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Error Analysis LectureCHM 335

Error Analysis LectureCHM 335 – p. 1/39

Reading Assignment

Text, 4-7, 18-23, 27-42 (Pages 29-32, 43-48and 52-66)

Syllabus, Pages 22-32


ExampleWhat is incomplete in the statement: “The distance between

Kingston and New York City is 150 miles?”

150 miles ± 1 mile

150 miles ± 100 miles

All future reports must provide results with errors. The error

is as important as the measured result.


Types of ErrorsSystematic errors

1. Definite algebraic sign

2. Accuracy

Random errors

1. No definite algebraic sign

2. Precision

Can control the precision by increasing the number of

independent measurements.


The Central LimitTheorem

The results of independent measurements fall ona Gaussian distribution (a bell curve)

p(x) =1√2πσ2

e−(x−µ)2/2σ2

x = what we are measuringµ = the exact valueσ = standard deviation, a measure of the width

p(σ + µ)

p(µ)=

1√e


GaussianDistribution

-2 0 2 40

0.1

0.2

0.3

0.4

µµ−σ µ+σ


Properties∫

∞

−∞

p(x) dx = 1

∫ µ+σ

µ−σp(x) dx ≈ 0.67

∫ µ+2σ

µ−2σp(x) dx ≈ 0.95

∫

∞

−∞

xp(x) dx = µ


Properties (cont.)∫

∞

−∞

x2p(x) dx− µ2 = σ2

Interpretation - There is a 95% chance that a particular

measurement of x will fall in the range µ± 2σ.


Finite Set of NMeasurements

Assume

µ = 〈x〉 = 1

N

N∑

i=1

xi

〈f〉 = 1

N

N∑

i=1

f(xi)

σ2 = 〈x2〉 − 〈x〉2

=1

N

N∑

i=1

x2i −(

1

N

N∑

i=1

xi

)2


Standard Deviation ofthe Mean

The deviation of the ith measurement from the exact result

is given by

εi = xi − µ

Then εi can be shown to be normally distributed with

standard deviation

σN =σ√

N − 1.

σN is called the standard deviation of the mean.


InterpretationThere is a ∼67% chance the correct result lies in the range

〈x〉 ± σN .

There is a ∼95% chance the correct result lies in the range

〈x〉 ± 2σN .

We report 〈x〉 ± 2σN .


ExampleMeasure the length of a book 5 times obtaining 9.30 in, 9.20

in, 9.30 in, 9.40 in and 9.30 in.

〈x〉 = 1

5(9.3 + 9.2 + 9.3 + 9.4 + 9.3) in = 9.3 in

〈x2〉 = 1

5(9.32 + 9.22 + 9.32 + 9.42 + 9.32) in2 = 86.494 in2

σ = (86.494− 9.32)1/2 in = 0.06 in


Report?

σN =σ√

N − 1=

0.06 in√5− 1

= 0.03 in

〈x〉 ± 2σN = (9.30± 0.06) in

In today’s experiment you will measure the flow times of

water and a polyvinyl alcohol solution using a viscosimeter.

For the flow times you make 10 measurements, and report

the flow times as 〈t〉 ± 2σN .


Error EstimationIn real laboratory research we always calculate statistical

errors by measuring the standard deviation of the mean. In

CHM 335 we do not have time, so we estimate random

errors using simple rules.

Rule of thumb: We assume an error of 2σN = ±2 in the last

digit of a measuring device.

Example: If we use a balance and determine a mass of

2.4354 grams, we assume the mass is 2.4354± 0.0002

grams.

Exceptions: Pipettes and volumetric flasks have estimated

statistical errors etched on the glassware. Alternatively, for

pipettes, see page 26 of the syllabus for a table.Error Analysis LectureCHM 335 – p. 14/39

Roundoff

> 5 increase

< 5 no change

= 5 increase if the result is even


Roundoff Examples24.43

24.43 = 24.4

24.46

24.46 = 24.5

24.45

24.45=24.4

24.55

24.55=24.6


Error Propagation

Suppose you measure variables x, y, z, . . . withassociated errors ǫ(x), ǫ(y), ǫ(z), . . .. Supposef = f(x, y, z, . . .). What is the error in f?


The DifferentialLet f(x) be a function of x. The differential of f is defined by

df =df

dxdx

The differential of f gives the infinitesimal variation of f

given infinitesimal variations dx.

Example: The radius of the earth at the equator is 6.356

×106 meters. Imagine a wire strung around the equator that

is exactly 2π × 6.356× 106 meters in length. The wire is then

raised on polls around the earth such that the radius of the

wire ring is increased by exactly 1 meter. What is the

increase in the circumference of the wire?


Solution

C(r) = 2πr

dC = 2πdr

∆C = 2π × 1 m ≈ 6.28318 meters.


The Total DifferentialLet f(x, y, z, . . .) be a function of x, y, z, . . .. The total

differential or equivalently the exact differential of f is

defined by

df =

(

∂f

∂x

)

y,z,...

dx+

(

∂f

∂y

)

x,z,...

dy +

(

∂f

∂z

)

x,y,...

dz + . . .

The total differential of f gives the infinitesimal variation of f

given infinitesimal variations of the arguments of f ; i.e.

dx, dy, dz, . . ..


Example

f(x, y) = x2y + yex

df = (2xy + yex)dx+ (x2 + ex)dy


Differentials andErrors

We can replace the differentials of the arguments

dx, dy, dz, . . . by the small errors in the arguments

ǫ(x), ǫ(y), ǫ(z), . . ..


Propagation ofSystematic Errors

df =

(

∂f

∂x

)

y,z,...

dx+

(

∂f

∂y

)

x,z,...

dy

+

(

∂f

∂z

)

x,y,...

dz + . . .

ǫ(f) ≈(

∂f

∂x

)

y,z,...

ǫ(x) +

(

∂f

∂y

)

x,z,...

ǫ(y)

+

(

∂f

∂z

)

x,y,...

ǫ(z) + . . .


ExampleThe expression for the gram molecular mass (molecularweight) G of an ideal gas at temperature T , pressure P ,volume V and mass m is given by

G =mRT

PV.

Suppose measurements are made with results

m = 10.0 g T = 269. K P = 1.00 bar

V = 0.50 L,

and suppose the volume has a systematic error ofǫ(V ) = +0.02 L. Find G.


Solution

G =mRT

PV=

(10.0 g)(0.08314 L bar/mol K)(269 K)(1. bar)(0.50 L)

= 447. g/mol

dG = −mRT

PV 2dV

dG

G= −mRT/PV 2

mRT/PVdV = −dV

V

ǫ(G) = −Gǫ(V )

V= −447. g/mol

(

0.02

0.50

)

= −18 g/mol


Report?

G = Gcalc + ǫ(G) = 447. g/mol − 18. g/mol = 429. g/mol

ǫ(G) = −Gǫ(V )

V

Both the errors in G and V have a definite sign.

The value of G decreases, because the volume V appears

in the denominator.


Random ErrorsRandom errors have no definite sign, so we square and

average.

ǫ2(f) ≈[

(

∂f

∂x

)

y,z,...

ǫ(x) +

(

∂f

∂y

)

x,z,...

ǫ(y)

+

(

∂f

∂z

)

x,y,...

ǫ(z) + . . .

]2

≈(

∂f

∂x

)2

ǫ2(x) +

(

∂f

∂y

)2

ǫ2(y) +

(

∂f

∂z

)2

ǫ2(z) + . . .

[Assume 〈ǫ(x)ǫ(y)〉 = 0]


Same Example

G =mRT

PV

m = (10.0± 0.2) g T = (269.± 1.) K

P = (1.0± 0.1) bar V = (0.50± 0.02) L

Assume no systematic error and R is exact (nottrue).


Example Propagation

G =mRT

PV

dG =

(

∂G

∂m

)

T,V,P

dm+

(

∂G

∂T

)

m,V,P

dT

+

(

∂G

∂V

)

T,m,P

dV +

(

∂G

∂P

)

T,V,m

dP

=

(

RT

PV

)

dm+

(

Rm

PV

)

dT

−(

mRT

PV 2

)

dV −(

mRT

V P 2

)

dP


Example (cont.)Divide both sides by G

dG

G=

RT/PV

mRT/PVdm+

Rm/PV

mRT/PVdT

−mRT/PV 2

mRT/PVdV − mRT/V P 2

mRT/PVdP

=dm

m+

dT

T− dV

V− dP

P

Replace differentials by errors, square and average

(

ǫ(G)

G

)2

=

(

ǫ(m)

m

)2

+

(

ǫ(T )

T

)2

+

(

ǫ(V )

V

)2

+

(

ǫ(P )

P

)2


Example (cont.)

G =mRT

PV

m = (10.0± 0.2) g T = (269.± 1.) K

P = (1.0± 0.1) bar V = (0.50± 0.02) L

G = 447. g mol−1

ǫ(G) = G

[

(

ǫ(m)

m

)2

+

(

ǫ(T )

T

)2

+

(

ǫ(V )

V

)2

+

(

ǫ(P )

P

)2]1/2


Example (cont.)

= 447. g mol−1

[

(

0.2

10.

)2

+

(

1.

269.

)2

+

(

0.02

0.50

)2

+

(

0.1

1.

)2]1/2

= 49 g mol−1

Report?

G = 450± 50 g mol−1

The first digit in the error defines the number of digits to be

reported in the final result.


Another ExampleThe concentration of hydrogen ions in a particular solution is

measured with the result [H+]± ǫ([H+]). What is the error in

the pH of the solution?

pH = − log10([H+]) = − 1

2.3ln([H+])

dpH = − d[H+]

2.3[H+]

ǫ(pH) =ǫ([H+])

2.3[H+]


Sums andDifferences Only

f = a+ b− c

df = da+ db− dc

ǫ(f) = [ǫ2(a) + ǫ2(b) + ǫ2(c)]1/2


Products andQuotients Only

f =ab

c

df =b

cda+

a

cdb− ab

c2dc

df

f=

b/c

ab/cda+

a/c

ab/cdb− ab/c2

ab/cdc

=da

a+

db

b− dc

c

ǫ(f) = f

[

(

ǫ(a)

a

)2

+

(

ǫ(b)

b

)2

+

(

ǫ(c)

c

)2]1/2


GraphsSome graphs are to be drawn by hand. When hand-drawn

graphs are required, do not use computer generated graphs.

Instructions associated with graphs will be given for each

experiment by the TA.

Give all graphs a title, and label the x and y axes.

Graphs should fill the entire page.

Include all points with associated error bars or error boxes.


Example Graph

200 250 300 350 400 450 500 550T(K)

20

25

30

35

40

45

50

PV

(L

atm

)

PV vs. T


Errors in GraphicalData

Error in slope m

ǫ(m) ≈ |m+ −m−|2

Error in intercept b

ǫ(b) ≈ |b+ − b−|2


Laboratory ReportsInclude all calculations of your final results, and include all

calculations of errors.

Find errors using a combination of error propagation and

graphical analysis (when required).

In intermediate calculations of results and errors, keep as

many figures as you wish.

Report the final results and errors with the proper number of

significant figures as dictated by the calculated errors.

Incorrect numbers of significant figures in the final results

are subject to significant penalties.

The error analysis represents between 30% and 40% of

your report grades.Error Analysis LectureCHM 335 – p. 39/39

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