Basics of the Labor Market
Participants are assigned motives:
• Workers look for the best job
• Firms look for profits
• Government uses regulation to achieve goals of public policy
– Minimum wages
– Occupational safety
Workers• The most important actor; without workers, there is no
“labor”• Desire to optimize (to select the best option from available
choices) to maximize well-being• Will want to supply more time and effort for higher payoffs,
causing an upward sloping labor supply curve
Firms
• Decide who to hire and fire
• Motivated to maximize profits
• Relationship between price of labor and the number of workers a firm is willing to hire generates the labor demand curve
Basics of the Labor Market
Government
• Imposes taxes
• Safety/environmental regulations
• Set minimum wages
• Force firms to shuttle workers from home to work (SF, CA)
• Mediate labor union disputes with firms
Why is there a shortage of math teachers?
Mathematics History
LS (mathematicians)
LS (historians)
LD
LD
w union
w*ind
w*ind
E*math E*
hist
shortage surplus
Why study Labor Economics?
How does increasing the minimum wage affect workers and firms?
Low skilled labor market
LS (workers)
LD (firms)
unemployment
unemployment
wmin
wmin
LFE LFE
Low skilled labor market
LS (workers)
LD (firms)
w*
E*
Why study Labor Economics?
Low skilled labor market
LS (workers)
LD (firms)
w*
E*
w*
E*
Is there a cost to immigration?
A flood of low skilled workers into
an economy…
Why study Labor Economics?
Rise = .018 – (-.012) = .03
Run = -0.053 – 0.037 = -0.09
= -.03/0.09 = -.333
Using data to confirm theory(Scatterplots and simple regression)
The equation that describes how the dependent variable y is related to the independent variables and error is called the multiple regression model
y = 0 + 1x1 + 2x2 + . . . + kxk +
The equation that describes how the mean value of y is related to the independent variables is called the multiple regression equation
E(y) = 0 + 1x1 + 2x2 + . . . + kxk
Using data to confirm theory
y = b0 + b1x1 + b2x2 + . . . + bkxk
The equation that describes how the predicted value of y is related to the independent variables is called the estimated multiple regression equation:
(multiple regression)
1. Formulate a research question:
How has welfare reform affected employment of low-income mothers?
Issue 1: How should welfare reform be defined?
Since we are talking about aspects of welfare reform that influence the decision to work, we include the following variables:
• Welfare payments allow the head of household to work less.
tanfben3 = real value (in 1983 $) of the welfarepayment to a family of 3 (x1)
• The Republican lead Congress passed welfare reform twice both of which were vetoed by President Clinton. Clinton signed it into law after the Congress passed it a third time in 1996. All states put their TANF programs in place by 2000.
2000 = 1 if the year is 2000, 0 if it is 1994 (x2)
Using data to confirm theory
1. Formulate a research question:
How has welfare reform affected employment of low-income mothers?
Issue 1: How should welfare reform be defined? (continued)
• Families receive full sanctions if the head of household fails to adhere to a state’s work requirement.
fullsanction = 1 if state adopted policy, 0 otherwise (x3)
Issue 2: How should employment be defined?
• One might use the employment-population ratio of Low-Income Single Mothers (LISM):
number of LISM that are employed
number of LISM livingepr
Using data to confirm theory
2. Use economic theory or intuition to determine what the true regression model might look like.
40
400
U0
55
550 U1
300
Leisure
Consumption
Receiving the welfare check
increases LISM’s leisure which decreases hours worked
Use economic graphs to derive testable hypotheses:
Economic theory suggests the following is not true:
Ho: 1 = 0
(Using theory to build testable hypotheses)Using data to confirm theory
2. Use economic theory or intuition to determine what the true regression model might look like.
Use a mathematical model to derive testable hypotheses:
Economic theory suggests the following is not true:
Ho: 1 = 0
max ( , )
. .
80
U C L C L
s t
H L
C P wH
The solution of this problem is:
* 402
PL
w
* 10
2
L
P w
1 0 *
0H
P
Using data to confirm theory(Using theory to build testable hypotheses)
3. Compute means, standard deviations, minimums and maximums for the variables.
state year epr tanfben3 fullsanction black dropo unemp
Alabama 1994 52.35 110.66 0 25.69 26.99 5.38
Alaska 1994 38.47 622.81 0 4.17 8.44 7.50
Arizona 1994 49.69 234.14 0 3.38 13.61 5.33
Arkansas 1994 48.17 137.65 0 16.02 25.36 7.50
West Virginia 2000 51.10 190.48 1 3.10 23.33 5.48
Wisconsin 2000 57.99 390.82 1 5.60 11.84 3.38
Wyoming 2000 58.34 197.44 1 0.63 11.14 3.81
(Model Specification in regression)Using data to confirm theory
3. Compute means, standard deviations, minimums and maximums for the variables.
1994Mean Std Dev Min Max
2000Mean Std Dev Min Max Diff
epr 46.73 8.58 28.98 65.64 53.74 7.73 40.79 74.72 7.01
tanfben3 265.79 105.02 80.97 622.81 234.29 90.99 95.24 536.00 -31.50
fullsanction 0.02 0.14 0.00 1.00 0.70 0.46 0.00 1.00 0.68
black 9.95 9.45 0.34 36.14 9.82 9.57 0.26 36.33 -0.13
dropo 17.95 5.20 8.44 28.49 14.17 4.09 6.88 23.33 -3.78
unemp 5.57 1.28 2.63 8.72 3.88 0.96 2.26 6.17 -1.69
(Model Specification in regression)Using data to confirm theory
0
10
20
30
40
50
60
70
80
0 2 4 6 8 10
unemp
epr
0
10
20
30
40
50
60
70
80
0 5 10 15 20 25 30
dropo
epr
0
10
20
30
40
50
60
70
80
0 10 20 30 40
black
epr
0
10
20
30
40
50
60
70
80
0 200 400 600 800
tanfben3
epr
4. Construct scatterplots of the variables. (1994, 2000)
(Model Specification in regression)Using data to confirm theory
5. Compute correlations for all pairs of variables. If | r | > .7 for a pair of independent variables, • multicollinearity may be a problem• it is not possible to determine the separate effect of any particular
independent variable on y.• Some say avoid including independent variables that are highly
correlated, but it is better to have multicollinearity than omitted variable bias.
epr fullsanction black dropo unemp
tanfben3 -0.03 -0.24 -0.53 -0.50 0.10
unemp -0.64 -0.51 0.16 0.47
dropo -0.44 -0.25 0.51
black -0.32 0.07
fullsanction 0.43
(Model Specification in regression)Using data to confirm theory
state year pmt pmt_ln
Alabama 1994 110.66 4.71
Alaska 1994 622.81 6.43
Arizona 1994 234.14 5.46
Arkansas 1994 137.65 4.92
West Virginia 2000 190.48 5.25
Wisconsin 2000 390.82 5.97
Wyoming 2000 197.44 5.29
Variable transformation
(Model Specification in regression)Using data to confirm theory
Least Squares Criterion: 2 2min ( ) min ( )i i iy y ˆe
Computation of Coefficient Values:
In simple regression:
You can use matrix algebra or computer software packages to compute the coefficients
In multiple regression:
11
1
cov( , )var( )
x yb
x
0 1 1b y bx
1( )b XX Xy
0
1
p
b
b
b
(Estimation)Using data to confirm theory
Regression Statistics
Multiple R 0.0279
R Square 0.0008
Adjusted R Square -0.0094
Standard Error 8.8978
Observations 100
ANOVA
df SS MS F
Regression 1 6.031 6.031 0.076
Residual 98 7758.733 79.171
Total 99 7764.764
Coefficients Standard Error t Stat P-value
Intercept 46.9192 12.038 3.897 0.000
pmt_ln 0.6087 2.206 0.276 0.783
r 2·100% of the variability in y
can be explained by the model.
.08%epr of LISM
Error
(Omitting variable bias)Using data to confirm theory
-1.984 1.984
.025
.2760t
.025
We cannot reject H0 at a 5% level of significance.
Do Not Reject RejectReject
1
1-statb
bt
s.276
.60872.2055
df = 100 – 1 – 1 = 98 (column) = .05 /2 = .025 (row)
H0: 1 = 0
(hypothesis testing)Using data to confirm theory
1 1$292 $266 1 1ˆ ˆ| | ln(292) ln(266) 0.6087 ln(1.10) .058x xepr epr b b .10 .058
• If estimated coefficient b1 was statistically significant, we would interpret its value as follows:
Increasing monthly benefit levels for a family of three by 10% would result in a .058 percentage point increase in the average epr
of LISM
• However, since estimated coefficient b1 is statistically insignificant, we interpret its value as follows:
Increasing monthly benefit levels for a family of three
has no effect on the epr of LISM.
Our theory suggests that this estimate has the wrong sign and is biased towards zero. This bias is called omitted variable bias.
(interpretation)Using data to confirm theory
R Square 0.166
Adjusted R Square 0.149
Standard Error 8.171
Observations 100
ANOVA
df SS MS F
Regression 2 1288.797 644.398 9.652
Residual 97 6475.967 66.763
Total 99 7764.764
Coefficients Standard Error t Stat P-value
Intercept
pmt_ln
2000
r 2·100% of the variability in y
can be explained by the model.
15%epr of LISM
Error
(Estimation)Using data to confirm theory
R Square 0.214
Adjusted R Square 0.190
Standard Error 7.971
Observations 100
ANOVA
df SS MS F
Regression 3 1664.635 554.878 8.732
Residual 96 6100.129 63.543
Total 99 7764.764
Coefficients Standard Error t Stat P-value
Intercept 31.544 11.204 2.815 0.006
pmt_ln 2.738 2.024 1.353 0.179
2000 3.401 2.259 1.506 0.135
full 5.793 2.382 2.432 0.017
r 2·100% of the variability in y
can be explained by the model.
19%epr of LISM
Error
(Estimation)Using data to confirm theory
R Square 0.517
Adjusted R Square 0.486
Standard Error 6.347
Observations 100
ANOVA
df SS MS F
Regression 6 4018.075 669.679 16.623
Residual 93 3746.689 40.287
Total 99 7764.764
Coefficients Standard Error t Stat P-value
Intercept
pmt_ln
2000
full
black
drop
unemp
r 2·100% of the variability in y
can be explained by the model.
49%epr of LISM
Error
(Estimation)Using data to confirm theory
R Square 0.517
Adjusted R Square 0.486
Standard Error 6.347
Observations 100
ANOVA
df SS MS F
Regression 6 4018.075 669.679 16.623
Residual 93 3746.689 40.287
Total 99 7764.764
Coefficients Standard Error t Stat P-value
Intercept
pmt_ln
2000
full
black
drop
unemp
(Estimation)
lnx1
x2
x3
x4
x5
x6
+
Error
1 2 3 4 5 6ˆ 104.529 5.709ln 2.821 3.768 0.291 0.374 3.023y x x x x x x
Using data to confirm theory
E() is probably equal to zero since E(e) = 0
1 2 3 4 5 6ˆ 104.529 5.709ln 2.821 3.768 0.291 0.374 3.023y x x x x x x
epry
pmt_ln 2000 full black drop unemp epr hat residuale
52.35 4.71 0 0 25.69 26.99 5.38 43.83 8.52
38.47 6.43 0 0 4.17 8.44 7.50 40.76 -2.29
49.69 5.46 0 0 3.38 13.61 5.33 51.19 -1.50
48.17 4.92 0 0 16.02 25.36 7.50 39.60 8.57
51.10 5.25 1 1 3.10 23.33 5.48 49.31 1.79
57.99 5.97 1 1 5.60 11.84 3.38 55.14 2.85
58.34 5.29 1 1 0.63 11.14 3.81 59.44 -1.10
Sum 0
1lnx 2x 3x 4x 5x 6x y
(A1: zero mean)Using data to confirm theory
Heteroscedasticity is likely present if scatterplots of residuals versus t, y, x1, x2 … xk are not a random horizontal band of points.
^
-15
-10
-5
0
5
10
15
20
30 40 50 60 70
predicted epr
resid
ual
-15
-10
-5
0
5
10
15
20
0 10 20 30 40
black
resid
ual
-15
-10
-5
0
5
10
15
20
4 5 6 7
pmt
resid
ual
-15
-10
-5
0
5
10
15
20
0 10 20 30
drop
resid
ual
-15
-10
-5
0
5
10
15
20
0 2 4 6 8 10
unemp
resid
ual
Non-constant variance in black?
Using data to confirm theory(A2: Constant variance)
Coefficients Standard Error t Stat P-value
Intercept -4681.00 3014.00 -1.55 0.125
pmt_ln 1526.70 972.20 1.57 0.121
2000 76.30 398.50 0.19 0.849
full -88.70 394.80 -0.22 0.823
black 28.53 28.72 0.99 0.324
drop 115.56 55.98 2.06 0.042
unemp -204.70 165.00 -1.24 0.219
pmt_ln2 -128.25 79.93 -1.60 0.113
black2 0.13 0.10 1.30 0.196
drop2 -0.75 0.44 -1.69 0.095
unemp2 -1.73 4.33 -0.40 0.690
pmt_lnX2000 -6.67 63.28 -0.11 0.916
Using data to confirm theory(A2: Constant variance)
To test for heteroscedasticity, perform White’s squared residual test by first squaring the residuals, and then using these as the “y” variable in a secondary regression:
pmt_lnXdrop -16.31 8.97 -1.82 0.073pmt_lnXfull 26.14 61.29 0.43 0.6712000Xunemp 30.33 17.53 1.73 0.088fullXblack 0.86 4.08 0.21 0.834fullXdrop 5.31 6.46 0.82 0.413fullXunemp -55.11 19.20 -2.87 0.005blackXdrop -0.56 0.33 -1.71 0.091blackXunemp 0.97 0.89 1.10 0.275dropXunemp 1.23 2.17 0.57 0.572
ANOVA
df SS MS F
Regression 25 81517 3261 1.24
Residual 74 194024 2622
Total 99 275541
If F-stat > F05 , we reject H0: no heteroscedasticity
25
Hence, 2 is probably constant
1.24
74 F.05 = 1.66
Using data to confirm theory(A2: Constant variance)
pmt_lnXblack -5.31 4.43 -1.20 0.234
If heteroscedasticity is a problem,
• Estimated coefficients aren’t biased
• Coefficient standard errors are wrong
• Hypothesis testing is unreliable
In our example, heteroscedasticity does not seem to be a problem.
If heteroscedasticity is a problem, do one of the following:
• Use Weighted Least Squares with 1/xj or 1/xj0.5 as weights where xj is the
variable causing the problem
• Compute “Huber-White standard errors”
1
1-statb
bt
s
Using data to confirm theory(A2: Constant variance)
Error is probably normally distributed if e is normally distributed
Histogram of residuals
0
5
10
15
20
25
30
1 2 3 4 5 6 7 8 9 10
residuals
freq
uen
cy
-20 -16 -12 -8 -4 0 4 8 12 16 20
(A3: Normality)Using data to confirm theory
There are a number of normality tests one can chose.
• The Jarque-Bera test involves using the skew and kurtosis of the residuals.
• The test statistic follows a chi-square distribution with 2 degrees of freedom:
kurtosis measures "peakedness" of the probability distribution. • High kurtosis → sharp peak, low kurtosis → flat peak.• involves raising standardized residuals to the 4th power • Excel: =kurt(A1:A100) → 0.0214
skewness measures asymmetry of the distribution. • 0 skew → symmetric distribution, negative skew → skewed left,
positive skew → skewed right• involves raising standardized residuals to the 3rd power• Excel: =skew(A1:A100) → 0.3276
2 22 2 2100-stat 1.791
6 4 6
.0214.3276
4
kurtw
nske
(A3: Normality)Using data to confirm theory
= .05 (column)
5.99
Do Not Reject H0 Reject H0: errors are normal
.05
2
There is no reason to doubt the assumption that the errors are normally distributed.
2 -stat
df = 2 (row) 2.05 5.99
1.791 2
(A3: Normality)Using data to confirm theory
If the errors are normally distributed,
• parameter estimates are normally distributed
• F and t significance tests are valid
If the errors are not normally distributed but the sample size is large,
• parameter estimates are approximately normally distributed (CLT) • F and t significance tests are valid
If the errors are not normally distributed and the sample size is small,
• parameter estimates are not normally distributed
• F and t significance tests are not reliable
(A3: Normality)Using data to confirm theory
The values of are probably independent if the autocorrelation residual plot or if the Durbin-Watson statistic (DW-stat) indicate the values of e are independent
The DW-stat varies when the data’s order is altered
• If there are multiple time periods, compute DW-stat after sorting by time periods
• If the data is cross-sectional, compute the DW-stat after sorting by geography (e.g., NE, NW, Central, SW, SE …)
• If the data is both, compute the DW-stat after sorting by time periods and geography
no autocorrelation if DW-stat = 2
perfect "" autocorrelation if DW-stat = 4
perfect "+" autocorrelation if DW-stat = 0
(A4: Independence)Using data to confirm theory
Observation region Residuals (ei - ei-1)2 ei2
2 0 -2.29 - 5.24
5 0 -0.56 3.00 0.31
11 0 -5.59 25.29 31.21
37 0 -4.42 1.36 19.55
47 0 -3.76 0.43 14.16
52 0 14.84 345.91 220.08
55 0 -4.80 385.57 23.05
61 0 11.91 279.27 141.86
87 0 3.51 70.62 12.30
98 5 1.79 161.93 3.19
99 5 2.85 1.12 8.11
sum 7620.63 3746.69
7620.63DW-stat
3746.69
DW-stat 2.03
There is no reason to doubt the assumption
that the errors are independent.
(A4: Independence)Using data to confirm theory
If autocorrelation (or serial correlation) is a problem,
• Estimated coefficients aren’t biased, but
• Their standard errors may be inflated
• Hypothesis testing is unreliable
In our example, autocorrelation does not seem to be a problem.
If autocorrelation is a problem, do one of the following:
• Change the functional form
• Include an omitted variable
• Use Generalized Least Squares
• Compute “Newey-West standard errors” for the estimated coefficients.
1
1-statb
bt
s
(A4: Independence)Using data to confirm theory
The true model is probably linear if the scatterplot of e versus y is a horizontal, random band of points
^
-15
-10
-5
0
5
10
15
20
30 40 50 60 70
predicted epr
resid
ual
2. There is no pattern in this scatter plot.1. The simple regression line’s slope = 0 and its height = 0
(A5: Linearity)Using data to confirm theory
If you fit a linear model to data which are nonlinearly related,
• Estimated coefficients are biased
• Predictions are likely to be seriously in error
In our example, nonlinearity does not seem to be a problem.
If the data are nonlinearly related, do one of the following:
• Rethink the functional form
• Transform one or more of the variables
All 5 model assumptions appear to be valid. Hence, the t and F tests are reliable
provided the “right” regressors are included.
1
1-statb
bt
s
(A5: Linearity)Using data to confirm theory
F
.05
Do not Reject H0 Reject H0
≈ 1
Hence, we reject H0.
There is insufficient evidence to conclude that the
coefficients are not all equal to zero simultaneously.
dfD = 93 and = .05 (row)
H0: 1 = 2 = . . . = 6 = 0
dfN = 6 (column)
16.6232.20
(Testing for Overall Significance)Using data to confirm theory
-1.986 1.986
.025
-2.3 0t
.025
Reject H0 at a 5% level of significance.
Do Not Reject RejectReject
1
1-statb
bt
s-2.32
-5.7092.461
df = 100 – 6 – 1 = 93 (column) = .05 /2 = .025 (row)
H0: 1 = 0
I.e., TANF welfare payments influence the decision to work.
(Testing for Coefficient Significance)Using data to confirm theory
-1.986 1.986
.025
-1.39 0t
.025
We cannot reject H0 at a 5% level of significance.
Do Not Reject RejectReject
2
2-statb
bt
s-1.39
-2.8212.029
df = 100 – 6 – 1 = 93 (column) = .05 (row) /2 = .025 (row)
H0: 2 = 0
I.e., welfare reform in general does not influence the decision to work.
(Testing for Coefficient Significance)Using data to confirm theory
-1.986 1.986
.025
1.960t
.025
Although we cannot reject H0 at a 5% level of significance,
we can at the 10% level (p-value = .054).
Do Not Reject RejectReject
3
3-statb
bt
s1.96
3.7681.927
df = 100 – 6 – 1 = 93 (column) = .05 (row) /2 = .025 (row)
H0: 3 = 0
I.e., full sanctions for failure to comply with work rules influence the decision to work.
(Testing for Coefficient Significance)Using data to confirm theory
-1.986 1.986
.025
-3.26 0t
.025
Reject H0 at a 5% level of significance.
Do Not Reject RejectReject
4
4-statb
bt
s-3.26
-0.2910.089
df = 100 – 6 – 1 = 93 (column) = .05 (row) /2 = .025 (row)
H0: 4 = 0
I.e., the share of the population that is black influences the decision to work.
(Testing for Coefficient Significance)Using data to confirm theory
-1.986 1.986
.025
-1.85 0t
.025
Do Not Reject RejectReject
5
5-statb
bt
s-1.85
-0.3740.202
df = 100 – 6 – 1 = 93 (column) = .05 (row) /2 = .025 (row)
H0: 5 = 0
Although we cannot reject H0 at a 5% level of significance,
we can at the 10% level (p-value = .068).
I.e., the share of the population that is high school droput influences the decision to work.
(Testing for Coefficient Significance)Using data to confirm theory
-1.986 1.986
.025
-4.89 0t
.025
Reject H0 at a 5% level of significance.
Do Not Reject RejectReject
6
6-statb
bt
s-4.89
-3.0230.618
df = 100 – 6 – 1 = 93 (column) = .05 (row) /2 = .025 (row)
H0: 6 = 0
I.e., the unemployment rate influences the decision to work.
(Testing for Coefficient Significance)Using data to confirm theory
• Since estimated coefficient b1 is statistically significant, we interpret its value as follows:
Increasing monthly benefit levels for a family of three by 10% would result in a .54 percentage point reduction in the average epr
of LISM
1 1$292 $266 1 1ˆ ˆ| | ln(292) ln(266) -5.709ln(1.10) .54x xepr epr b b .10 .54
• Since estimated coefficient b2 is statistically insignificant (at levels greater than 15%), we interpret its value as follows:
Welfare reform in general
had no effect on the epr of LISM.
(Interpreting Coefficients)Using data to confirm theory
• Since estimated coefficient b3 is statistically significant at the 10%
level, we interpret its value as follows:
33
yb
x
3.768 +3.768
+1
The epr of LISM is 3.768 percentage points higher in states that adopted full sanctions for families that fail to comply with work rules.
• Since estimated coefficient b4 is statistically significant at the 5%
level, we interpret its value as follows:
44
yb
x
-0.291 -0.291
+1
Each 10 percentage point increase in the share of the black population in states is associated with a 2.91 percentage point
decline in the epr of LISM.
10
10
-2.91
+10
(Interpreting Coefficients)Using data to confirm theory
• Since estimated coefficient b5 is statistically significant at the 5%
level, we interpret its value as follows:
55
yb
x
-0.374 -0.374
+1
Each 10 percentage point increase in the high school droput rate is associated with a 3.74 percentage point decline in the
epr of LISM.
10
10
-3.74
+10
• Since estimated coefficient b6 is statistically significant at the 5%
level, we interpret its value as follows:
66
yb
x
-3.023 -3.023
+1
Each 10 percentage point increase in the unemployment rate is associated with a 30.23 percentage point decline in the epr
of LISM.
10
10
-30.23
+10
(Interpreting Coefficients)Using data to confirm theory