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Past exam from 2019...

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UNIVERSITY OF MELBOURNE DEPARTMENT OF ECONOMICS SUMMER SEMESTER ASSESSMENT, 2019 ECON20003 QUANTITATIVE METHODS 2 Time Allowed: TWO hours Reading Time: 15 minutes

This exam contributes 70% to the assessment in ECON20003.

Answer BOTH questions from Part A and at least TWO (of three) from Part B. (If three Part B questions are answered, the Part B answer with the lowest mark will not be counted toward the final exam mark.) Part A is worth 30 marks: Question A1 comprises four true/false questions worth 2.5 marks each (10 marks); Question A2 comprises four multiple choice questions worth five marks each (20 marks). Part B is worth 40 marks: Each question is worth 20 marks; each part (a, b, c, d) of each question is worth five marks.

Each question starts on a separate page.

The following items are authorized in the exam room: •

Foreign language/English dictionaries;

•

Casio FX82 calculator (any suffix).

This exam paper has 17 (seventeen) pages (including this one). You are welcome to detach the following pages for convenience: •

pages 9–11: Formula Sheet;

•

pages 13–17: Statistical Tables.

This paper will not be held in the Baillieu Library.

Page 1 of 17

PART A (Answer BOTH questions; A1 and A2)

A1.

True/False

Note: •

Answer each question below (1, 2, 3, 4) by writing (in full) either “True” or “False” in the answer booklet; no marks will be given to answers written on this question paper.

•

No explanation of your choice (True or False) is required. However, if you have made the wrong choice, partial credit may be given for a worthy explanation or useful working.

1.

If a null hypothesis is rejected at the 10% level of significance, it must be rejected at the 5% level of significance.

2.

A primary school teacher of five-year-old children wants to test whether boys have a longer attention span than girls on average. The best approach to conducting the test would involve taking a matched-pairs sample.

3.

Based on a variance equality test statistic of 𝐹 = (𝑠1 /𝑠2 )2 = 0.5 ~𝐹(18,24) , where 𝑠 denotes the sample standard deviation, one should not reject the null hypothesis against the alternative of variance inequality at the 5% significance level.

4.

Regressing by OLS individual wages on an intercept, 𝑀𝑖 —which is a dummy variable

equal to 1 for men and 0 for women—and 𝐹𝑖 —which is a dummy variable equal to 1 for women and 0 for men—should yield a larger estimated coefficient on 𝑀𝑖 than on 𝐹𝑖 if men earn higher wages than women on average.

Page 2 of 17

A2.

Multiple Choice

•

Answer each question below (1, 2, 3, 4) by writing either A, B, C or D in the answer booklet; no marks will be given to answers written on this question paper.

•

No explanation of your choice (A, B, C or D) is required. However, if you have made the wrong choice, partial credit may be given for a worthy explanation or useful working.

1.

A winemaker asks 100 customers to rate each of his shiraz and merlot red wines on the following scale: 1 = Bad; 2 = OK; 3 = Good. Of the 100 customers, 36 rated the shiraz higher, and the rest were equally split between rating the wines the same and rating the

merlot higher. Based on these sample results, the 𝑝-value for testing the null hypothesis

that the wines are equally preferred against the alternative that the merlot is preferred is: A. less than 25% B. greater than 25% but less than 50% C. greater than 50% D. none of the above 2.

Suitable tests for comparing independent populations using ordinal data include: A. the Kruskal–Wallis test B. the sign test C. both of the above D. none of the above

3.

In single-factor analysis of variance (ANOVA), the null hypothesis of equal population means is unlikely to be rejected if: A. the ‘within’ variation is high relative to the ‘between’ variation B. the mean square for the treatments is low relative to the mean square for the error C. both of the above D. none of the above

Page 3 of 17

4.

Given the (partially hidden) EViews output below, Dependent Variable: Y Included observations: 8000 Variable

Coeff.

Std. Error

Z-Statistic

C X

0.1132 0.0766

0.3017

0.3752

Prob. 0.7075 0.1010

the 95% confidence interval for the slope coefficient (i.e., the one on X) excludes: A. 0.05 B. 0.10 C. 0.15 D. none of the above

Page 4 of 17

PART B (Answer at least TWO questions; only the two best answers will count)

B1.

The gender wage gap

Consider the simple regression model, ln𝑊𝑖 = 𝛽0 + 𝛽1 𝑀𝑖 + 𝜀𝑖 ,

(B1.1)

where ln denotes the natural log, 𝑊𝑖 is the hourly wage rate of worker 𝑖, 𝑀𝑖 = 1 for men (0 for

women) and 𝜀𝑖 is an error term.

Using OLS on a sample comprising 51 men and 51 women to estimate (B1.1) yielded 󰆹𝛽0 =

3.611 and 𝛽󰆹1 = 0.07.

The (partially hidden) descriptive statistics in Table 1.1 below were obtained from the same

sample, with 𝑋𝑖 denoting individual 𝑖’s years of work experience. Table 1.1. Descriptive statistics ln𝑊𝑖

Sample statistic

Men

Mean

𝑋𝑖

Women

ln𝑊𝑖

10.50

Standard deviation Observations

𝑋𝑖

6.25

0.26

1.45

0.32

1.20

51

51

51

51

(a)

Interpret 𝛽󰆹0 and 𝛽󰆹1 .

(b)

Given the regression results above, what are the two numerical values missing from Table 1.1?

(c)

Test whether men earn more per hour than women on average at the 5% significance level and interpret the result.

(d)

Consider the multiple regression model, ln𝑊𝑖 = 𝛾0 + 𝛾1 𝑀𝑖 + 𝛾2 𝑋𝑖 + 𝜀𝑖 .

(B1.2)

Given the information above and stating any assumption(s) you make, would you

󰆹1 , and why? expect 𝛾1 to be larger, smaller or no different than 𝛽

Page 5 of 17

B2.

Reoffending among the paroled

A researcher wants to investigate why some individuals released from prison on parole reoffend whereas others do not. As a starting point, the researcher considers the following probit model: Pr(𝑅𝑖 = 1) = Φ(𝛽0 + 𝛽1 𝑀𝑖 + 𝛽2 ln𝑌𝑖 +𝛽3 (𝑀𝑖 × ln𝑌𝑖 )),

(B2.1)

where Φ denotes the standard normal cumulative distribution function, ln denotes the natural log and:

𝑅𝑖 = 1 if paroled individual 𝑖 reoffends within three years of being paroled (0 otherwise);

𝑀𝑖 = 1 if individual 𝑖 is male (0 otherwise);

𝑌𝑖 is the number of years since individual 𝑖 was paroled. The above model was estimated using data on a sample of convicted individuals who were recently released from prison on parole, whose reoffending behaviour was subsequently monitored. The (rounded) parameter estimates obtained (with standard errors in brackets) were

𝛽󰆹0 = 0.00 (0.12), 𝛽󰆹1 = 0.7 (0.02), 𝛽󰆹2 = −0.5 (0.07), 𝛽󰆹3 = −0.5 (0.10). (a)

Robin Banks was paroled a year ago. What is his estimated probability of reoffending?

(b)

Courtney Act has an estimated probability of reoffending of 50%. How long ago was she paroled?

(c)

Based on a hypothesis test at the 5% significance level, could Courtney’s probability of reoffending be as high as 60%?

(d)

Estimate the number of years a man must be on parole to be equally likely to reoffend as a woman paroled 12 months ago.

Page 6 of 17

B3.

Time series

Consider the time series model, 𝑌𝑡 = 𝛽𝑋𝑡 + 𝛾𝑌𝑡−1 + 𝜀𝑡 ,

where 𝑇 is the sample size. (a)

(b)

𝑡 = 1, 2, … , 𝑇,

(B3.1)

If the estimated impact multiplier is 1.5 and the estimated total multiplier is 6, what are

the estimated values of 𝛽 and 𝛾?

Let 𝑋𝑡 follow an AR(1) process with no intercept and a first-order autocorrelation

coefficient of 0.5. Based on a current value of one (𝑋𝑇 = 1), calculate the two-period

ahead forecast, 𝑋𝐹𝑇+2, where the 𝐹 superscript denotes forecast. (c)

Assuming a current value of one (𝑌𝑇 = 1), use your answers to (a) and (b) above to

forecast the dependent variable of (B3.1) two periods ahead; i.e., compute 𝑌𝐹𝑇+2, where

the 𝐹 superscript denotes forecast. (d)

Based on the residual correlogram for (B3.1) below, should the forecasting model be improved, and why (or why not)? If it should be improved, how could this be done?

END OF QUESTIONS

Page 7 of 17

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Page 8 of 17

FORMULA SHEET (Three pages)

Independent Samples: Comparing Two Populations t=

t=

2

( s2 n ) + ( s 22 n 2 )  df . =  1 1  ( s12 n1)2 (s 22 n 2 ) 2  +   n2 − 1   n1 − 1

( X1 − X 2 ) − (μ1 − μ 2 ) s12 s22 + n1 n2

( X 1 − X 2 ) − (μ1 − μ 2 )

s2p =

1 1  s  +   n1 n2  2 p

F=

s12 s22

F(1− 2, v , v ) = 1 2

Z=

T1 − E (T1 ) σT

1 F(  2, v2 , v1 )

(n1 − 1)s12 + (n2 − 1)s 22 n1 + n 2 − 2

Z=

E ( T1 ) =

df . = n1 + n2 − 2

( pˆ 1 − pˆ 2 ) − ( p1 − p2 )

pˆ =

1 1 pˆ (1 − pˆ )  +   n1 n2 

n1(n1 + n2 +1) 2

σT =

n1 pˆ1 + n 2 pˆ 2 n1 + n 2

n1n2 (n1 + n2 + 1) 12

Independent Samples: Comparing Two or More Populations

F=

SST ( k − 1) MST = SSE ( n − k ) MSE

k

k

SST =  n j (x j − x )2

SSE =  (nj − 1) s 2j

j =1

j =1

k

k

x = 1  nj x j n j =1

n = n j j =1

k T2   12 H =  j  − 3(n + 1)  n(n +1) j =1 n j 

k

n = nj j =1

Matched Pairs

t=

X D − D sD nD

Z=

Page 9 of 17

X − 0.5n 0.5 n

Correlation and Linear Regression 𝑠𝑋𝑌 =

1

𝑛−1

𝑛 ) = ∑ 𝑖=1 (𝑋𝑖 − 𝑋 )(𝑌𝑖 − 𝑌

1

𝑛−1

(∑ 𝑛𝑖=1 𝑋𝑖 𝑌𝑖 − 𝑛𝑋 𝑌 ).

n n 2 sX2 = 1  ( X i − X ) = 1   X i2 − nX 2  n − 1 i =1 n − 1 i=1 

t=

r−ρ sr

2

ˆ =

n

(

i=1

R 2 = 1−

)

2

t=

( )

var βˆ1 =

ˆ j − j se ˆ

σ2 ( n − 1) sX2

( )

( )

se ˆ j = var ˆ j

( ) j

n

(

SSE =  Yi − Yˆi i=1

SSE SST

sXY s X sY

Z = rS n − 1

ˆ 0 = Y −ˆ 1 X

n 1 2  ei n − k − 1 i=1

SSR =  Yˆi − Y

1−r2 n −2

sr =

s ˆ 1 = XY2 sX

r=

F=

) = e 2

n

i=1

SST = SSR + SSE

2 i

SSR k SSE ( n − k − 1)

t=

Y0 − Yˆ0

(

se Y0 − Yˆ0

)

Linear Probability Model, Probit and Logit Models 𝐼𝑖 = 𝛽0 + 𝛽1 𝑋1𝑖 + 𝛽2 𝑋2𝑖 + 𝛽3 𝑋3𝑖 + ⋯ + 𝛽𝑘 𝑋𝑘𝑖 𝑃𝑖 = 𝛽0 + 𝛽1 𝑋1𝑖 + 𝛽2 𝑋2𝑖 + 𝛽3 𝑋3𝑖 + ⋯ + 𝛽𝑘 𝑋𝑘𝑖 = 𝐼𝑖 𝜕𝑃𝑖 𝜕𝐼𝑖 = = 𝛽𝑗 𝜕𝑋𝑗𝑖 𝜕𝑋𝑗𝑖

𝑃𝑖 = Φ (𝛽0 + 𝛽1 𝑋1𝑖 + 𝛽2 𝑋2𝑖 + 𝛽3 𝑋3𝑖 + ⋯ + 𝛽𝑘 𝑋𝑘𝑖 ) = Φ(𝐼𝑖 ) 𝐼𝑖2 1 𝜕𝐼𝑖 𝜕𝑃𝑖 exp (− ) ) × (𝜙 (𝐼𝑖 )) = 𝛽𝑗 × =( 𝜕𝑋𝑗𝑖 𝜕𝑋𝑗𝑖 2 √2𝜋

𝑃𝑖 =

1

1 + exp{−(𝛽0 + 𝛽1 𝑋1𝑖 + 𝛽2 𝑋2𝑖 + 𝛽3 𝑋3𝑖 + ⋯ + 𝛽𝑘 𝑋𝑘𝑖 )} 𝜕𝑃𝑖 exp{−𝐼𝑖 } = 𝛽𝑗 × 𝜕𝑋𝑗𝑖 (1 + exp{−𝐼𝑖 })2 Page 10 of 17

1 = 1 + exp{−𝐼 } 𝑖

Time Series Regression T

rs =

 (Yt − Y )( Yt− s − Y )

Z = Trs

t= s+ 1

 (Yt − Y ) T

2

f j = YT + j − YˆT + j

t =1

AR(1) model

 1− 1

Yt =  + 1Yt−1 +  t

E (Yt ) =

ˆ se ( f 1 ) = 

se( f 2 ) = ˆ  ˆ 12 + 1

var ( Yt ) =

 2 1 − 21

 s = 1s

ˆ 2 +1 se( f3 ) = ˆ  ˆ 14 +  1

Testing for unit roots p−1

p −1

Yt =  +  t + Yt −1 + i Yt −i +  t

 Yt =  +  Yt −1 +  i  Yt −i +  t i =1

i =1

Yt = Yt − Yt−1

=

ˆ se ( ˆ )

Finite distributed lag model 𝑌𝑡 = 𝛼 + 𝛽0 𝑋𝑡 + 𝛽1 𝑋𝑡−1 + 𝛽2 𝑋𝑡−2 + ⋯ + 𝛽𝑞 𝑋𝑡−𝑞 + 𝜀𝑡

s =

Yt Y = t +s Xt −s X t

Autoregressive distributed lag models ARDL(𝑝, 𝑞 ) ARDL(1, 0)

𝑌𝑡 = 𝛿 + 𝜃1 𝑌𝑡−1 + ⋯ + 𝜃𝑝 𝑌𝑡−𝑝 + 𝛿0 𝑋𝑡 + 𝛿1 𝑋𝑡−1 + ⋯ + 𝛿𝑞 𝑋𝑡−𝑞 + 𝜀𝑡 𝑌𝑡 = 𝛿 + 𝜃1 𝑌𝑡−1 + 𝛿0 𝑋𝑡 + 𝜀𝑡

𝛿 2 3 = ( 1−𝜃1 ) + 𝛿0 𝑋𝑡 + 𝜃1 𝛿0 𝑋𝑡−1 + 𝜃1 𝛿0 𝑋𝑡−2 + 𝜃1 𝛿0 𝑋𝑡−3 + ⋯ + 𝑣𝑡

Page 11 of 17

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Page 12 of 17

STATISTICAL TABLES (Five pages)

Page 13 of 17

Page 14 of 17

Page 15 of 17

Critical Values of the F-distribution

α = 0.025

upper (right) tail only

ν1 ν2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 24 26 28 30 35 40 45 50 60 70 80 90 100 120 140 160 180 200 ∞

1 648 38.5 17.4 12.2 10.0 8.81 8.07 7.57 7.21 6.94 6.72 6.55 6.41 6.30 6.20 6.12 6.04 5.98 5.92 5.87 5.79 5.72 5.66 5.61 5.57 5.48 5.42 5.38 5.34 5.29 5.25 5.22 5.20 5.18 5.15 5.13 5.12 5.11 5.10 5.03

2 799 39.0 16.0 10.6 8.43 7.26 6.54 6.06 5.71 5.46 5.26 5.10 4.97 4.86 4.77 4.69 4.62 4.56 4.51 4.46 4.38 4.32 4.27 4.22 4.18 4.11 4.05 4.01 3.97 3.93 3.89 3.86 3.84 3.83 3.80 3.79 3.78 3.77 3.76 3.69

3 864 39.2 15.4 10.0 7.76 6.60 5.89 5.42 5.08 4.83 4.63 4.47 4.35 4.24 4.15 4.08 4.01 3.95 3.90 3.86 3.78 3.72 3.67 3.63 3.59 3.52 3.46 3.42 3.39 3.34 3.31 3.28 3.26 3.25 3.23 3.21 3.20 3.19 3.18 3.12

4 900 39.2 15.1 9.60 7.39 6.23 5.52 5.05 4.72 4.47 4.28 4.12 4.00 3.89 3.80 3.73 3.66 3.61 3.56 3.51 3.44 3.38 3.33 3.29 3.25 3.18 3.13 3.09 3.05 3.01 2.97 2.95 2.93 2.92 2.89 2.88 2.87 2.86 2.85 2.79

5 922 39.3 14.9 9.36 7.15 5.99 5.29 4.82 4.48 4.24 4.04 3.89 3.77 3.66 3.58 3.50 3.44 3.38 3.33 3.29 3.22 3.15 3.10 3.06 3.03 2.96 2.90 2.86 2.83 2.79 2.75 2.73 2.71 2.70 2.67 2.66 2.65 2.64 2.63 2.57

Denominator degrees of freedom (ν 2) in left hand column

6 937 39.3 14.7 9.20 6.98 5.82 5.12 4.65 4.32 4.07 3.88 3.73 3.60 3.50 3.41 3.34 3.28 3.22 3.17 3.13 3.05 2.99 2.94 2.90 2.87 2.80 2.74 2.70 2.67 2.63 2.59 2.57 2.55 2.54 2.52 2.50 2.49 2.48 2.47 2.41

7 948 39.4 14.6 9.07 6.85 5.70 4.99 4.53 4.20 3.95 3.76 3.61 3.48 3.38 3.29 3.22 3.16 3.10 3.05 3.01 2.93 2.87 2.82 2.78 2.75 2.68 2.62 2.58 2.55 2.51 2.47 2.45 2.43 2.42 2.39 2.38 2.37 2.36 2.35 2.29

8 957 39.4 14.5 8.98 6.76 5.60 4.90 4.43 4.10 3.85 3.66 3.51 3.39 3.29 3.20 3.12 3.06 3.01 2.96 2.91 2.84 2.78 2.73 2.69 2.65 2.58 2.53 2.49 2.46 2.41 2.38 2.35 2.34 2.32 2.30 2.28 2.27 2.26 2.26 2.19

Numerator Degrees of Freedom 9 10 11 963 969 973 39.4 39.4 39.4 14.5 14.4 14.4 8.90 8.84 8.79 6.68 6.62 6.57 5.52 5.46 5.41 4.82 4.76 4.71 4.36 4.30 4.24 4.03 3.96 3.91 3.78 3.72 3.66 3.59 3.53 3.47 3.44 3.37 3.32 3.31 3.25 3.20 3.21 3.15 3.09 3.12 3.06 3.01 3.05 2.99 2.93 2.98 2.92 2.87 2.93 2.87 2.81 2.88 2.82 2.76 2.84 2.77 2.72 2.76 2.70 2.65 2.70 2.64 2.59 2.65 2.59 2.54 2.61 2.55 2.49 2.57 2.51 2.46 2.50 2.44 2.39 2.45 2.39 2.33 2.41 2.35 2.29 2.38 2.32 2.26 2.33 2.27 2.22 2.30 2.24 2.18 2.28 2.21 2.16 2.26 2.19 2.14 2.24 2.18 2.12 2.22 2.16 2.10 2.21 2.14 2.09 2.19 2.13 2.07 2.19 2.12 2.07 2.18 2.11 2.06 2.11 2.05 1.99

12 977 39.4 14.3 8.75 6.52 5.37 4.67 4.20 3.87 3.62 3.43 3.28 3.15 3.05 2.96 2.89 2.82 2.77 2.72 2.68 2.60 2.54 2.49 2.45 2.41 2.34 2.29 2.25 2.22 2.17 2.14 2.11 2.09 2.08 2.05 2.04 2.03 2.02 2.01 1.95

13 980 39.4 14.3 8.71 6.49 5.33 4.63 4.16 3.83 3.58 3.39 3.24 3.12 3.01 2.92 2.85 2.79 2.73 2.68 2.64 2.56 2.50 2.45 2.41 2.37 2.30 2.25 2.21 2.18 2.13 2.10 2.07 2.05 2.04 2.01 2.00 1.99 1.98 1.97 1.90

14 983 39.4 14.3 8.68 6.46 5.30 4.60 4.13 3.80 3.55 3.36 3.21 3.08 2.98 2.89 2.82 2.75 2.70 2.65 2.60 2.53 2.47 2.42 2.37 2.34 2.27 2.21 2.17 2.14 2.09 2.06 2.03 2.02 2.00 1.98 1.96 1.95 1.94 1.93 1.87

15 985 39.4 14.3 8.66 6.43 5.27 4.57 4.10 3.77 3.52 3.33 3.18 3.05 2.95 2.86 2.79 2.72 2.67 2.62 2.57 2.50 2.44 2.39 2.34 2.31 2.23 2.18 2.14 2.11 2.06 2.03 2.00 1.98 1.97 1.94 1.93 1.92 1.91 1.90 1.83

16 987 39.4 14.2 8.63 6.40 5.24 4.54 4.08 3.74 3.50 3.30 3.15 3.03 2.92 2.84 2.76 2.70 2.64 2.59 2.55 2.47 2.41 2.36 2.32 2.28 2.21 2.15 2.11 2.08 2.03 2.00 1.97 1.95 1.94 1.92 1.90 1.89 1.88 1.87 1.80

17 989 39.4 14.2 8.61 6.38 5.22 4.52 4.05 3.72 3.47 3.28 3.13 3.00 2.90 2.81 2.74 2.67 2.62 2.57 2.52 2.45 2.39 2.34 2.29 2.26 2.18 2.13 2.09 2.06 2.01 1.97 1.95 1.93 1.91 1.89 1.87 1.86 1.85 1.84 1.78

18 990 39.4 14.2 8.59 6.36 5.20 4.50 4.03 3.70 3.45 3.26 3.11 2.98 2.88 2.79 2.72 2.65 2.60 2.55 2.50 2.43 2.36 2.31 2.27 2.23 2.16 2.11 2.07 2.03 1.98 1.95 1.92 1.91 1.89 1.87 1.85 1.84 1.83 1.82 1.75

19 992 39.4 14.2 8.58 6.34 5.18 4.48 4.02 3.68 3.44 3.24 3.09 2.96 2.86 2.77 2.70 2.63 2.58 2.53 2.48 2.41 2.35 2.29 2.25 2.21 2.14 2.09 2.04 2.01 1.96 1.93 1.90 1.88 1.87 1.84 1.83 1.82 1.81 1.80 1.73

20 993 39.4 14.2 8.56 6.33 5.17 4.47 4.00 3.67 3.42 3.23 3.07 2.95 2.84 2.76 2.68 2.62 2.56 2.51 2.46 2.39 2.33 2.28 2.23 2.20 2.12 2.07 2.03 1.99 1.94 1.91 1.88 1.86 1.85 1.82 1.81 1.80 1.79 1.78 1.71

Critical Values of the F-distribution

α = 0.025

upper (right) tail only

ν1 ν2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 24 26 28 30 35 40 45 50 60 70 80 90 100 120 140 160 180 200 ∞

22 995 39.5 14.1 8.53 6.30 5.14 4.44 3.97 3.64 3.39 3.20 3.04 2.92 2.81 2.73 2.65 2.59 2.53 2.48 2.43 2.36 2.30 2.24 2.20 2.16 2.09 2.03 1.99 1.96 1.91 1.88 1.85 1.83 1.81 1.79 1.77 1.76 1.75 1.74 1.67

24 997 39.5 14.1 8.51 6.28 5.12 4.41 3.95 3.61 3.37 3.17 3.02 2.89 2.79 2.70 2.63 2.56 2.50 2.45 2.41 2.33 2.27 2.22 2.17 2.14 2.06 2.01 1.96 1.93 1.88 1.85 1.82 1.80 1.78 1.76 1.74 1.73 1.72 1.71 1.64

26 999 39.5 14.1 8.49 6.26 5.10 4.39 3.93 3.59 3.34 3.15 3.00 2.87 2.77 2.68 2.60 2.54 2.48 2.43 2.39 2.31 2.25 2.19 2.15 2.11 2.04 1.98 1.94 1.91 1.86 1.82 1.79 1.77 1.76 1.73 1.72 1.70 1.69 1.68 1.61

28 1000 39.5 14.1 8.48 6.24 5.08 4.38 3.91 3.58 3.33 3.13 2.98 2.85 2.75 2.66 2.58 2.52 2.46 2.41 2.37 2.29 2.23 2.17 2.13 2.09 2.02 1.96 1.92 1.89 1.83 1.80 1.77 1.75 1.74 1.71 1.69 1.68 1.67 1.66 1.59

30 1001 39.5 14.1 8.46 6.23 5.07 4.36 3.89 3.56 3.31 3.12 2.96 2.84 2.73 2.64 2.57 2.50 2.44 2.39 2.35 2.27 2.21 2.16 2.11 2.07 2.00 1.94 1.90 1.87 1.82 1.78 1.75 1.73 1.71 1.69 1.67 1.66 1.65 1.64 1.57

Denominator degrees of freedom (ν 2) in left hand column

35 1004 39.5 14.1 8.43 6.20 5.04 4.33 3.86 3.53 3.28 3.09 2.93 2.80 2.70 2.61 2.53 2.47 2.41 2.36 2.31 2.24 2.17 2.12 2.08 2.04 1.96 1.90 1.86 1.83 1.78 1.74 1.71 1.69 1.67 1.65 1.63 1.62 1.61 1.60 1.52