Download Calculation of Pooled Standard Deviations:

Lab 2
You are asked to report the molar concentration of
the unknown acid, Cu, the standard deviation, su, and
the 95% confidence interval.
Assuming no uncertainty associated with CNaOH and
Vu, the overall precision is determined by the
precision associated with the VNaOH measurement.
Example: CNaOH = 0.1114 M; Vu = 25.00 mL
V NaOH C NaOH
Cu =
Vu
Unknown
V(NaOH, mL)
18.36
18.29
18.41
V NaOH
18.35
s Cu
0.00027
sCu 0.00016 (0.0002)
sC u =
∑ (C
i
− C )2
N −1
Cu(M)
0.08181
0.08150
0.08203
0.08178 (0.0818)
sCu =
sCu
N
For lab write-up, assume no error in Vu, CNaOH in
determining Cu.
Cu = 0.0818 ± 0.0002 M
%relative standard deviation, %RSD, or coefficient
of variation, CV:
RSD (%) =
(0.0002 ) (100) = 0.24%
s
(100) =
0.0818
C
Confidence interval:
Ctrue = C ± t × sC
t2,95 = 4.3
Ctrue = 0.0818M ± ( 4.3)(0.00016)
Ctrue = 0.0818M ± 0.0007
Random or indeterminant error (cont.):
Coin flipping is analogous to the example we
discussed about measurements with sources of
uncertainty with equally probably values of +U and –
U. That is, each “heads” is analogous to + or –U and
each “tails” is analogous to – or + U.
The equation for a Gaussian curve:
− ( x − μ )2
y=
e
2σ 2
σ 2π
Since we know the mathematical form in which
random errors are manifest, we can develop a
statistical treatment of random error.
Terminology:
Population: the collection of all measurements of
interest (often a set that is too large to
actually make all the measurements)
Sample:
the subset of measurements selected
from a population
(note the difference between a statistical “sample” and a chemical “sample”. n chemical
samples typically make up a statistical sample of the population of interest)
μ: the average or mean of the population
⎛ N ⎞
⎜ ∑ xi ⎟
μ = ⎝ i =1 ⎠ where N is the number of measurements
N
for the entire population
x : the sample mean (arithmetic average of limited
set of data)
⎛ N ⎞
⎜ ∑ xi ⎟
x = ⎝ i=1 ⎠
N
where N is the number of
measurements for N replicates
x is an estimate of μ. The accuracy of the estimate
improves as the number of measurements approaches
the population N. (i.e. as Nsample → Npopulation)
The precision of a population of data is reflected by
the population standard deviation, σ, given as:
N
σ=
2
(
)
x
−
μ
∑ i
i =1
N
The precision of a subset of data is reflected by the
sample standard deviation, s, given as:
∑ (x − x )
N
s=
i =1
2
i
N −1
s approximates σ, just as x approximates μ, as Nsample
→ Npopulation, i.e., when (N-1) →N
The population standard deviation:
N
σ=
2
(
)
x
−
μ
∑ i
i =1
N
The sample standard deviation:
∑ (x − x )
N
s=
i =1
2
i
N −1
Differences: μ vs. x ; N vs. (N-1)
N, N-1 are the respective number of degrees of
freedom for determination of σ, s respectively
# degrees of freedom is the number of independent
data that goes into the computation of st. dev.
e.g.
Assume we have determined three values of a
quantity, x; N=3
We compute an average, x
x1 + x2 + x3
x=
= 3.31
3
If x1 = 3.17 and x2 = 3.33, can we assign and
independent value to x3?
x3 = 3(3.31)-(3.17+3.33) = 3.43
Hence, we lose one degree of freedom when we
calculate the x in the equation for the sample
standard deviation. Degrees of freedom = 3-1 = 2
Overall, one degree of freedom lost for every
parameter determined from a data set:
x
y = mx + b
y = a + bx + cx2
average
linear equation
quadratic equation
N-1
N-2
N-3
Two reasons for taking the average of multiple
measurements:
1. The average of several measurements is more
likely to give a better estimate of the true mean
than any single measurement.
i.e., x is a better estimate of μ than any xi
2. There is a smaller standard deviation associated
with the mean of i measurements than for any
individual measurement, xi.
Standard error of the mean, sm:
The standard deviation, s, indicates the precision
associated with a single measurement. The precision
associated with the average of N measurements is
better by 1 N , that is,
sm = s
N
This is why we like to make replicate measurements
and use the average. The average is more precise
than any individual measurement.
Simulated Data for Lactose in Milk
Symbol
conc.
(mg/250 mL)
xi- x
(xi- x )2
x1
11.21
+0.02
0.0004
x2
11.45
+0.26
0.0676
x3
10.91
-0.28
0.0784
__________________________________________
N
∑
33.57
0.56*
0.1464
x
11.19
0.187
sx=0.27
i =1
*sum of unsigned values
∑ (x
N
sx =
sm =
i =1
i
−x
N −1
sx
N
=
)
2
=
0.27
3
0.1464
= 0.27
3 −1
= 0.16