急问一道求sample variance的SAS程序题-经管之家官网！

分别求下面的x1和x2的sample variance S11 AND S22. 用SAS程序COMPUTE these two sample variance.
X1: 1 2 3 3 4 5 6 8 9 11
X2： 18.9519.0017.95 15.54 14 12.95 8.94 7.49 6 3.99
下面是我写的程序，而且手算出来的s11与s22与sas程序相同，均为9.56和27.77，但答案显示是3.09和5.27，刚好是前者的平方根，求指教！
以下是我的程序：
proc iml;
X={1 18.95,2 19,3 17.95,3 15.54,4 14,5 12.95,6 8.94,8 7.49,9 6,11 3.99};
n=10;
print X;
meanx1=mean(X[*,1]);
meanx2=mean(X[*,2]);
varx1=((n‐1)/n)*var(X[*,1]);
varx2=((n‐1)/n)*var(X[*,2]);
jwvar1 = ((X[1,1]‐meanx1)**2 + (X[2,1]‐meanx1)**2 + (X[3,1]‐
meanx1)**2 + (X[4,1]‐meanx1)**2)+((X[5,1]‐meanx1)**2 + (X[6,1]‐meanx1)**2 + (X[7,1]‐
meanx1)**2 + (X[8,1]‐meanx1)**2)+(X[9,1]‐meanx1)**2 + (X[10,1]‐meanx1)**2)/10;
jwvar2 = ((X[1,2]‐meanx2)**2 + (X[2,2]‐meanx2)**2 + (X[3,2]‐
meanx2)**2 + (x[4,2]‐meanx2)**2)+(X[5,2]‐meanx2)**2 + (X[6,2]‐meanx2)**2 + (X[7,2]‐
meanx2)**2 + (x[8,2]‐meanx2)**2)+((X[9,2]‐meanx2)**2 + (X[10,2]‐meanx2)**2)/10;
S = cov(X);
Sn = ((n‐1)/n)*S;
R = corr(X);
x1 = X[*,1];
x2 = X[*,2];
print x1,x2;
print meanx1, meanx2;
print varx1, varx2, jwvar1, jwvar2;
print jwvar1, jwvar2;
print Sn, R;
quit;
run;