Transcript Short Course on Data Analysis and Errors of Observation

Summary Part 1
Measured Value
= True Value + Errors
Errors = Random Errors + Systematic
Errors
How to minimize RE and SE:
(a) RE – by taking MORE measurements!
(b) SE – by SPOTTING!
Precision  Accuracy
How to present the measured value ?
In general, the result of any measurement
of a quantity x is stated as following
( x   x ) units
(Best estimate  Uncertainty) units
If possible  x is represented by the
standard error!
How to present the measured value ?
Rule for stating Uncertainties:
Experimental uncertainties should almost
always be rounded to one significant figure.
Example:
 x = 0.000153 = 0.0002 m
How to present the measured value ?
RuleExample:
for stating Answers:
x = 12.23452 m
The last significantfigure
in any stated
x = 0.0002
m
answer should usually be of the same order
x same
±  xorder
= (12.2345
± 0.0002)
m
of the
of magnitude
(in the same
decimal position) as uncertainty.
Statistical Formulas
Mean
Standard Deviation
n
1
x   xi
n i 1
 
Standard Error
x 

n
1 n
2
( xi  x )

n  1 i 1
Error Propagation
How to estimate the error of the quantity R
from the errors associated with the primary
measurement of x, y, z, …, respectively?
R 
 R 2 2  R 2 2  R 2 2

   x     y     z  ...
 z 
 x 

 y 
Summary Part 2
Presentation of final answer: (1.2  0.3) m
Mean: average of the data
Standard deviation: measures the spread
of the data about the mean
Standard error: standard deviation of the
mean
Use propagation of error formula to
compute the combined error!
Example 1
In the determination of gravity, g by means of
a simple pendulum (i.e. T = 2 (ℓ /g) ), the
following data are obtained:
Length of the string, ℓ: 99.40, 99.50, 99.30,
99.45, 99.35 cm.
Time for 20 oscillations, t: 40.0, 39.8, 40.2,
39.9, 40.1 sec.
Calculate g and its error.
Solution for Example 1
For ℓ, n = 5
Mean:
1 5
 
5 i1
i
 99.40 cm
Standard deviation of ℓ:
 
Standard error:
1 5
2
( i  )  0.07906 cm

5  1 i 1
 

5
 0.03536 cm
Solution for Example 1
For time of an oscillation Ti = ti/20, n =5
1 5
Mean: T   Ti  2.00sec
5 i 1
Standard deviation:
T 
Standard error:
1 5
2
(
T

T
)
 0.007906 sec

i
5  1 i 1
T 
T
5
 0.003536 sec
Solution for Example 1
   (99.40  0.04) cm T   T  (2.000  0.004)sec
g  4 2
T
2
 981.039 cm / sec2
g  9.81039 m / sec2
The estimated uncertainty of g:
 g  2  g  2
        T
 
 T 
2
2
g
2
Solution for Example 1

From this function g  4
2
T
2
Take the partial derivative of g w.r.t. ℓ and T
g
8 
 3
T
T
2
g 4
 2
 T
2
Solution for Example 1
g (, T )
8 2 
 3
T
T
g (, T ) 4 2
 2

T
 g  2  g  2
 g         T  3.486 cm/sec 2
 
 T 
2
2
g   g  (9.81  0.04) m / sec
2
Which is the BEST line?
Least Squares Method (LSM)
Finding the best straight line
ybest = mbest x + cbest
to fit a set of measured points (x1, y1), (x2, y2),
…, (xn, yn).
The following assumptions are made to find
the best line:
1. The uncertainty in our measurements of x
is negligible but not in y
2. The uncertainty in our measurements of y
is the same (i = )
Least Squares Method (LSM)
Let coordinates (xi , yi) is the i-th data point
that you have plotted in the graph and
n
 
i 1
The best fit of a straight line takes the
following form:
ybest(xi) = mbest xi + cbest
Least Squares Method (LSM)
cbest
1
2
   xi  yi   xi  xi yi 
'
mbest
1
  n xi yi   xi  yi 
'
 '  n x   xi 
2
i
2
Least Squares Method (LSM)
c
best


2
x

'
2
i
m
best
 n

2
'
1
2

 yi  cbest  mbest xi 

n2
Least Squares Method (LSM)
ybest(x) = mbest x + cbest
Example 2
In the determination of g by means of a simple
pendulum (i.e. T = 2 (ℓ /g)), the following data are
obtained:
ℓ1 = 40.00 cm
t1 = 26.3, 25.5, 25.9 sec
ℓ2 = 60.00 cm
t2 = 31.8, 32.3, 32.7 sec
ℓ3 = 80.00 cm
t3 = 36.9, 36.5, 37.1 sec
ℓ4 = 100.00 cm t4 = 42.6, 41.5, 41.7 sec
ℓ5 = 120.00 cm t5 = 43.8, 44.8, 45.3 sec
ℓ6 = 140.00 cm t6 = 48.4, 48.9, 49.1 sec
ℓ7 = 160.00 cm t7 = 51.2, 51.5, 51.9 sec
ℓ8 = 180.00 cm t8 = 54.5, 54.8, 54.3 sec
where ti is the time for 20 oscillations. Determine g and
its error.
Solution for Example 2
2
 x T  y n8
 x  880.00 cm/sec  x  113600.00 cm /sec
2
i
2
i
i
i
y
i
 37.075475 sec
-2
x y
i
i
 4765.85778 cm/sec
i
i
  y( x )  y
i
i
2
best
( xi )  0.05645254 sec
2
-4
2
2
Solution for Example 2
slope, mbest = 0.040926 sec2/cm
y-intercept, cbest = 0.132583 sec2
m_best = 0.000748 sec2/cm
c_best = 0.089178 sec2/cm
mbest m_best = (0.0409  0.0008) sec2/cm
cbest c_best = (0.13  0.09) sec2
The best fit of a straight line,
T2best = 0.0409 ℓ + 0.13
Solution for Example 2
42 / g = Tbest2 / ℓ
42 / g = mbest = 0.0409 sec2/cm
 gbest = g = 965.2425 cm/sec2
How To Compute, g ?
Solution for Example 2
How To Compute, g :
From 42 / g = mbest  g = 42 / mbest

g
4

where

2
mbest
mbest
 g = 18.88005 cm/sec2
 g = 6.67510 cm/sec2
 g  g = (9.65  0.07) m/sec2
2
 g  2
 
  mbest
 mbest 
2
g
2