Transcript Data Analysis - Pomona College

Precision & Accuracy
• Precision
– How close are the measurements to each other?
– Reproducibility
• Accuracy
– How close is a measurement to the true value?
Could be affected by the following:
Systematic errors – all higher or lower than actual value
(lack of accuracy)
Random errors – some high and some low
(lack of precision)
Significant Figures
• Presenting
measurements and
calculated results with
the appropriate
significant figures and
units is an indication
of the precision of
values.
Rules for Sig Figs
•
•
•
•
All nonzero digits are significant
Trapped zeros are significant
Trailing zeros are significant if there is a decimal point
Leading zeroes are NOT significant
# sig figs
300
300.
300.0
0.003
0.0030
0.00300
1
3
4
1
2
3
How would one write 300 with 2 sig figs?
Scientific Notation
3.0 x 10 2 has two sig figs
Written as a number between 1-10 x a power of ten
Unambiguously displays the precision of the value making it easier to
make comparisons
300
3 x 102
300.
3 .00 x 102
300.0
3.000 x 102
0.003
3 x 10-3
0.0030
3.0 x 10-3
0.00300
3.00 x 10-3
Making Measurements - Thermometer
The number of significant figures in your measurement
depends on the measuring device.
87.5±.1°C
The bottom of the meniscus is between 87 and 88 ° C.
This can be read to 1 digit more precision than indicated by the calibration.
The last estimated digit can vary from person to person, but each should
record a value to the tenth’s place.
There are 3 sig figs and the last digit is the uncertain digit.
Generally, measurements are uncertain by ± 1 in that last digit unless
otherwise indicated by your measuring device. Usually, 1/10 of an increment.
Beaker vs. Graduated Cylinder
Each contains the same amount of water.
Beaker
10. ± 1mL
Graduated Cylinder
10.05 ± .05 mL
The Analytical Balance
All digits should be recorded as given, precision is to the
0.1 mg, & the accuracy is determined by the calibration.
Calculations & Sig Figs
• Multiplication & Division
The total number of sig figs in the answer is equal to the
same number of sig figs in the measurement used in the
calculation with the smallest number of sig figs.
Ex:
5.1 cm x 2.01 cm = 10.0701 cm2 = 10. cm2
• Round the final answer using the number to the right of the last sig fig.
• Avoid round off errors by keeping extra digits beyond the last sig fig
when calculating intermediate values.
Calculations & Sig Figs
• Addition & Subtraction
The final answer should be rounded to the right-most
filled column (according to the value with the biggest
uncertain digit – the weakest link).
Ex:
6.5 cm
100.01 cm
+
.044 cm
106.554 cm
= 106.6 cm
“Scientific notation” can make it
easier…..
• What is the sum of 4.5 x 10-6, 3.2 x10-5,
and 15.2 x 10-7?
.45 x 10-5
3.2 x 10-5
.152 x 10-5
3.802 x 10-5
= 3.8 x 10-5
SI Prefixes
Prefix Symbol
Mega
M
Kilo
k
Deci
d
Centi
c
Milli
m
Micro
μ
Nano
n
Femto
f
Atto
a
Meaning
Power of 10
1,000,000
106
1,000
103
0.1
10-1
0.01
10-2
0.001
10-3
0.000001
10-6
0.000000001 10-9
0.000000000000001 10-15
0.000000000000000001
10-18
Fundamental SI Units
Physical Quantity Unit
Mass
Length
Time
Temperature
kilogram
meter
second
kelvin
Abbreviation
kg
m
s
K
Dimensional Analysis
• Use conversion factors (definitions, ratios) to convert
from one unit to another.
• Conversion factors are exact numbers that have no
uncertainty.
• Ex. Convert 6.4 weeks to hours.
6.4 weeks x 7 days x 24 hours = 1100 hrs
1 week 1 day
Group Problems
•
Convert 47 hours to weeks.
47 hours x 1 day x 1 week = 0.28 weeks
24 hours 7 days
The same conversions were used as in the previous example. The top equals the bottom.
Round off answers at the end. Keep additional sig figs for intermediate answers.
• Calculate the sum of 2.5 + 3.5 + 4.5 +5.5.
2.0
3.5
4.5
5.5
14.5
• The tread on a certain automobile tire wears 0.00100 inches per 2,600 miles driven. If
the car is driven 45 miles a day, how many months ( 1mo = 30 days) can a tire w/ 0.010
in of treat be used before it wears down and needs to be replaced?
.010in x 2,600 mi x 1 day x 1 month = 19.25 = 19 months
0.00100in 45 mi 30 days
• In a displacement of water by gas experiment the initial volume of water in a burette is
45.50 mL and the final volume is 37.50 mL. What is the total volume of water displaced?
In mL? in L?
45.50
(4sf)
- 37.50 mL (4sf)
8.00 mL
(3sf) = 0.00800 L (still 3sf)
Statistical Analysis and Expression of Data
Reading: Lab Manual 29 - 40
Today: Some basics that will help you the entire year
The mean or average
x
x
i
i
n
=  true value,  measurements
≈  true value, finite # of measurements

2


x


 i
i
n 1
Uncertainty given by
standard deviation
For finite # of measurements
Standard deviation: S
(Calculators can calculate & σ quite easily!!!
Learn how to do this on your calculator.)
For small number of measurements σ ≈ S is
very poor. Must use Student t value.
σ ≈ tS; where t is Student t
Usually use 95% Confidence Interval
So, 95% confident that if we make a
measurement of x it will be in the range
x ± t95s
Uncertainty of a SINGLE MEASUREMENT
σ ≈ S = [(xi-x)2/(n-1)]1/2
Usually interested in mean (average) and its uncertainty
Standard Deviation of the mean
S
Sm 
n
Then average and uncertainty is expressed as
x ± t95Sm
Often want to know how big uncertainty is compared to the mean:
Relative Confidence Interval (C.I.)
= (Sm / x )(t95)
Expression of experimental results:
1. Statistical Uncertainties (S, Sm, t95S, Sm (t95) / x)
always expressed to 2 significant figures
2. Mean (Average) expressed to most significant digit in Sm
(the std. dev. of the mean)
Example
Measure 3 masses: 10.5763, 10.7397, 10.4932 grams
Average = 10.60307 grams
Std. Dev. S = .125411 = .13 grams
Sm = .125411 / 3 = .072406 = .072 grams
Then average = ?
= 10.60 grams
t95 for 3 measurments
95% C.I. = t95Sm = 4.303 * .072406 = .311563 = .31 grams
of the mean
Relative 95% C.I.of the mean = ?
= 95 % C.I. / Average = .311563/10.60307 = .029384 = .029
of the mean
Usually expressed at parts per thousands (ppt)
= .029 * 1000 parts per thousand = 29 ppt = Relative 95% C.I. of the mean
What if measure 10.5766, 10.5766, 10.5767 grams? Ave. = 10.57663; Sm = .000033
Ave. = ?
Ave. = 10.5766, not 10.57663 because limited by measurement to
.0001 grams place
Now work problems.
Sally
5 times, average value of 15.71635% ; standard deviation of 0.02587%.
Janet
7 times, average value of 15.68134% ; standard deviation of 0.03034%.
(different technique)
Express the averages and standard deviations to the correct number of significant figures.
Must use Sm.
Sally: Sm = 0.02587/5 = 1.157 x 10-2 = 1.2 x 10-2%
Janet: Sm = 0.03034/7 = 1.147 x 10-2 = 1.1 x 10-2%
Sally: 15.72%; S = 0.026%
Janet: 15.68%; S = 0.030%
Using the proper statistical parameter, whose average value is more precise?
Must use Sm.
Sm(Janet) < Sm(Sally) so Janet’s average value is more precise.
95% confidence intervals of the mean, relative 95% confidence intervals of the mean
Sally: 95% C.I. = ±t95Sm = ±2.776 (1.157 x 10-2) = ±3.21 x 10-2% = ±3.2 x 10-2%
Range = 15.69 – 15.75%
Relative 95% C.I. = 3.21 x 10-2 / 15.71635 * 1000 ppt = 2.04 = 2.0 ppt
Janet: 95% C.I. = ±t95Sm = ±2.447(1.147 x 10-2) = ±2.81 x 10-2% = ±2.8 x 10-2%
Range = 15.65 – 15.71%
Relative 95% C.I. = 2.81 x 10-2 / 15.68134 * 1000 ppt = 1.79 = 1.8 ppt
Are the two averages in agreement at this confidence level?
Because the 95% C.I. for both measurements overlap, the two averages are in agreement
If you owned a chemical company and had to choose between Sally’s and Janet’s technique,
whose technique would you choose and why?
Choose Sally’s techniques because the uncertainty in a single measurement based on S is better
than that using Janet’s technique.