Ec 395 Assignment 4
Q1
Mean
minscore
-7.5
person_years
2792.435
avgearnings
13004.6
receivehsd
.5404829
n
816.7609
Note: person_y = #ppl * #years they in study
Q2
Standard deviation
13.42262
2295.43
2172.684
.2905733
670.6678
Q3
Yes, passing the last chance exam substantially increase the probability of getting a diploma
Q4
Treatment effect for 1 = 0. 5428506
Treatment effect for 2 = 0. 4016485
Treatment effect for 3 = 0. 4317979
Q5
Linier function
̂ 𝑖 = .3779684 + .5428506𝑝𝑎𝑠𝑠𝑖 + .0028436𝑚𝑖𝑛𝑠𝑐𝑜𝑟𝑒 𝑖
𝑟𝑒𝑐𝑒𝑖𝑣𝑒ℎ𝑠𝑑
− .0019075 𝑝𝑎𝑠𝑠_𝑚𝑖𝑛𝑠 + 𝑢𝑖
Square function
̂
𝑟𝑒𝑐𝑒𝑖𝑣𝑒ℎ𝑠𝑑
𝑖 = .5029239 + .4016485 𝑝𝑎𝑠𝑠𝑖 + .0262728𝑚𝑖𝑛𝑠𝑐𝑜𝑟𝑒 𝑖
+ .0007558𝑚𝑖𝑛𝑠𝑐𝑜𝑟𝑒2 − .0183738 𝑝𝑎𝑠𝑠_𝑚𝑖𝑛𝑠𝑖 − .00122 𝑝𝑎𝑠𝑠_𝑚𝑖𝑛𝑠2 + 𝑢𝑖
Cubic function
̂
𝑟𝑒𝑐𝑒𝑖𝑣𝑒ℎ𝑠𝑑
𝑖 = .4893236 + .4317979 𝑝𝑎𝑠𝑠𝑖 + .0214028𝑚𝑖𝑛𝑠𝑐𝑜𝑟𝑒 𝑖
+ .0003694𝑚𝑖𝑛𝑠𝑐𝑜𝑟𝑒2 − 8.31𝑒 − 06𝑚𝑖𝑛𝑠𝑐𝑜𝑟𝑒3 − .0293497 𝑝𝑎𝑠𝑠_𝑚𝑖𝑛𝑠𝑖
+ .0018943 𝑝𝑎𝑠𝑠_𝑚𝑖𝑛𝑠2 − .0001129 𝑝𝑎𝑠𝑠_𝑚𝑖𝑛𝑠3 + 𝑢𝑖
H0: 𝑚𝑖𝑛𝑠𝑐𝑜𝑟𝑒3 = 𝑝𝑎𝑠𝑠_𝑚𝑖𝑛𝑠3=0
H1: they don’t equal 0
In conclusion, we need the cubic terms because the p value of .18 56is greater than the level of
significance 5% so we do not reject the null
Q6
Q7
H0: 𝑝𝑎𝑠𝑠𝑖 = 0
H1: 𝑝𝑎𝑠𝑠𝑖 ≠0
So we reject the null because it is not significant