Econometrics Exam

Page 1 of 15

Office Use Only

 

 

 

 

Semester One 2019 Examination Period

Faculty of Business and Economics

EXAM CODES: ETC2410-ETW2410-BEX2410 TITLE OF PAPER: Introductory Econometrics – PAPER 1 EXAM DURATION: 2 hours writing time READING TIME: 10 minutes

THIS PAPER IS FOR STUDENTS STUDYING AT: (tick where applicable)

 

 Caulfield  Clayton  Parkville  Peninsula  Monash Extension  Off Campus Learning  Malaysia  Sth Africa  Other (specify) During an exam, you must not have in your possession any item/material that has not been authorised for your exam. This includes books, notes, paper, electronic device/s, mobile phone, smart watch/device, calculator, pencil case, or writing on any part of your body. Any authorised items are listed below. Items/materials on your desk, chair, in your clothing or otherwise on your person will be deemed to be in your possession. No examination materials are to be removed from the room. This includes retaining, copying, memorising or noting down content of exam material for personal use or to share with any other person by any means following your exam. Failure to comply with the above instructions, or attempting to cheat or cheating in an exam is a discipline offence under Part 7 of the Monash University (Council) Regulations, or a breach of instructions under Part 3 of the Monash University (Academic Board) Regulations. AUTHORISED MATERIALS OPEN BOOK  YES  NO CALCULATORS  YES  NO Only HP 10bII+ or Casio FX82 (any suffix) calculator permitted SPECIFICALLY PERMITTED ITEMS  YES  NO if yes, items permitted are: one A4 sheet of paper with hand written notes on both sides

Candidates must complete this section if required to write answers within this paper STUDENT ID: __ __ __ __ __ __ __ __ DESK NUMBER: __ __ __ __ __

 

 

INSTRUCTIONS TO STUDENTS

• Answer all FOUR questions. All questions are of equal value (15 marks). This paper is worth 60 marks in total and constitutes 60% of the final assessment.

• For multiple choice questions write the question number and only one letter (a), (b), (c), (d) or (e) for each question in your answer book (not on the question

sheet).

• When testing a hypothesis, to obtain full marks you need to specify the null and the alternative hypotheses, the test statistic and its distribution under the null,

and then perform the test and state your conclusion.

• If a question does not specify the level of significance of a hypothesis test explicitly, use 5%.

• Statistical tables are provided after Question 4.

Question 1 (15 marks)

This question has 15 multiple choice questions. Make sure that you clearly specify the question

number and only one letter for each multiple choice question in your answer book (not on the

question sheet).

1. Consider two datasets. In dataset A, we have data on consumption expenditure, income and

hours of work for every year from 2000 to 2017 for a group of individuals who were randomly

selected in the year 2000. In data set B, we have data on consumption per capita, income

per capita and unemployment rate for Australia, Indonesia, Malaysia, New Zealand, Thailand

and Vietnam for every year from 2000 to 2017.

(a) Both datasets are examples of time series data.

(b) Both datasets are examples of cross-sectional data.

(c) Both datasets are examples of panel data.

(d) Dataset A is an example of panel data, dataset B is an example of time series data.

(e) Dataset A is an example of cross-sectional data, dataset B is an example of time-series

data.

(1 mark)

2. Which of the following statements is NOT true?

(a) Randomised controlled trials are the best means for measuring causal relationships.

(b) In predictive modelling, the variables that are used as predictors need not cause the

variable that they try to predict.

(c) Correlation is not causation.

(d) Time series observations are always i.i.d.

(e) Time series data are ordered whereas cross section data are not.

(1 mark)

Page 2 of 15

 

 

3. Let  denote the weight of a newborn baby immediately after birth.  is a random variable

with mean , i.e. () =  and variance 2 i.e.  (−)2 = 2. We denote weights of 5 newborn babies selected at random by 12 3 4 and 5, and their sample average by ̄

Which of the following statements is NOT true

(a) P5

=1( − ̄) = 0 (b)

P5 =1  = 5̄

(c) (̄) = 

(d) ̄ = 

(e) ̄ is a linear combination of 12 3 4 and 5 (1 mark)

4. Let  and  denote returns to two risky assets. We are told that  () =  () =  and

 () =   () = 2 If we invest half of our savings in one of these assets and the other

half in the other asset, then the variance of the return to our investment will be

(a)  2

4 if  and  are uncorrelated

(b)  2

2 if  and  are uncorrelated

(c)  2

2 always

(d) 2 always

(e) (−)2+(−)2

4 if  and  are uncorrelated

(1 mark)

Questions 5 and 6 refer to the following p.d.f.: According to an expert, the annual growth

rate of the real GDP and the inflation rate for Malaysia in 2019 are governed by the following joint

probability density function:

Inflation rate ↓ , GDP growth rate → 4% 5% 6% 1% 0.1 0.1 0.0

2% 0.1 0.2 0.0

3% 0.1 0.1 0.1

4% 0.0 0.1 0.1

5. The expected growth rate of real GDP in Malaysia in 2019 according to this expert is:

(a) a random variable

(b) 500% because 4+5+6 3

= 5

(c) 490% because 4×03+5×05+6×02 = 49 (d) 250% because 1×02+2×03+3×03+4×02 = 25 (e) 492% because

1

4 ×{(4× 01

01+01 +5× 01

01+01 )+(4× 01

01+02 +5× 02

01+02 )+

(4× 01 01+01+01

+5× 01 01+01+01

+6× 01 01+01+01

)+

(5× 01 01+01

+6× 01 01+01

)} = 492

(1 mark)

Page 3 of 15

 

 

6. Conditional on 5% GDP growth rate, the expected inflation rate in Malaysia in 2019 according

to this expert is:

(a) a random variable

(b) 250% because 1+2+3+4 4

= 25

(c) 250% because 1×02+2×03+3×03+4×02 = 25 (d) 120% because 1×01+2×02+3×01+4×01 = 12 (e) 240% because 1× 01

05 +2× 02

05 +3× 01

05 +4× 01

05 = 24

(1 mark)

Questions 7, 8 and 9 refer to the multiple regression model

 = 0 + 11 + 22 + · · ·+  +   = 12      (1)

which in matrix notation is

y ×1

= X ×(+1)

β (+1)×1

+ u ×1

7.  (u | X) = 0 implies that

(a) (X0u) = 0

(b) X0bu = 0 where bu is the vector of OLS residuals of regression of y on X (c)  (u | X) = 2I where I is the identity matrix of order  (d) X0X is invertible

(e) Columns of X are linearly independent

(1 mark)

8. Which one of the following statements is correct?

(a) Xy is an  ×1 vector (b) X0X is an  ×  matrix (c) X0u = 0

(d) X0u is a ( +1)×1 vector (e) X0β is a ( +1)×1 vector

(1 mark)

9. Assuming that this model satisfies all assumptions of the Classical Linear Model (CLM) and

denoting the OLS estimator of β by bβ, which of the following statements is NOT correct? (a) bβ is an unbiased estimator of β (b) bβ is a consistent estimator of β (c) Conditional on X bβ is normally distributed (d) bβ is the best linear unbiased estimator of β (e) bβ is equal to β

(1 mark)

Page 4 of 15

 

 

10. We have chosen a random sample of 100 publicly listed companies and recorded their average

share price, profits, revenues and total costs in 2017-2018 financial year. Note that profits =

revenue – total cost. In a regression model with the share price as the dependent variable and

a constant, profit, revenue and total cost as independent variables, the OLS estimator

(a) cannot be computed because X0X matrix is not invertible

(b) will be biased because share price is not normally distributed

(c) will be unbiased

(d) will be BLUE

(e) will be unbiased but not BLUE

(1 mark)

Questions 11 to 13 refer to the following problem: We would like to model the relationship

between the price of an apartment with its area and its number of bedrooms. We postulate the

following population regression model

 = 0 + 1 + 2 + 

Suppose all assumptions of the Classical Linear Model applies to this model. We have collected

data on price (in 1000 dollars), area (in square metres) and number of bedrooms for 120 randomly

selected apartments and estimated the parameters of this models using OLS. This resulted in 31899

135 and 6237 for estimates of 0, 1 and 2 respectively.

11. Which of the following equations reports the results appropriately?

(a) d = 31899+135  +6237  (b) d = 31899+135  +6237  + ̂ (c) d = 31899+135  +6237  +  (d)  = 31899+135  +6237  + 

(e)  ( |  ) = 31899+135  +6237  (1 mark)

12. Which of the following statements is correct?

(a)  ( |  ) = 31899+135  +6237  (b)  ( |  ) = 31899+135  +6237  +  (c)  ( |  ) = 31899+135  +6237  + ̂ (d)  ( |  ) = 0 + 1 + 2 (e)  ( |  ) = 0 + 1 + 2 + 

(1 mark)

13. The null hypothesis for testing that given the area of an apartment, its number of bedrooms

is not a significant predictor of its price, is:

(a) 0 :  = 0

(b) 0 : ( | ) = 0 (c) 0 : b2 = 0 (d) 0 : 2 = 0

(e) 0 : b2 6= 0 (1 mark)

Page 5 of 15

 

 

Questions 14 and 15 relate to the following econometric model: Some economists believe

that the relationship between greenhouse gas emission and income is nonlinear. Denote a country’s

emission of CO2 per capita by 2 and its GDP per capita by  and consider the

following model:

2 = 0 + 1 + 2 2 +  (2)

14. The hypothesis that the relationship between 2 and  is linear versus the al-

ternative that it is an inverted U shape relationship can be written as:

(a) 0 : 2 = 0 against 1 : 2  0

(b) 0 : 2 = 0 against 1 : 2  0

(c) 0 : 1 = 0 against 1 : 1  0

(d) 0 : 1 = 0 against 1 : 1  0

(e) 0 : 1 = 2 = 0 against 1 : at least one of 1 or 2 not equal to zero

(1 mark)

15. If we know that in the model shown in equation (2)   ( | ) = 2, but all other assumptions of the Classical Linear Model are satisfied, then

(a) we can still use the OLS estimator because it is unbiased, and we can use the usual OLS

standard errors to perform  tests

(b) we can still use the OLS estimator because it is unbiased, but we need to use heteroskedas-

ticity robust standard errors to perform  tests

(c) we cannot use the OLS estimator because the OLS estimator is biased in this case

(d) we can still use the OLS estimator because it is the best linear unbiased estimator in this

case

(e) we can still use the OLS estimator because the OLS estimator is the same as the “weighted

least squares” estimator in this case

(1 mark)

Question 2 (15 marks)

2.a. Suppose we have a sample of  observations on a variable . Show that if we run a regression

of  on a constant only, the OLS estimate of the constant will be the sample average of 

(3 marks)

2.b. From the World Development Indicators database, we have extracted data on the following

variables for 121 countries in 2015:

Variable Definition Range

UNDER5 Mortality rate in children under 5 (per 1000 live births) 2.4 – 130.9

GDPPC GDP per capita in PPP adjusted dollars (as defined in assignment 1) 626 – 80892

SANITATION People using basic sanitation services (% of population) 7 – 100

WATER People using basic drinking water services (% of population) 0 – 100

The “Range” column provides the range of these variable in our sample.

Page 6 of 15

 

 

From these 121 countries, 35 are in sub-Saharan Africa. We have created a dummy variable called

SUBSAHARA which is equal to 1 if the country is a sub-Saharan country and 0 otherwise. Using

this data set, we have estimated the following regressions using OLS (standard errors are provided

in parentheses below parameter estimates)

d5 = 172 (21)

+596 (38)

 (3)

d5 = 1590 (145)

− 72 (22)

log()− 06 (01)

 − 02 (01)

 (4)

i. From the information provided, compute the average under-5 mortality rate (a) for the 35

sub-Saharan countries, (b) for the remaining 86 countries, and (c) for all 121 countries in this

sample.

(3 marks)

ii. Explain the estimated coefficients of log() in equation (4) in a way that a person

with no econometric training would understand.

(2 marks)

iii. Suppose we want to test the hypothesis that after controlling for log(), a 1 percentage

point increase in the proportion of population with access to basic sanitation has the same

effect on under-5 mortality as a 1 percentage point increase in the proportion of population

with access to drinking water, against the alternative that these effects are not equal, at the

5% level of significance. Explain how we could do that. For full marks, you need to state

the null, the alternative, the test statistic and its distribution under the null, any additional

regressions that we may have to estimate to calculate the test statistic, and how to come up

with a conclusion using this procedure. All of these need to be explained in the context of this

question where appropriate.

(4 marks)

iv. We have added  to equation (4) and re-estimated it and obtained the following

equation:

d5 = 1354 (145)

− 74 (22)

log()− 04 (01)

 − 01 (01)



+182 (58)

 (5)

Use this information to test the hypothesis that after controlling for GDP per capita and access

to sanitation and water services, there is no difference between the mean of under-5 mortality

in sub-Saharan countries and the rest of the world, against the alternative that sub-Saharan

countries have a higher mean, at the 5% level of significance. Remember that you need to

state all steps of hypothesis testing to obtain full marks.

(3 marks)

Page 7 of 15

 

 

Question 3 (15 marks)

3.a. In predictive modelling, when we want to find the best subset of  explanatory variables

{1 2     } to predict a target variable  we do not use 2 to compare models. Explain why, and provide the formula of an alternative statistic (only one) that we can use for selecting

the best predictive model, highlighting specifically how this statistic overcomes the deficiency

of 2 for model selection.

(3 marks)

3.b. We have randomly selected a sample of 249 employed men and collected the following infor-

mation: Variable Definition Range Median

WAGE hourly wage in dollars 7.5 – 125 30

EDUC years of education 2 – 18 12

EXPER years of experience 0 – 38 13

The “Range” and “Median” columns show the range and the median of each variable within

our sample, and zero years of experience means people who have less than 6 months experience.

Consider the following population regression model for the logarithm of wage given education

and experience:

log() = 0 + 1 ( −12)+ 2  + 3 2 +  (6) We have estimated the following regression using OLS:

dlog() = 2837 (0066)

+ 0095 (0010)

( −12)+ 0055 (0009)

− 0001 (00003)

2 (7)

2 = 0394 standard error of the regression = 0420  = 249

Note that we have subtracted 12 from years of education in order to make the results more

readily interpretable.

i. Interpret the estimated coefficients in this regression, including its intercept.

(4 marks)

ii. Can we interpret the coefficient of ( −12) as the estimate of the “return to education”, i.e. proportional increase in wage caused by an extra year of education? Explain.

(2 marks)

iii. In order to test the hypothesis that the errors of this model are homoskedastic against a specific

alternative, we have estimated the following auxiliary regression:

̂2 = 0096 (0031)

+ 0004 (0006)

( −12)+ 0005 (0002)



2 = 0039 standard error of the regression = 0262  = 249

where ̂ is the estimated residual of equation (7). Use this information to perform the test

at the 5% level of significance. Remember that you need to write down the null and the

alternative and all steps of hypothesis testing to obtain full marks.

(4 marks)

iv. Suppose we are told that the conditional variance of the error in model (6) is proportional to

experience, i.e.  ( | ) = 2 × . Explain how we can use this information to transform model (6) in such a way that the transformed model will have the

same parameters but no heteroskedasticity.

(2 marks)

Page 8 of 15

 

 

Question 4 (15 marks)

4.a. Infinance, it is generally believed thatassetpricesarenon-stationary timeseries. The following

graph shows a time series plot of the Standard and Poor’s U.S. Composite Stock Price Index

(S&P500) from January 1985 to July 2017.

i. Define a covariance stationary time series.

(3 marks)

ii. With reference to the graph, explain why S&P500 is not a covariance stationary time series.

(1 mark)

4.b. In order to forecast the number of international visitors to Victoria, we postulated the following

model:

log( ) = 0 + 1  + 2 1 + 3 2 + 4 3 + 5 −1 +  (8)

where

  : number of short term international visitors arriving in Victoria at time 

1 : dummy variable =1 if period  is the first quarter in a year, 0 otherwise

2 : dummy variable =1 if period  is the second quarter in a year, 0 otherwise

3 : dummy variable =1 if period  is the third quarter in a year, 0 otherwise

−1 : the value of 1 Australian dollar in terms of US dollar at time −1

and estimated it using quarterly data from 1991Q1 to 2008Q2:

dlog( ) = 1071 (0037)

+ 0017 (00002)

− 0027 (0017)

1 − 0364 (0017)

2 − 0300 (0017)

3 − 0370 (0050)

−1

 = 109 2 = 0987

i. Explain what we can learn from the signs of the estimated coefficients of the three dummy

variablesandthe signof thecoefficientof −1 (noneedto interpret theirmagnitudes). (2 marks)

Page 9 of 15

 

 

ii. A tourism expert tells us that “as far as seasonal variation of tourism around its trend is

concerned, there are only two seasons in Victoria, the cold season comprising 2 + 3,

and the hot season comprising 1 + 4”. Explain how you would use this data set to

test this hypothesis. To obtain full marks, you need to specify the null and alternative

hypotheses, the test statistic and its distribution under the null, and the regression or

regressions that you need to run to test this hypothesis.

(4 marks)

iii. Since we are using time series data, we suspect that the errors of equation (8) might be

serially correlated. To investigate that, we have obtained the correlogram of the residuals

of the estimated model.

iii.a. What are the consequences of serial correlation in errors for the OLS estimator of the

parameters and their usual OLS standard errors reported by statistical packages?

(2 marks)

iii.b. With reference to the correlogram of the residuals, explain how you would improve

model (8). Justify your answer.

(3 marks)

––––– END OF EXAMINATION –––––

Page 10 of 15

 

 

TABLE G.2 Critical Values of the tDistribution

1-Tailed: .10 .05

2-Tailed: .20 .10

1 3.078 6.314

2 1.886 2.920

3 1.638 2.353

4 1.533 2.132

5 1.476 2.015

6 1.440 1.943

7 1.415 1.895

8 1.397 1.860

9 1.383 1.833

10 1.372 1.812

11 1.363 1.796

12 1.356 1.782 e

g 13 1.350 1.771

r 14 1.345 1.761 e

15 1.341 1.753 e

s 16 1.337 1.746

17 1.333 1.740 0

18 1.330 1.734

19 1.328 1.729

F 20 1.325 1.725 r

21 1.323 1.721 e

e 22 1.321 1.717

d 23 1.319 1.714

0 24 1.318 1.711

25 1.316 1.708

26 1.315 1.706

27 1.314 1.703

28 1.313 1.701

29 1.311 1.699

30 1.310 1.697

40 1.303 1.684

60 1.296 1.671

90 1.291 1.662

120 1.289 1.658

00 1.282 1.645

Significance Level

.025 .01 .005

.05 .02 .01

12.706 31.821 63.657

4.303 6.965 9.925

3.182 4.541 5.841

2.776 3.747 4.604

2.571 3.365 4.032

2.447 3.143 3.707

2.365 2.998 3.499

2.306 2.896 3.355

2.262 2.821 3.250

2.228 2.764 3.169

2.201 2.718 3.106

2.179 2.681 3.055

2.160 2.650 3.012

2.145 2.624 2.977

2.131 2.602 2.947

2.120 2.583 2.921

2.110 2.567 2.898

2.101 2.552 2.878

2.093 2.539 2.861

2.086 2.528 2.845

2.080 2.518 2.831

2.074 2.508 2.819

2.069 2.500 2.807

2.064 2.492 2.797

2.060 2.485 2.787

2.056 2.479 2.779

2.052 2.473 2.771

2.048 2.467 2.763

2.045 2.462 2.756

2.042 2.457 2.750

2.021 2.423 2.704

2.000 2.390 2.660

1.987 2.368 2.632

1.980 2.358 2.617

1.960 2.326 2.576

Examples: The 1 % critical value for a one-tailed test with 25 df is 2.485. The 5% critical value for a two-tailed test with large (> 120) dfis 1.96.

Source: This table was generated using the Stata® function invttail.

STATISTICAL TABLES

Page 11 of 15

 

 

TABLE G.3a 10% Critical Values of the F Distribution

Numerator Degrees of Freedom

1 2 3 4 5 6

10 3.29 2.92 2.73 2.61 2.52 2.46

D 11 3.23 2.86 2.66 2.54 2.45 2.39

e 12 3.18 2.81 2.61 2.48 2.39 2.33

n 13 3.14 2.76 2.56 2.43 2.35 2.28

0 14 3.10 2.73 2.52 2.39 2.31 2.24

m

i 15 3.07 2.70 2.49 2.36 2.27 2.21

n 16 3.05 2.67 2.46 2.33 2.24 2.18

a 17 3.03 2.64 2.44 2.31 2.22 2.15

0 18 3.01 2.62 2.42 2.29 2.20 2.13

r 19 2.99 2.61 2.40 2.27 2.18 2.11

20 2.97 2.59 2.38 2.25 2.16 2.09 D

.

e 21 2.96 2.57 2.36 2.23 2.14 2.08

g 22 2.95 2.56 2.35 2.22 2.13 2.06

r 23 2.94 2.55 2.34 2.21 2.11 2.05 e

24 e

2.93 2.54 2.33 2.19 2.10 2.04

s 25 2.92 2.53 2.32 2.18 2.09 2.02

26 2.91 2.52 2.31 2.17 2.08 2.01

0 27 2.90 2.51 2.30 2.17 2.07 2.00

28 2.89 2.50 2.29 2.16 2.06 2.00

F 29 2.89 2.50 2.28 2.15 2.06 1.99 r

30 2.88 2.49 2.28 2.14 2.05 1.98 e

e 40 2.84 2.44 2.23 2.09 2.00 1.93

d 60 2.79 2.39 2.18 2.04 1.95 1.87

0 90 2.76 2.36 2.15 2.01 1.91 1.84

m

120 2.75 2.35 2.13 1.99 1.90 1.82

00 2.71 2.30 2.08 1.94 1.85 1.77

Example: The 10% critical value for numerator df = 2 and denominator df = 40 is 2.44.

Source: This table was generated using the Stata® function invFtail.

7

2.41

2.34

2.28

2.23

2.19

2.16

2.13

2.10

2.08

2.06

2.04

2.02

2.01

1.99

1.98

1.97

1.96

1.95

1.94

1.93

1.93

1.87

1.82

1.78

1.77

1.72

8 9 10

2.38 2.35 2.32

2.30 2.27 2.25

2.24 2.21 2.19

2.20 2.16 2.14

2.15 2.12 2.10

2.12 2.09 2.06

2.09 2.06 2.03

2.06 2.03 2.00

2.04 2.00 1.98

2.02 1.98 1.96

2.00 1.96 1.94

1.98 1.95 1.92

1.97 1.93 1.90

1.95 1.92 1.89

1.94 1.91 1.88

1.93 1.89 1.87

1.92 1.88 1.86

1.91 1.87 1.85

1.90 1.87 1.84

1.89 1.86 1.83

1.88 1.85 1.82

1.83 1.79 1.76

1.77 1.74 1.71

1.74 1.70 1.67

1.72 1.68 1.65

1.67 1.63 1.60

Page 12 of 15

 

 

TABLE G.Jb 5% Critical Values of the f Distribution

Numerator Degrees of Freedom

1 2 3 4 5 6 7

D 10 4.96 4.10 3.71 3.48 3.33 3.22 3.14

e 11 4.84 3.98 3.59 3.36 3.20 3.09 3.01

n 12 4.75 3.89 3.49 3.26 3.11 3.00 2.91

0 13 4.67 3.81 3.41 3.18 3.03 2.92 2.83 m 14 4.60 3.74 3.34 3.11 2.96 2.85 2.76

15 4.54 3.68 3.29 3.06 2.90 2.79 2.71 n

a 16 4.49 3.63 3.24 3.01 2.85 2.74 2.66

t 17 4.45 3.59 3.20 2.96 2.81 2.70 2.61

0 18 4.41 3.55 3.16 2.93 2.77 2.66 2.58

r 19 4.38 3.52 3.13 2.90 2.74 2.63 2.54

20 4.35 3.49 3.10 2.87 2.71 2.60 2.51 D

21 4.32 3.47 3.07 2.84 2.68 2.57 2.49

g 22 4.30 3.44 3.05 2.82 2.66 2.55 2.46

r 23 4.28 3.42 . 3.03 2.80 2.64 2.53 2.44

e 24 4.26 3.40 3.01 2.78 2.62 2.51 2.42 e 25 4.24 3.39 2.99 2.76 2.60 2.49 2.40

26 4.23 3.37 2.98 2.74 2.59 2.47 2.39

0 27 4.21 3.35 2.96 2.73 2.57 2.46 2.37

f 28 4.20 3.34 2.95 2.71 2.56 2.45 2.36

29 4.18 3.33 2.93 2.70 2.55 2.43 2.35

F 30 4.17 3.32 2.92 2.69 2.53 2.42 2.33 r

40 4.08 3.23 2.84 2.61 2.45 2.34 2.25 e

60 4.00 3.15 e

2.76 2.53 2.37 2.25 2.17

d 90 3.95 3.10 2.71 2.47 2.32 2.20 2.11

0 120 3.92 3.07 2.68 2.45 2.29 2.17 2.09

m 00 3.84 3.00 2.60 2.37 2.21 2.10 2.01

Example: The 5% critical value for numerator df = 4 and large denominator df( oo) is 2.37.

Source: This table was generated using the Stata® function invFtail.

8 9 10

3.07 3.02 2.98

2.95 2.90 2.85

2.85 2.80 2.75

2.77 2.71 2.67

2.70 2.65 2.60

2.64 2.59 2.54

2.59 2.54 2.49

2.55 2.49 2.45

2.51 2.46 2.41

2.48 2.42 2.38

2.45 2.39 2.35

2.42 2.37 2.32

2.40 2.34 2.30

2.37 2.32 2.27

2.36 2.30 2.25

2.34 2.28 2.24

2.32 2.27 2.22

2.31 2.25 2.20

2.29 2.24 2.19

2.28 2.22 2.18

2.27 2.21 2.16

2.18 2.12 2.08

2.10 2.04 1.99

2.04 1.99 1.94

2.02 1.96 1.91

1.94 1.88 1.83

Page 13 of 15

 

 

TABLE G.3c 1 % Critical Values of the F Distribution

Numerator Degrees of Freedom

1 2 3 4 5 6

10 10.04 7.56 6.55 5.99 5.64 5.39

D 11 9.65 7.21 6.22 5.67 5.32 5.07

e 12 9.33 6.93 5.95 5.41 5.06 4.82

n 13 9.07 6.70 5.74 5.21 4.86 4.62

0

m 14 8.86 6.51 5.56 5.04 4.69 4.46

i 15 8.68 6.36 5.42 4.89 4.56 4.32

n 16 8.53 6.23 5.29 4.77 4.44 4.20

17 8.40 6.11 5.18 4.67 4.34 4.10 t

0 18 8.29 6.01 5.09 4.58 4.25 4.01

r 19 8.18 5.93 5.01 4.50 4.17 3.94

20 8.10 5.85 4.94 4.43 4.10 3.87 D

21 8.02 5.78 4.87 4.37 4.04 3.81 e

. g 22 7.95 5.72 4.82 4.31 3.99 3.76

r 23 7.88 5.66 4.76 ·4.26 3.94 3.71

e 24 7.82 5.61 4.72 4.22 3.90 3.67

e 25 7.77 5.57 4.68 4.18 3.85 3.63

s 26 7.72 5.53 4.64 4.14 3.82 3.59

0 27 7.68 5.49 4.60 4.11 3.78 3.56

f 28 7.64 5.45 4.57 4.07 3.75 3.53

F 29 7.60 5.42 4.54 4.04 3.73 3.50

r 30 7.56 5.39 4.51 4.02 3.70 3.47

e 40 7.31 5.18 4.31 3.83 3.51 3.29

e 60 7.08 4.98 4.13 3.65 3.34 3.12

0 90 6.93 4.85 4.01 3.54 3.23 3.01

m 120 6.85 4.79 3.95 3.48 3.17 2.96

00 6.63 4.61 3.78 3.32 3.02 2.80

Example: The I% critical value for numerator df = 3 and denominator df = 60 is 4.13.

Source: This table was generated using the Stata® function invFtail.

7

5.20

4.89

4.64

4.44

4.28

4.14

4.03

3.93

3.84

3.77

3.70

3.64

3.59

3.54

3.50

3.46

3.42

3.39

3.36

3.33

3.30

3.12

2.95

2.84

2.79

2.64

8 9 10

5.06 4.94 4.85

4.74 4.63 4.54

4.50 4.39 4.30

4.30 4.19 4.10

4.14 4.03 3.94

4.00 3.89 3.80

3.89 3.78 3.69

3.79 3.68 3.59

3.71 3.60 3.51

3.63 3.52 3.43

3.56 3.46 3.37

3.51 3.40 3.31

3.45 3.35 3.26

3.41 3.30 3.21

3.36 3.26 3.17

3.32 3.22 3.13

3.29 3.18 3.09

3.26 3.15 3.06

3.23 3.12 3.03

3.20 3.09 3.00

3.17 3.07 2.98

2.99 2.89 2.80

2.82 2.72 2.63

2.72 2.61 2.52

2.66 2.56 2.47

2.51 2.41 2.32

Page 14 of 15

 

 

TABLE G.4 Critical Values of the Chi-Square Distribution

Significance Level

.10 .05 .01

1 2.71 3.84 6.63

2 4.61 5.99 9.21

3 6.25 7.81 11.34

4 7.78 9.49 13.28

5 9.24 11.07 15.09

6 10.64 12.59 16.81

D 7 12.02 14.07 18.48

e 8 13.36 15.51 20.09

g 9 14.68 16.92 21.67

r 10 15.99 18.31 23.21

e 11 17.28 19.68 24.72 e 12 18.55 21.03 26.22 s

13 19.81 22.36 27.69

0 14 21.06 23.68 29.14

f 15 22.31 25.00 30.58

16 23.54 26.30 32.00

F 17 24.77 27.59 33.41

r 18 25.99 28.87 34.81

e 19 27.20 30.14 36.19 e

20 28.41 31.41 37.57

21 29.62 32.67 38.93 0

22 30.81 33.92 40.29

23 32.01 35.17 41.64

24 33.20 36.42 42.98

25 34.38 37.65 44.31

26 35.56 38.89 45.64

27 36.74 40.11 46.96

28 37.92 41.34 48.28

29 39.09 42.56 49.59

30 40.26 43.77 50.89

Example: The 5% critical value with df = 8 is 15.51.

Source: This table was generated using the Stata® function invchi2tail.

Page 15 of 15

Do you need a similar assignment written for you from scratch? We have qualified writers to help you. You can rest assured of an A+ quality paper that is plagiarism free. Order now for a FREE first Assignment! Use Discount Code "FREE" for a 100% Discount!

NB: We do not resell papers. Upon ordering, we write an original paper exclusively for you.

Order New Solution