variance of the difference of two independent variables is the sum of variances It is said that variance of the difference of two independent variables is the sum of variances (topic - Confidence intervals. Two means. Independent samples)
I tried to experiment with small dataset of two variables. i found that sum of variance of 2 independent variable is not equal to variance of difference. the experiment is below:-
new york apple price [$3.80, $3.76, $3.87, $3.99, $4.02, $4.25,$4.13, $3.98] its variance = 0.027
LA apple price - [$3.02, $3.22, $3.24, $3.02, $3.06, $3.15, $3.81, $3.44], variance = 0.071
Their difference in prices = [0.78, 0.54, 0.63, 0.97, 0.96, 1.10, 0.32, 0.54] - variance = 0.0715
sum of LA apple price variance and NY apple price variance = 0.098 which not equal to variance of difference = 0.0715
can someone explain me why is it so please?

The relevant equation is $V(X-Y) = V(X)+V(Y)-2Cov(X,Y).$
The correlation may seem small to you, but @whuber's Comment
is exactly correct.
Computations in R for your apple prices:
ny = c(3.80, 3.76, 3.87, 3.99, 4.02, 4.25, 4.13, 3.98)
la = c(3.02, 3.22, 3.24, 3.02, 3.06, 3.15, 3.81, 3.44)
var(ny)
[1] 0.02711429
var(la)
[1] 0.07177143

cor(ny,la)
[1] 0.310884
cov(ny,la)/(sd(ny)*sd(la))
[1] 0.310884

plot(ny, la, pch=10)


Finally:
var(ny-la)
[1] 0.07145714
var(ny)+var(la)-2*cov(ny,la)
[1] 0.07145714

Note: One couldn't possibly have just $V(X + Y) = V(X) + V(Y),$ without restriction,
because, if $X \equiv Y,$ then we'd have $V(X + Y) = V(2X) = 4V(x) \ne V(X) + V(X) = 2V(X).$
