1. Suppose that the wage distribution for Riverside is given below: a. Compute t
ID: 1194777 • Letter: 1
Question
1. Suppose that the wage distribution for Riverside is given below:
a. Compute the level of wage inequality for the town using the 90th / 10th percentile ratio and the 90th / 50th and 50th /10th ratios. (Yes, this is as easy as it looks)
b. Assume a minimum wage law is passed that doesn’t affect the market in the high-wage sector A but boosts wages to $7/hour in sector B, the covered sector, while reducing employment to 20 in that sector. Displaced workers in sector B move into the uncovered sector, C, where wages fall to $4.50 per hour as employment grows to 40. Has wage inequality risen or fallen? Does your answer change depending on the measure of wage inequality you use?
SECTOR NUMBER OF WORKERS WAGE A 40 $10 (per hour) B 30 $5 C 30 $5Explanation / Answer
1. Suppose that the wage distribution for Riverside is given below:
SECTOR
NUMBER OF WORKERS
WAGE
A
40
$10 (per hour)
B
30
$5
C
30
$5
a. Compute the level of wage inequality for the town using the 90th / 10th percentile ratio and the 90th / 50th and 50th /10th ratios. (Yes, this is as easy as it looks)
SECTOR
NUMBER OF WORKERS
WAGE
Total Wage
90th Percentile
50th Percentile
10th Percentile
90th to 10th percentile
90th to 50th percentile
50th to 10th percentile
A
40
$10 (per hour)
400
350.00
150
150
2.33
2.33
1
B
30
$5
150
C
30
$5
150
b. Assume a minimum wage law is passed that doesn’t affect the market in the high-wage sector A but boosts wages to $7/hour in sector B, the covered sector, while reducing employment to 20 in that sector. Displaced workers in sector B move into the uncovered sector, C, where wages fall to $4.50 per hour as employment grows to 40. Has wage inequality risen or fallen? Does your answer change depending on the measure of wage inequality you use?
Sector
Number of workers (1)
Wage(1)
Number of workers(2)
Wage(2)
A
40
$10 (per hour)
40
$10 (per hour)
B
30
$5
20
$7
C
30
$5
40
$4.5
Case I
Mean in the first case
.40*10+.30*5+.30*5
7
Standard deviation = Sum of (X-Xbar)2 /N-1
= {40*(10-7)2+30*(5-7)2+30*(5-7)2}/99
= {40*(3)2+30*(-2)2+30*(-2)2}/99
= {40*9+30*4+30*4}/99
= 600/99
=6.06
So, in the first case, the mean is $7./hour and the standard deviation is $6.06. The range is $5 (10-5).
Case II
Mean in the second case = .40*10+ .20*7+.40*4.5
= 7.2
Standard deviation = Sum of (X-Xbar)2 /N-1
= {40*(10-7.2)2+20*(7-7.2)2+40*(4.5-7.2)2}/99
= {40*(2.8)2+20*(-.2)2+40*(-2.7)2}/99
= {40*7.8+20*0.04+40*7.29}/99
= 604.4/99
=6.10
In the second case, the mean is $7.2/hour and the standard deviation is $6.10. The range is $5.5 (10-4.5).
Notice that the spread (range) increased with the rise in the minimum wage, and the standard deviation fell. The difference between highest and lowest wage increased, but more workers earned wages clustered around the mean. That produced a rise in the standard deviation. I would conclude that the minimum wage reduced wage inequality, based on the notion that more workers earned pay similar to the average.
SECTOR
NUMBER OF WORKERS
WAGE
A
40
$10 (per hour)
B
30
$5
C
30
$5
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