UNIVERSITY OF VICTORIA Midterm - Web hostingbettyj/246/midterm246_summer2015_solutions.pdf · standard deviation of $18.23. Assuming these values are representative of all high school

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: 25 marks

Q2: 6 marks

Q3: 6 marks

Part II:

Q4: 5 marks

Q5: 5 marks

Q6: 5 marks

Part III:

Q7: 8 marks

Part IV:

Q8: 6 marks

Q9: 10 marks

Bonus Question: 5 marks

Materials allowed: Non-programmable calculator

Econ 246 Summer 2015 CRN#:31188

Page 2

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: A

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: B

3. 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


Page 3

4. 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: C

5. 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) 5.

B) 15.

C) 20.

D) 25.


Page 4

ANSWER: D

9. 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: A

14. 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?


Page 6

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 2/z

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: D

22. 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:


Page 7

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: B

25. 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:

4.1 4.1

0.95 .950.4111.3/ 10

4.1 1.645 3.42 minutes

0.411

a aP X a P Z P Z

aa


Page 8 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:

P( 2s > 0.5)= 2

2

1 29 0.5

0.3

n sP

=P( 2 >48.33)<0.01

b) The probability is 0.10 that for a sample of 30 bottles, the sample variance is less

than what number?

ANSWER:

2

2

2

1 29( ) 0.10 0.10

0.3

n s kP s k P

96.67 19.768k

2 96.67 0.10P k

0.2045k

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:


Page 9

“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 is a population parameter to be estimated,

and is an estimator where

( , , , ); X X Xn1 2

is said to be an unbiased estimator of if: E .

Example 1) )(XE 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 (

~)

E( ) E(~)

,~

Bias ~

[ (~) ] E

Two estimators:

~

is unbiased.

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.”

f ( )


Page 10

(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 X is

approximately normal as the sample size increases.

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 s

E n

andn s

n sn

E sn

n

E s

2 2

2

22

2

2

2

2 2

2 2

1

1

11

1

1

Since

if you take the expectation:

E

( )

( )

( )

(ii) Prove that E X( ) .

Let Xi ~( X, 2) for all i.

Since : (i)

Xn

X X X n 1

1 2( )


Page 11

and

(ii) E (Xi) = µ ,

we can apply the rules of expectation:

E X En

X

nE X

nE X X X

nE X E X E X E X

n

nn

ii

n

ii

n

n

n

X

( )

. #

1

1

1

1

1

1

1

1

1 2

1 2 3


Page 12

Part IV: SHORT ANSWER

Question 8: (6 marks)

Suppose we have two estimators of the population parameter :

E n

V n

( ) /

( ) /

17

96

4

2 3

and

E n

V n

(~

) /

(~

) /

93

84

2

2 4

(i) Determine the bias, if any, of each estimator.

Bias E n n ( ) / /17 174 4

bias E n n (~

) / /93 932 2

(ii) Determine the MSE. Which estimator is preferred?

MSE V bias n n ( ) / ( / ) 2 2 3 4 296 17

MSE V bias n n (~

) / ( / ) 2 2 4 2 284 93

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

3140 150 160 450 3 150

1 1

367700 3 22500

1

367700 67500 200 3 66 672 2 2

/

[ ] [ ( )( )] [ ] / .n

x nXi

2 2

1

2

1

21 1

N

XN

Xii

N

ii

N

( )

(ii) Determine the sampling distribution of the sample mean for a sample of size 2. Graph this distribution with

a simple bar graph.


Page 13

Total marks:4

X1,X2 X

140, 140 140

140, 150 145

140, 160 150

150, 140 145

150, 150 150

150, 160 155

160, 140 150

160, 150 155

160, 160 160

X P( X )

140 1/9

145 2/9

150 3/9

155 2/9

160 1/9

(iii) Determine the variance of X ? Total Marks:3

66.67/2=33.335

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?

P( X )

X

140 145 150 155 160


Page 14

Formulae

Central Location:

Population mean 1

Nxi

Grouped Population Mean

x f

f Nx f

i i

i

i i

1

Sample Mean Xn

X ii

n

1

1

Sample Mean for frequency distribution: Xn

X fi ii

n

1

1

Mean of the Sample Mean ( X ) E X X P XX i i

i

k

( ) ( )

1

where: i= 1,2,...,k, and k is the number of distinct possible values of X .

Dispersion:

Population variance 2 21

Nxi

(Grouped data 2 21

Nx fi i

Sample variance for frequency distribution: sn

x x fi i

2 21

1

( )

Sample variance sn

x xi

2 21

1

( )

Sample Standard Deviation s s 2

Variance of the Sample Mean X X

X

V X n X P X22

2 ( ) ( ) ( )

Standard Error of the mean:

X n n

2

.

Distributions:

Standard Normal:

ZX

( )

;

The Standardization of X:

ZX

n

t-distribution

tX

sn

; Chi-square distribution

n

s n s

1

2

2

2

2

2

1( ) ( )


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UNIVERSITY OF VICTORIA Midterm - Web hostingbettyj/246/midterm246_summer2015_solutions.pdf · standard deviation of $18.23. Assuming these values are representative of all high school