Cronbach’s Alpha
& Maximized 4
SAS
proc corr nosimple nocorr nomiss alpha;
var q1-q10; run;
Cronbach Coefficient Alpha
Variables
Alpha
Raw
0.743825
Cronbach Coefficient Alpha with Deleted Variable
Deleted
Variable
Raw Variables
Correlation with Total
Alpha
q1
0.443795
0.718348
q2
0.441119
0.717466
q3
0.604168
0.690420
q4
0.531520
0.704966
q5
0.622641
0.694956
q6
0.336597
0.732539
q7
0.246974
0.748538
q8
0.308285
0.736302
q9
0.405834
0.724557
q10
0.200549
0.755307
The Idealism Scale
• 7. Deciding whether or not to perform an
act by balancing the positive
consequences of the act against the
negative consequences of the act is
immoral.
• 10. Moral actions are those which closely
match ideals of the most "perfect" action.
Alpha from Item Variances
If nmiss (of Q1-Q10) > 0 then delete;
TOT=Q1+Q2+Q3+Q4+Q5+Q6+Q7+Q8+Q9
+Q10;
proc means var; var q1-q10 tot;
OUTPUT OUT=VarOut VAR=VQ1-VQ10
VTot;
n(1   /  )

n 1
2
I
2
T
Variable Variance
q1
0.7279195
q2
0.9817002
q3
1.0462640
q4
0.8286801
q5
0.6897539
q6
0.5516331
q7
1.1234452
q8
0.7986130
q9
1.3523490
q10
1.1051902
TOT
27.8485906
title 'Computing alpha from item variances.';
DATA _NULL_; FILE PRINT; SET VAROUT;
SUMVAR = SUM(OF VQ1-VQ10);
Alpha = (10/9)*(1-SUMVAR/VTOT);
put;
put Alpha = ;
run;
Alpha=0.7438254943
Working with Only First 4 Items
• Create all three possible split halves
A1=Q1+Q2; B1=Q3+Q4; A2=Q1+Q3;
B2=Q2+Q4; A3=Q1+Q4; B3=Q2+Q3;
• Obtain alpha the easy way
proc corr nosimple nocorr nomiss alpha;
var Q1-Q4;
• Alpha = 0.716882
Find the 3 Split-Half Corrs
proc corr nosimple nomiss; var A1; with B1;
proc corr nosimple nomiss; var A2; with B2;
proc corr nosimple nomiss; var A3; with B3;
r1 = 0.49581
r2 = 0.66198
r3 = 0.53911
Obtain 4 for Each Split Half
• Apply Spearman-Brown correction
2r
4 
1 r
• The three values are .6629, .7966, and
.7005.
• The mean of these three is .72,
Cronbach’s alpha.
• The highest of these three is maximized
4
• Osburn (2000) showed that Cronbach
alpha is a conservative estimate of
reliability.
• Maximized 4 is a better estimate.
• But for 2n items, there are .5(2n)!
2
(
n
!
)
pairs of split halves
• Ouch. For 10 items, 126 split halves.
• There are ways to estimate, if you have
FORTRAN.