Re: Another Error in Afonso's Confidence Interval Code

Let's try again to get Afonso to explain his Basic
code for simulating
confidence intervals.
The complete code copied directly from an earlier
post is posted down
below the line of &&&&&&&&&.

The error I have cited (which he refuses to address)
is shown here. I
have added <--- with my comments. This is copied
directly from
Afonso's code (see the entire code posted further
down.)

FOR i = 1 TO n
a = SQR(-2 * LOG(RND))
x(i) = a * COS(2 * pi * RND) <---- 1
md = md + x(i) / n <---- 2
soma2 = soma2 + x(i) * x(i) <---- 3
NEXT i
sqd = soma2 - n * (nmd) ^ 2 <---- 4
: sd = SQR(sqd / n) <---- 5
FOR ii = 1 TO n
x(ii) = (x(ii) - md) / sd <---- 6
NEXT ii

1. x(i) is random normal numbers; simulated data. OK
2. md is the average of n random normal numbers. OK
3. soma2 is the raw sum of squares of those numbers
(square, then sum) OK
4. This seems to be an attempt to go from the raw sum
of squares to the usual corrected sum of squares.
However, since nmd was never defined anywhere above
then nmd is always "0". Thus n * (nmd)^2 is "zero"
for each and every sample of n data. This is the
error
I have tried repeatedly to get Afonso to address, but
he has not.
5. sd = SQR(sqd/n) is an attempt to calculate the
usual sample standard deviation. But since sqd is the
raw
sum of squares, this is not a meaningful calculation.
In addition, there is a second error here. (I hoped
Afonso
would find this and correct it, but he has not.) He
should
divide the corrected sum of squares by n - 1, not by
n, to
get an unbiased estimate of sigma.

So now we have two errors for Afonso to address.
Based
on the cockroach theory (when we find one cockroach,
that means there are more) I'll bet there are more
errors
beyond these two.

6. Here he is "standardizing" the data by dividing
the individual data by subtracting the average and
dividing
by sd. But sd is not the usual unbiased estimate of
sigma.

The failure to reduce the raw sum of squares to the
corrected
sum of squares is a major coding error. (Or is it a
conceptual
error? Afonso will tell us soon.) So sd is simply
incorrect
throughout. Hence the standardardized values of x(i)
are
also incorrect.

Dividing by n instead of n - 1 to obtain an unbiased
estimate
of sigma is another error. This one seems to be
conceptual,
but perhaps we'll learn more later.

Here's yet another problem...

Attempts to get this software code to produce the
numerical
results that Afonso cited earlier have failed.

This is an opportunity for Afonso to explain his
software code
in detail... step-by-step... so we can all follow
what he is
claiming about simulating confidence intervals. This
should
not be difficult since he wrote the code himself.

The complete code... copied precisely from an earlier
Afonso post...
is shown below. OMU

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Afonso wrote....

Check (below) what was the algorithm I used.
What is extraordinary is your claim that you
performed similar job years ago. Changing
your mind since that date? If the values are
worthless
why did you performed the calculations? This is no
sense, AT ALL! (I am not surprised , even a bit).

______licas (Luis A. Afonso)

REM "LiLLI"
CLS
PRINT " LILLI(EFORS) "
INPUT " n = "; n
INPUT " all = "; all
pi = 4 * ATN(1): c = 1 / SQR(2 * pi)
DIM x(n), xx(n), F(n)
DIM w(9001)
DEF fng (z, j) = -.5 * z ^ 2 * (2 * j + 1) /
((j + 1) * (2 * j + 3))
F(0) = 0
FOR ji = 1 TO n: F(ji) = ji / n: NEXT ji
FOR k = 1 TO all: RANDOMIZE TIMER
LOCATE 5, 50:
PRINT USING "##########"; all - k
mmaior = -1: md = 0: soma2 = 0
FOR i = 1 TO n
a = SQR(-2 * LOG(RND))
x(i) = a * COS(2 * pi * RND)
md = md + x(i) / n
soma2 = soma2 + x(i) * x(i)
NEXT i
sqd = soma2 - n * (nmd) ^ 2
: sd = SQR(sqd / n)
FOR ii = 1 TO n
x(ii) = (x(ii) - md) / sd
NEXT ii
FOR ii = 1 TO n: u = x(ii): w = 1
FOR jj = 1 TO n
IF x(jj) < u THEN w = w + 1
NEXT jj: xx(w) = u
NEXT ii
FOR tt = 1 TO n: z = xx(tt)
REM calcula FI(z)
IF z > 0 THEN kw = 0
IF z < 0 THEN kw = 1
zu = ABS(z): s = c * zu: antes = c * zu
FOR j = 0 TO 1000
xx = antes * fng(zu, j)
s = s + xx
antes = xx
IF ABS(xx) < .00005 THEN GOTO 20
NEXT j
20 IF kw = 0 THEN ff = .5 + s
IF kw = 1 THEN ff = .5 - s
b = ABS(ff - F(tt - 1))
bb = ABS(F(tt) - ff)
maior = b
IF bb > b THEN maior = bb
IF maior > mmaior THEN mmaior = maior
NEXT tt
mm = INT(1000 * mmaior + .5)
IF mm > 9000 THEN mm = 9000
w(mm) = w(mm) + 1
fff = INT(k / 50000): ff = k / 50000
IF ff <> fff THEN GOTO 1000
cc(1) = .95 * k: cc(2) = .99 * k
FOR iji = 1 TO 2
ciji = cc(iji): s = 0
FOR iij = 0 TO 9000
s = s + w(iij)
IF s > ciji THEN GOTO 100
NEXT iij
100 PRINT USING "##.### #.#### ";
iij / 1000; s / k
NEXT iji
1000 NEXT k
END


OMU, what Afonso is doing is generating a large number of samples of n Xi from N(0,1). He then tries to replicate what a user of Lilliefors would do. He calculates the sample mean and standard deviation, fits a normal cdf, and then finds the distance between the empirical cdf and the fitted normal cdf as the maximum absolute difference.

Now, you've found an error in Afonso's code in calculating the sample standard deviation. Apparently he meant md instead of nmd. Along with dividing by n instead of n-1, the effect is that his formula for the variance becomes Sumsq(Xi)/n rather than Sumsq(Xi-xbar)/(n-1).

His actual Sumsq(Xi)/n is an unbiased estimate of the population sigma^2, as is also Sumsq(Xi-xbar)/(n-1). The difference is that he has n degrees of freedom instead of n-1 degrees of freedom for the estimated standard deviation. The user would not know the population mean in advance and thus has only n-1 degrees of freedom in the estimate.

This would have a very small effect in his tables. His tables should give slightly smaller erroneous critical values for the Lilliefors statistic, maybe in the third or fourth decimal. However, since he is comparing his simulation results with those of 40 years ago, we are looking at those small differences.

Jack
.