Rayleigh nitrogen statistics
Solutions
All means and standard deviations have units of grams.
1) The mean is:
mavg = (2.2978 + 2.2980 + 2.2982 + 2.2994)/4 = 2.2985 .
The standard deviation is:
=
(m1  mavg) 2  (m2  mavg) 2  (m3  mavg )2  (m4  mavg) 2
n 1
where mavg is the mean and n is the number of measurements. So
,
(2.2978  2.2985)2  (2.2980  2.2985) 2  (2.2982  2.2985) 2  (2.2994  2.2985)2
,
4 1
= 0.0007 .
So the last digit in the mean is highly uncertain.
=
2a) For atmospheric nitrogen:
mavg = (2.3103 + 2.3100 + 2.3102)/3 = 2.3102 .
The standard deviation is:
=
(m1  mavg) 2  (m2  mavg) 2  (m3  mavg )2
n 1
,
(2.3103  2.3102) 2  (2.3100  2.3102)2  (2.3102  2.3102)2
So
=
,
3 1
= 0.0002 .
For chemical nitrogen:
mavg = (2.3001 + 2.2990 + 2.2987 + 2.2985 + 2.2987)/5 = 2.2990 .
The standard deviation is:
=
(m1  mavg) 2  ... (m5  mavg )2
n 1
,
(2.3001 2.2990)2  ... (2.2987  2.2990) 2
= 0.0006 .
5 1
measured value - true value
2.3102 2.2990
b)
% error =
100% 
100% ,
true value
2.2990
% error = 0.487% .
c) The error is indeed a small one. Most of us would be quite pleased to have our
determinations turn out to be within 1/2% of the actual value in most of our experiments! But
that fact is not relevant here. What is relevant is the precision and reproducibility of the data in
this case.
The standard deviations computed in (2a) above give some indication of experimental
error. Standard assumptions (in particular, the assumption that experimental errors are normally
distributed around the true value and that the mean is our best estimate of the true value) would
tell us that if one made many measurements by the same methods, about two-thirds of the
So
=
measurements would be within ± of the mean, and that about 95% of them would be within
±2 of the mean. Thus, measurements of chemical nitrogen ought to fall in the range 2.29772.3003 95% of the time and measurements of atmospheric nitrogen in the range 2.3099-2.3105 .
These ranges do not overlap; in fact, the distance between the ranges is much greater than the
size of either range. One more way of looking at the data takes note that the difference between
means (0.0112) is more than ten times even the larger standard deviation (0.0006). Any way you
look at it, the difference between chemical nitrogen and atmospheric nitrogen may be small
(compared to the true value), but it is much larger than experimental error (which is roughly
represented by ). On the basis of the data, then, the conclusion that there is a real physical
difference between chemical nitrogen and atmospheric nitrogen is much more likely than the
conclusion that the measurements are different because of experimental error.
3) The distribution function is:
-(m - m )2 
av g
exp 

2
2s


f(m) =
.
s 2
The plot below shows both distributions. (To download a spreadsheet containing the calculation
of the distribution functions, go to
http://webserver.lemoyne.edu/faculty/giunta/classicalcs/rayleighN2.xls) The variable m in our
case corresponds to mass measurements. It is evident that the distributions are well separated.
(If this were a chromatogram, we would say that the two peaks are well resolved.) It is therefore
not likely that the two peaks represent measurements of the same thing.
distributions
relative probability
3000
2000
chemical
atmospheric
1000
0
2.295
2.300
2.305
mass (g)
2.310
2.315