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yields
Fcalc =
1.0762
Fcrit =
1.6741
Since the F value calculated from the data is smaller than the critical F
value, we cannot reject the null hypothesis without another cause. We
conclude therefore that the variances are identical at a 5% signii cance level.
Alternatively, we can apply the function vartest2(x,y,alpha) to the two
independent samples corg1 and corg2 at an alpha=0.05 or a 5% signii cance
level. MATLAB also provides a one-sample variance test vartest(x,variance)
analogous to the one-sample t -test discussed in the previous section. h e
one-sample variance test, however, virtually performs a ˇ 2 -test of the
hypothesis that the data in the vector x come from a normal distribution
with a variance dei ned by variance . h e ˇ 2 -test is introduced in the next
section. h e command
[h,p,ci,stats] = vartest2(corg1,corg2,0.05)
yields
h =
0
p =
0.7787
ci =
0.6429
1.8018
stats =
fstat: 1.0762
df1: 59
df2: 59
h e result h=0 means that we cannot reject the null hypothesis without
another cause at a 5% signii cance level. h e p -value of 0.7787 or ~78%
(which is much greater than the signii cance level) means that the chances
of observing more extreme values of F than the value in this example from
similar experiments would be 7,787 in 10,000. A 95% coni dence interval
is [-0.6429,1.8018], which includes the theoretical (and hypothesized) ratio
var(corg1)/var(corg2) of 1.2550 2 /1.2097 2 =1.0762.
We now apply this test to two distributions with very dif erent standard
deviations, 1.8799 and 1.2939.
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