UNIVERSITY OF VICTORIA Midterm EXAM

SUMMER 2012: Solutions Course Name & No.: Economics 246

Sections(s): A01

CRN: 30868

Instructor: Betty Johnson

Duration: 1 hour 50 minutes NAME: STUDENT NUMBER: V00 This exam has a total of _12__ pages including this cover page and __0_ separation handouts(s). Students must count the number of pages and report any discrepancy immediately to the Invigilator. This exam is to be answered:

_X_ In Booklets provided Marking Scheme: Q1: 13 marks Q2: 5 marks Q3: 5 marks Q4: 5 marks Q5: 5 marks Q6: 5 marks Q7: 5 marks Q8: 6 marks Q9: 3 marks Q10: 3 marks Q11: 3 marks Q12: 11 marks Q13: 6 marks

Materials Allowed: Non-programmable calculator Answer ALL QUESTIONS (Total Marks = 75).

DEPARTMENT OF ECONOMICS

UNIVERSITY OF VICTORIA ECONOMICS 246: STATISTICAL INFERENCE

Solutions Term Test : Summer 2012 Instructor: Betty J. J. Johnson Instructions: Answer ALL questions. This exam has 13 questions. One formula sheet and 3 statistical tables are attached. Total pages: 12 All answers should be in the answer book. Total Marks =75; Time allowed =110 minutes. Question 1: Multiple Choice: Choose the best answer. (1 MARK EACH) 13 marks 1) If the variance of X is equals 25 and the sample size n=6, the population variance is: a) 30 b) 43.30 c) 25 *d) 150 e) cannot be determined. 2) Suppose ‘W’ is the width of all toilet seat covers used by all hotels in Victoria. Suppose W ~N(15, 9). A random sample of 36 covers is drawn from this

population. What are the standard normal z-value and the probability that the sample average width is more than 14 cm?

*a) -2 and 0.9772 b) 2 and 0.0228 c) 0.6667 and 0.7486 d) 12 and 0.6293 e) Cannot be determined since the distribution is undefined. 3) Suppose ‘H” is the height of all seven year olds in an early morning gymnastics class which is normally distributed, H~(42, 4). A random sample of 12 children is drawn from this population. What is the probability that the sample standard deviation is greater than 1.2 inches?

a) between 97.5% and 99% . *b) between 95% and 97.5% c) between 10 % and 90%. d) between 0.5% and 0.01%. e) none of the above.

4) If the variance of the sample mean is 10, and the sample size is 55, what is the standard deviation of the population of X values? *a) 23.4521 b) 550 c) 74.162 d) 8.6117 e) cannot be determined without the population mean 5) What is not a property of cluster sampling?

a) It suffers from geographical bias. b) little inter-cluster variation c) *little intra-cluster variation d) uses prior knowledge to generate clusters e) sampling consists of choosing a few clusters and using all elements in the clusters.

6) Response bias occurs when: a) the survey is poorly worded b) the respondent misunderstand the question c) interviewer misleads or influences the survey question such that the response does

not reflect the actual or true response. d) The respondent wants to impress the interviewer. e) *All of the above. 7 In examining the invoices issued by a company, an auditor finds that the dollar amounts of invoices have a mean of $1,732 and a standard deviation of $298. What is the probability that for a sample of 45 invoices, the average invoice is greater than $1,800?

a) 0.563 b) 0.063 c) 0.437 d) 0.937

e) cannot be determined without the population mean ANSWER: B

8) 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)

9) 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 e) The sample size

ANSWER: D

10) Which of the following statements is true regarding the standard error of the mean?

a) It is less than the standard deviation of the population. b) It decreases as the sample size increases. c) It measures the variability of the mean from sample to sample. d) All of the above e) None of the above

ANSWER: D

11) 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 e) 13.33 and 0.5 ANSWER: C

12) 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 e) mean of the standard deviation ANSWER: B

1. 13) According to the central limit theorem, the sampling distribution of the sample mean can be approximated by the normal distribution as the

A) number of samples gets "large enough" B) sample size gets "large enough" C) population standard deviation increases D) sample standard deviation decreases

ANSWER: B Question 2: Why does the sample size play such an important role in reducing the standard error of the mean? What are the implications of increasing the sample size? (5 marks) ANSWER: The standard error is the standard deviation of the population you are sampling from divided by the standard deviation of the sample size. So, mathematically as the sample size increases, the standard error naturally decreases. But there is more to this, because the standard error is the standard deviation of the population of sample means. So, as the sample size increases, the sample means are deviating less and less from the true population mean. Hence, as we sample more, we get statistics which are closer to the true parameters and our inference methods will improve. This is true for sampling distributions of mean, proportions, and variances. Question 3: Why is the Central Limit Theorem so important to statistical analysis? ( 5 marks)

ANSWER: Although the normal distribution occurs frequently and describes many populations; it is by no means the only probability distribution in existence. The value of the Central Limit Theorem is found in its conclusion that regardless of the shape of the population distribution (i.e. samples can come from any type of distribution), the sampling distribution of the mean will form a normal distribution. As a result, we can use the normal distribution when we are sampling from populations that we know nothing about and still know the distribution the sample means. Much of statistical inference is based not on the population we are sampling from, but on the distribution of the sample mean or sample proportion.

Question 4: In examining the invoices issued by a company, an auditor finds that the dollar amount of invoices have a mean of $1,732 and standard deviation of $298. Which pair of symmetric numbers around the mean make the statement P(a< X

Question 5: Given that σ 2=25, n=25, use the Chi-squared distribution to determine the

probability that the sample variance is less than 12. Use the Chi-square table to solve.

[ ] [ ]χ

σχ

112

2

2

2242

1 24 1225

1152

12 11520 010 0 025

=−

= =

≤ = ≤

( ) ( )( ) .

.

. . .

n s

P s Pto

Using the Chi-square table there is no specific value for 11.52. But, it is between 1% and 2.5%. (Total marks 5) Question 6: 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 stata. Question 7: Describe the concept of Cluster sampling. Total Marks: 5

See notes Question 8: Total marks:6 (i) Using the fact that the mean of the chi-squared distribution is (n-1), prove that

E S( )2 2= σ

( )

[ ][ ]

E s

E n

and n s

n s n

E s nn

E s

2 2

2

22

2

2

2

2 2

2 2

11

1 1

11

=

= −

=−

−⎡

⎣⎢

⎤

⎦⎥ = −

=−−

=

σ

χ

χσ

σ

σ

σ

Since

if you take the expectation:

E

( )( )

( )

(ii) Prove that

E X( ) = μ.

Let Xi ~(μ X, σ 2) for all i. Since : (i)

Xn

X X X n= + + +1

1 2( )L

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

L

L

Lμ μ μ μ

μ μ μ

(iii) Prove that E n( ) .χν

2 1= − .

( )[ ]

χσ

σ σ

σ

χσ

σ

22

2

2

2 22

2 2

22

2

1

1 1

1 1

=−

−⎡

⎣⎢

⎤

⎦⎥ =

−

=

=−

= −

( )

( ) ( ( ))

n s

n s n E s

Since E s

E n n

if you take the expectation:

E

Question 9: Determine an interval (a, b) such that [ ]P a t b≤ ≤ = 0 80. , and that 10% of the area is on each side of a and b, assuming that the sample is of size 21.

Total marks: 3 a=-1.325 b=1.325 Question 10: Determine an interval (a, b) such that [ ]P a b≤ ≤ =χ 2 0 99. , and that 0.5% of the area is on each side of a and b, assuming that the sample is of size 21.

Total marks: 3 a= 7.43 b= 40 Question 11: Suppose ‘G’ is the length of three-day facial hair growth of all male professional drivers on the college team. Suppose G is normally distributed with a mean

of 4mm, but an unknown variance. A random sample of 9 drivers is drawn from this population. What is the probability that the sample average is more than 5 with a sample variance of 9? Total marks:3 Between 10% and 25%

t Xsn

=−

=−

=μ 5 4

33

1

P(t>1) for 8 degrees of freedom greater than 10%. Question 12: (11 marks) Consider the following population of data: {34, 36, 38}.

(i) Determine the mean and variance of the population. Total marks:4

( )μ

σ

= + + = =

= − = − = − =∑

13

34 36 38 108 3 36

1 13

3896 3 1296 13

3896 3888 8 32 2 2

/

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

x nXi

σ μ μ2 21

2

1

21 1= − =⎡

⎣⎢

⎤

⎦⎥ −

= =∑ ∑N X N 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.

Total marks:5 X1,X2 X 34, 34 34 34, 36 35 34, 38 36 36, 34 35 36, 36 36 36, 38 37 38, 34 36 38, 36 37 38, 38 38 X P( X )

34 1/9 35 2/9 36 3/9 37 2/9 38 1/9

(iii) Determine the variance of X ? [8/3]/2=1.333

Total marks:2 Question 13:EViews: Answer the following using the EViews windows:

Total marks:6 Figure 1:

P( X )

X34 35 36 37 38

i) What is the variance of this sample of ten observations? 0.27177

(ii) What is the sample size (Figure 2)?10 (iii) What does the 0.219197 represent? Prob(t>0.8108)) Figure 3:

(iv) What does the 0.003830? The Z value.

Figure 4:

(v) What is the expected value of the distribution in figure 4? 10 (vi) What is the variance of the distribution in figure 4? 20 End of Exam

Formulae Central Location:

Population mean [ ]μ = ∑1N xi

Grouped Population Mean [ ]μ = =∑∑ ∑x ff N

x fi ii

i i1

Sample Mean Xn

Xii

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 ii

k

( ) ( )= ==∑μ

1

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

Population variance ( )σ μ2 21= ⎛⎝⎜⎞⎠⎟

−∑N xi

(Grouped data ( )[ ]σ μ2 21= ⎛⎝⎜ ⎞⎠⎟ −∑N x fi i Sample variance for frequency distribution: ( )s

nx x fi i

2 211

=−

−∑( )

Sample variance ( )sn

x xi2 21

1=

−−∑( )

Sample Standard Deviation s s= 2 Variance of the Sample Mean σ σ μX X

XV X n X P X

2 2 2= = = −∑( ) ( ) ( )

Standard Error of the mean: σσ σ

X n n= =

2

.

Distributions:

Standard Normal:

ZX

=−( )

;μ

σ The Standardization of X :

ZX

n=

− μσ

t-distribution

tXs

n=

− μ; Chi-square distribution χ

νσ σns n s

− = =−

12

2

2

2

2

1( ) ( )