# How to do Chow test for constancy of parameters across 2 groups

I am a beginner in statistics and poor in mathematics. I am trying to to assess effect of intervention in one state versus another using annual data.

My data are
State 1         State 2
Cases   Deaths  Cases   Deaths
2004    1125    5   2024    254
2005    1213    5   1978    209
2006    1003    4   2294    217
2007    1425    6   2312    249
2008    1172    4   1528    197
2009    1092    3   1683    204
2010    1316    4   2024    218


When I was stuck for the correct statistical procedure, one senior member "irishstat" advised me the following very convincing analysis for which I am ever grateful: X1 is the number of cases and Y is the number of deaths. X2 is the empirically identified point of anomaly; (2009 .. period 6 for State1 and 2004 .. period 1 for State2. Outlier detection led to identifying one anomalous data point for each state reflecting some unknown background variable thus yielding a more robust estimate of the mortality rates.

Analysis of State1

State1 Y(T) = -.65649
+[X1(T)][.0046)] CASES +[X2(T)][-1.3608)] PULSE6 I~P00006STATE1 + [A(T)]
Suggesting an unusually low mortality rate for 2009
Analysis of State2
State2 Y(T) = 123.55
+[X1(T)][(+ .0468)] CASES +[X2(T)][(+ 35.7590)] PULSE1 I~P00001STATE2 + [A(T)]
Suggesting an unusually high mortality rate for 2004


This leads to estimating two cleansed data points

STATE    YEAR   Y OBSERVED   Y ADJUSTED
STATE1   2009   3            4.36
STATE2   2004   254          218.24


Replacing these two observed possibly errant values possibly due to some unspecified concomitant factor (“lurking Variable”) one computes a rate of.0046 for STATE1 and .0468 for STATE2. My problem now is how to do The Chow Test for constancy of parameters across groups to check for rejection of the null hypothesis of equal coefficients. I have SPSS v19. Kindly advise me step by step.

## 1 Answer

Using the suggested corrected data we have:

The model to be tested is : Y(T)=B0 + B1*X(T) + A(T) The null hypothesis is that the set B0 and B1 are the same over the two states

step 1 : Estimate this for STATE1 obtaining an error sum of squares SOS1 =.789

step 2 : Estimate this for STATE2 obtaining an error sum of squares SOS2 = 548.272

step 3 : Estimate this for all of the data (12 pairs) obtaining an error sum of squares SOS3 = 23920.4

step 4 : Compute NUM= [SOS3-(SOS2+SOS3)]/2 = 11685

step 5 : Compute MSE for composite analysis =23920.4/10 = 2392

step 5 : F value = NUM / MSE = 11685/2392 = 4.9

STEP 6 : An F OF 4.9 with 2 and 10 degrees of freedom is .03 Thus the hypothesis of equality is rejected at alpha < .03 ; accepted otherwise

• Thank you very much for the clarification. Now I am able to recover lost data from a crashed hard disk. I have posted complete data in a new question as I felt the statistical approach and calculations are likely to change but increased sample size will increase the power of the study. The revised question is at stats.stackexchange.com/questions/8358/… This new question has additional data from 1994 to 2003. I apologise for the inconvenience. – DrWho Mar 17 '11 at 4:05
• AFS university book at autobox.com/AFSUniversity/afsuFrameset.htm is excellent for a beginner like me. Entertainment section is a great collection. Can it be downloaded? I wish it can be so that I can read it even when offline. – DrWho Mar 17 '11 at 4:43
• @DrWho: Simply right click on autobox.com/AFSUniversity/afsuFrameset.htm and then select "open link in new window" and then left click on "Entertainment" which will bring up the text. Higlight the text and copy to the clipboard and then paste to a word doc and you are good to go. I will look at your revised data and try to answer your "new question" – IrishStat Mar 17 '11 at 6:06
• Thank you very much. It is appreciable that you are working late night, – DrWho Mar 17 '11 at 8:12