re sf and rounding

[John Whakamoe]

HI All

Here are a few ideas from me and below a few ideas from a colleague of  
mine, Derek Christie. I hope they help

There is no doubt that by taking a “larger” sample you do decrease the  
spread or variance of your estimate of the value being measured. I  
would note however that you are not taking a sample of 10 by taking  
the value of 10 each individual sheets, but a total of 10 sheets  
finding the variance of this set of ten and then dividing by the  
sample size 10. It still gives the same results
        s

SD = -------

     Sqrt(n)



s = s.d. of sheets measured on their own

SD = s.d. of sheets deduced from the sample measurement

n  = # of sheets in the sample: 10


ie you have reduced the  sd of  “the total width of ten sheets  
divided10” by a factor of sqr(10) or about 3 ie the sd of our estimate  
for the value has been reduced by a factor of 3

The estimate of the value is now 0.2mm and if the sd of each  
individual sheet was (“sort of” see below**) +_ 0.01mm then this has  
now been reduced to 0.2 +_ 1/3x0.1mm or .03mm. Thus our estimated  
value is 0.20mm with a sd of +-0.03mm. However should we increase the  
number of sfs?

**In fact this is a major problem. Somehow there has to be an estimate  
of the sd of each sheet and to measure this would require the use of  
an instrument. The reason why we used this method in the first place  
(take 10, find average, divide by 10) is because we cannot use our  
instrument to measure this. Ie the problem is the instrument more than  
the sd

What happens if we go to the extreme? If we keep on increasing the  
sample size then the ”accuracy” (sd)  will improve so 0.2 could become  
0.20000mm based on the argument given (increasing the sample size).

For instance: if we went to the ultimate extreme and did a census, in  
our case “an infinite sample size”, then we know the value exactly (no  
variation). Based on the argument given, the value would be  
0.20000000000…..mm Obviously not! The limit will be set by the limit  
of the measuring instrument, which brings us back to what is  
acceptable as the number of sf.

Another way of saying this is to say the Normal Curve is getting  
“thinner and thinner” as we increase the sample size and is coming  
closer to being a “delta function” shape , or a vertical line. Before  
that is reached the width (sd) of the Normal Curve will be the same as  
the SD of the measuring instrument. After this, reducing the SD  
(increasing sample size) will not be the important factor in  
determining the number of sf but the measuring instrument will be.

So the answer seems to be (at level 3 physics) not to get too wound up  
about it and accept both as you cannot tell with out more information  
(data)

John Whakamoe


 From Derek Christie

The question probably isn't well enough defined to have a single  
answer so your final comments are very
likely right, but my sympathies are would lie with the teachers who  
say 0.2
at this level.  I think the standard error versions are irrelevant here
because the variance in paper thickness will be very small compared  
with the
apparent instrument accuracy.  The standard error gives the accuracy  
of the
estimate of the average thickness of all sheets given a sample of true
values.  We are concerned about "the" thickness of a sheet assuming  
they are
all the same.  So we are really talking about the how the "divide by 10"
part of our calculation affects the accuracy.

What does "10 sheets = 2 mm" actually mean in this example?  Unless  
you ask
the kid "2.what?",  it probably means that s/he is measuring to the  
nearest
mm.  Then 10 sheets are between 1.5 and 2.5 mm (ignoring the zeroing  
issue)
so 1 sheet is between 0.15 and 0.25 so we say 0.2 mm.  (I always  
taught 1/5
of a division, so perhaps this ruler has 1 cm divisions.)

On the other hand, if they measure 100 sheets = 20 mm to 1 mm  
accuracy, then
the answer would be 0.20 but then I suppose that we are dealing with 2sf
even if it isn't so obvious.

I would give the s/root(n) stuff a miss, myself, ingenious as it is.

My view - the exemplar has it wrong.  They should give a clearer, better
thought out example.

Cheers

Derek
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