainment each week was normally distributed with a mean of $52.30 and a
standard deviation of $18.23. Assuming these values are representative of all high school
students, what is the probability that for a sample of 25, the average amount spent by
each student exceeds $60?
A) 0.3372
B) 0.0174
C) 0.1628
D) 0.4826
ANSWER:
3.
A
B
If a sample of size 100 is taken from a population whose standard deviation is equal to
100, then the standard error of the mean is equal to
A) 10
B) 100
C) 1,000
D) None of the above
ANSWER: A
4.
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The time it takes to complete the assembly of an electronic component is normally
distributed with a standard deviation of 4.5 minutes. If we randomly select 20
components, what is the probability that the standard deviation for the time of assembly
of these units is less than 3.0 minutes?
A) <0.05
B) <0.005
C) <0.025
D) <0.01
ANSWER:
5.
C
What is the name of the parameter that determines the shape of the chi-square
distribution?
A) The mean
B) The variance
C) The proportion
D) The degrees of freedom
ANSWER: D
6.
Random samples of size 36 are taken from an infinite population whose mean is 80 and
standard deviation is 18. The mean and standard error of the sampling distribution of
sample means are, respectively,
A) 80 and 18
B) 80 and 2
C) 80 and 3
D) 36 and 2
ANSWER: C
7.
The standard deviation of the sampling distribution of the sample mean is also called the
A) central limit theorem
B) standard error of the mean
C) finite population correction factor
D) population standard deviation
ANSWER: B
8.
If all possible samples of size n are drawn from an infinite population with a mean of 20
and a standard deviation of 5, then the standard error of the sampling distribution of
sample means is equal to 1.0 only for samples of size
A)
B)
C)
D)
5.
15.
20.
25.
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ANSWER:
9.
D
If a random sample of size n is drawn from a normal population, then the sampling
distribution of sample means will be:
A) normal for all values of n
B) normal only for n > 30
C) approximately normal for all values of n
D) approximately normal only for n > 30
ANSWER: A
10.
Why is the Central Limit Theorem important in statistics?
A) Because for a large sample size n, it says the population is approximately normal.
B) Because for any population, it says the sampling distribution of the sample mean is
approximately normal, regardless of the shape of the population.
C) Because for a large sample size n, it says the sampling distribution of the sample
mean is approximately normal, regardless of the shape of the population.
D) Because for any sample size n, it says the sampling distribution of the sample mean
is approximately normal.
ANSWER: C
11.
As the sample size, n, increases, what happens to the shape of the sampling distribution
of the means?
A) Approaches a normal distribution
B) Positively skewed
C) Negatively skewed
D) Approaches the exponential distribution
ANSWER: A
12.
If all possible random samples of size n are taken from a population, and the mean of
each sample is determined, what can you say about the mean of the sample means?
A) It is larger than the population mean.
B) It is exactly the same as the population mean.
C) It is smaller than the population mean.
D) None of the above.
ANSWER: B
13.
If the standard deviation of the sampling distribution of sample means is 5.0 for samples
of size 16, then the population standard deviation must be
A) 20.
B) 5.0.
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C) 3.2.
D) 80.
ANSWER:
14.
A
The amount of material used in making a custom sail for a sailboat is normally
distributed. For a random sample of 15 sails, the mean amount of material used is 912
square feet, with a standard deviation of 64 square feet. Which of the following
represents a 99% confidence interval for the population mean amount of material used in
a custom sail?
A) 912  49.2
B) 912  42.6
C) 912  44.3
D) 912  46.8
ANSWER: A
15. Which of the following is not a property for evaluating point estimators?
A) Effectiveness
B) Efficiency
C) Consistency
D) Unbiasedness
ANSWER: A
16.
Which of the following statements is correct?
A) A point estimate is an estimate of the range of a population parameter
B) A point estimate is a single value estimate of the value of a population parameter
C) A point estimate is an unbiased estimator if its standard deviation is the same as
the actual value of the population standard deviation
D) All of the above
ANSWER: B
17. Which of the following statements is correct?
A) An interval estimate describes a range of values that is likely not to include the actual
population parameter
B) An interval estimate is an estimate of the range for a sample statistic
C) An interval estimate is an estimate of the range of possible values for a population
parameter
D) All of the above
ANSWER: C
18. Which of the following is not a property of the student's t distribution?
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A) It is symmetric.
B) Its shape is characterized by the degrees of freedom.
C) As the sample size increases, it gradually approaches the normal distribution.
D) All of the above are properties of the t distribution.
ANSWER: D
19. The larger the level of confidence (e.g., .99 versus .95) used in constructing a confidence
interval estimate of the population mean, the:
A) smaller the probability that the confidence interval will contain the population mean
B) wider the confidence interval
C) smaller the value of z / 2
D) narrower the confidence interval
ANSWER: B
20. A husband and wife, both are statisticians, decided to construct a 90% confidence
intervals for an unknown population mean. The husband selected a random sample of 50
observations while his wife's sample size was 30 observations. Which of the following is
true?
A) The wife's confidence interval has a greater degree of confidence.
B) The husband’s confidence interval has a greater degree of confidence.
C) The husband’s confidence interval is narrower.
D) The husband’s confidence interval is wider.
ANSWER: C
21.
If a sample has 20 observations and a 90% confidence estimate for  is needed, the
appropriate t-score is:
A) 2.12
B) 1.746
C) 2.131
D) 1.729
ANSWER:
22.
D
If a sample of size 30 is selected, the value of A for the probability P(t  A) = 0.01 is:
A) 2.247
B) 2.756
C) 2.462
D) 2.750
ANSWER:
C
23. The t- distribution approaches the normal distribution as the:
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A) degrees of freedom increase
B) degrees of freedom decrease
C) sample size decreases
D) population size increases
ANSWER: A
24. Which of the following symbols would be used as a point estimate for the population
mean  ?
A) 
B) x
C) p̂
D) s / n
ANSWER:
25.
B
Which of the following statement(s) is (are) correct about the t - distribution?
A) Has a mean of zero
B) Is a symmetrical distribution
C) Is based on degrees of freedom
D) All of the above are correct
ANSWER: D
Question 2: (6 marks)
The length of time it takes to fill an order at a local sandwich shop is normally distributed with a
mean of 4.1 minutes and a standard deviation of 1.3 minutes.
a) What is the probability that the average waiting time for a random sample of ten
customers is between 4.0 and 4.2 minutes?
ANSWER:
P(4.0< X <4.2) = P(Z < 0.24) – P(Z < -0.24) = 0.5948 - 0.4052 = 0.1896
b) The probability is 95% that the average waiting time for a random sample of ten
customers is greater than how many minutes?
ANSWER:
a  4.1 
a  4.1 


P  X  a   0.95  P  Z 
  P  Z  0.411   .95
1.3/ 10 



a  4.1

 1.645  a  3.42 minutes
0.411
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Question 3: (6 Marks)
The filling machine at a bottling plant is operating correctly when the variance of the fill amount
is equal to 0.3 ounces. Assume that the fill amounts follow a normal distribution.
a)
What is the probability that for a sample of 30 bottles, the sample variance is
greater than 0.5?
ANSWER:
  n  1 s 2
P( s 2 > 0.5)= P 

b)

2

 29  0.5  =P(
0.3


 2 >48.33)<0.01
The probability is 0.10 that for a sample of 30 bottles, the sample variance is less
than what number?
ANSWER:
  n  1 s 2 29k 
P( s 2  k )  0.10  P 

  0.10
2
0.3 
 
 96.67k  19.768
 P   2  96.67k   0.10
 k  0.2045
Part II: Concepts
Question 4: Describe the concept of stratified sampling. Illustrate the technique with an
example.
Total marks:5
“The use of stratified sampling requires that a population be divided into homogeneous
groups called strata. Each stratum is then sampled according to certain specified criteria.”
Under sampling with prior knowledge.
Divide population into strata.
Each strata is different.
Elements in the strata are the same.
Sample each strata to replicate the same socio-economic situation as the population.
Sampling is random within each strata.
Question 5: Describe the concept of Unbiasness with respect to estimator properties. Total
Marks: 5
On average, the value of the estimate should equal the population parameter being estimated.
If the average value of the estimator does not equal the actual parameter value, the estimator is a
biased estimator.
Ideally, an estimator has a bias of zero if it is said to be unbiased:
Econ 246 Summer 2015 CRN#:31188
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“An estimator is said to be unbiased if the expected value of the estimator is
equal to the true value of the parameter being estimated.”
Generally, if
and



is a population parameter to be estimated,
is an estimator where
  ( X1 , X 2 ,  , X n );
is said to be an unbiased estimator of
Example 1)
E (X )  


if:

E    .
under simple random sampling
(Topic 1); so X is an unbiased estimator.
The sample variance is an unbiased estimator of the population variance.
~
f (  )
f (f( ))
~
f ( )
E( )  
Bias

Two estimators:  is unbiased.
~
E(  )  
~
 , 
~
~
  [ E(  )   ]
~
 is biased.
Question 6: Define and illustrate the concept of Central Limit Theorem Total Marks: 5
Regardless of the form probability distribution of the population, as the sample size increases, the sampling
distribution will be approximately normal. A normal population will generate a normal sampling distribution.
“Regardless of the distribution of the parent population, as long as it has a finite mean µ and variance σ2, the
distribution of the means of the random samples will approach a normal distribution, with mean μ and variance
σ2/n, as the sample size n, goes to infinity.”
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(I) When the parent population is normal, the sampling distribution of
X
is exactly normal.
(II) When the parent population is not normal or unknown, the sampling distribution of
approximately normal as the sample size increases.
X
Part III: Proofs
Question 7: Total marks:8
(i) Using the fact that the mean of the chi-squared distribution is (n-1), prove that E( S 2 )   2
 
E s2   2
Since E (  2 )  n  1
and  2 
(n  1) s2
2
if you take the expectation:
 (n  1) s2 
E
  n 1
2
 

n 1 2
E s2 

n 1
E s2   2
 
 
(ii)
Prove that E ( X )   .
Let Xi ~(  X,  2) for all i.
Since :
(i)
X
1
( X  X 2   X n )
n 1
is
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and
(ii) E
(Xi) = µ ,
we can apply the rules of expectation:
1 n

E( X )  E  Xi 
 n i 1 


1  n
E  Xi 
n  i 1 
1
E  X 1  X 2   X n 
n
1

E  X 1   E  X 2   E  X 3   E  X n 
n
1
          
n
1
 X   n   .
#
n



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Part IV: SHORT ANSWER
Question 8: (6 marks)
Suppose we have two estimators of the population parameter  :
E ()    17 / n 4
V ()   2 / 96n 3
and
~
E ( )    93 / n 2
~
V ( )   2 / 84n 4
(i)
Determine the bias, if any, of each estimator.
Bias    E ()      17 / n4  17 / n4
~
bias    E ( )      93 / n2  93 / n2
(ii)
Determine the MSE. Which estimator is preferred?
MSE  V ()  bias2   2 / 96n3  (17 / n4 ) 2
~
MSE  V ( )  bias2   2 / 84n4  (93 / n2 ) 2
First estimator is preferred
(iii)
Determine if the estimators are consistent. Explain.
The bias and variance go to zero as n gets large.
Yes Consistent
Question 9: Consider the following population of data: {140, 150, 160}.
(i)
Determine the mean and variance of the population.
Total marks:4
1
  140  150  160  450 / 3  150
3
1
1
1
 2  [ xi2  nX 2 ]  [67700  (3)(22500)]  [67700  67500]  200 / 3  66.67
n
3
3
N
N

1
1
 2   ( X i  ) 2    X i2    2
N i 1
N  i 1

(ii) Determine the sampling distribution of the sample mean for a sample of size 2. Graph this distribution with
a simple bar graph.
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Total marks:4
X1,X2
140, 140
140, 150
140, 160
150, 140
150, 150
150, 160
160, 140
160, 150
160, 160
X
140
145
150
145
150
155
150
155
160
P( X )
1/9
2/9
3/9
2/9
1/9
X
140
145
150
155
160
P( X )
X
140
145
150
(iii)
Determine the variance of X ?
66.67/2=33.335
155
160
Total Marks:3
Bonus: Question (5 Marks) The mean selling price of new homes in a city over a year was $120,000.
The population standard deviation was $28,000. A random sample of 100 new home sales from this
city was taken. What is the probability that the sample mean selling price was between $119,000 and
$121,000?
Econ 246 Summer 2015 CRN#:31188
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Formulae
Central Location:
 x 
1
N
Population mean

Grouped Population Mean

i
x f
f
i

i
i
 x f 
1
N
i
i
1 n
X   Xi
n i 1
Sample Mean
1 n
X   Xi fi
n i 1
Sample Mean for frequency distribution:
E( X )  X 
Mean of the Sample Mean ( X )
where: i= 1,2,...,k, and k is the number of distinct possible values of
k
 X P( X )
i 1
i
i
X.
Dispersion:
2
 1
   xi  
 N
2  
Population variance


2
 1
   xi   f i
 N
1
Sample variance for frequency distribution: s2 
 xi  x 2 f i

(n  1)
1
Sample variance
s2 
 xi  x 2

(n  1)
(Grouped data
2  
Sample Standard Deviation
s
2
 X2  V ( X )   n 
Variance of the Sample Mean
Standard Error of the mean:
s2
X 
2
n


n
 (X  
X
) 2 P( X )
X
.
Distributions:
Standard Normal: Z 
t-distribution t 
( X  )

;
The Standardization of
X
; Chi-square distribution
s
n
X:
Z
X

n
n21 
 ( s 2 ) (n  1) s 2

2
2
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