Download 2/6 Lecture Slides

Chem. 31 – 2/6 Lecture
Announcements I
• Returning graded material in lab (diagnostic
quizzes, quiz 1, lab procedures quiz)
• Q1 scores were not great
– (ave = 6.2 vs. 9.3 for Summer, ’16 which had part a)
– It is your responsibility to learn material covered in
class and in text homework problems (H2O2 example
and 1.34/1.37)
• SacCT has been updated (has quiz scores and
quiz solutions)
• Pipet Calibration Lab Report due next Monday
Announcements II
• Additional Problem 1 – due 2/15
• Today’s Lecture
– Error and Uncertainty (Chapter 3)
• Propagation of Uncertainty
– Mixed operation problems
– Gaussian Statistics and Calibration (Chapter 4)
•
•
•
•
•
Mean value and standard deviation
Gaussian distributions for populations
Application to measurement statistics
Z-value problems - % between limits
Confidence Intervals (if time)
Propagation of Uncertainty
• Calculation of the density of a
liquid:
Density = mliquid/Vliquid with mass
measured as weight gain in a
container (mfull – mempt)
Chapter 4
Calculation of Average and Standard Deviation
• Average
x

x
i
n
• Standard Deviation
 x  x 
2
Sx 
i
n 1
Note: You are welcome to use function keys on your calculator to
calculate average and standard deviation
Chapter 4 – Gaussian Distributions
•
•
Gaussian Distributions are often
observed when sample size gets
large
Example 1: repeated measurement
of MMP conc.
μ
Sample vs. Population:
Sample mean
x
Sample standard
deviation
S
population mean
μ
population standard
deviation
σ
x
MMP Conc.
6.00
5.80
5.60
MMP Conc. (ppm)
•
MMP = methylmannopyranoside = interal
standard added at a conc. of 5.00 ppm
5.40
5.20
S
5.00
4.80
4.60
4.40
4.20
4.00
1
3
5
7
9
11
13
15
17
Sample Number
•
As sample size increases:
x →μ
Also note that mean values vary less than individual
measurements (see 4 measurement means)
and S → σ
19
21
23
25
Chapter 4 – Gaussian Distributions
MMP Concentration Distribution
8
7
Number of Values
6
5
Actual
Expected from Data
4
3
2
1
0
4.0-4.25 4.25-4.5 4.5-4.75 4.75-5.0 5.0-5.25 5.25-5.5 5.5-5.75
Conc. Range
5.75-6
Chapter 4 – Gaussian Distributions
• Second Example: Mass Spectrometer
Measurements (x-axis is mass to charge
ratio or mass for +1 ions)
(1) Spec #1 * [BP = 234.1, 52707]
100
1343.9877
90
80
70
% Intensity
60
50
40
1344.9770
30
20
10
0
1 3 4 3 .1 2 3 6 0
1 3 4 3 .9 2 6 3 6
1 3 4 4 .7 2 9 1 3
1 3 4 5 .5 3 1 8 9
m/z
1 3 4 6 .3 3 4 6 6
Chapter 4 – Gaussian Distributions
Idealized distribution occurs as n →∞
• Math for Gaussian
Distribution:
1
e ( x   )
 2
0.45
0.4
/ 2
Normalized Gaussian
Distribution:
μ = 0 and σ = 1
Use of Normalized Gaussian
Distribution:
0.35
0.3
Frequency
y 
2
Normal Distribution
0.25
0.2
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
0
1
2
X or Z value
Area under curve gives probability of finding value between limits
3
4
5
Chapter 4 – Gaussian Distributions
Examples:
Normal Distribution
-∞<Z<∞
Area = 1
0.45
0.4
Since curve is symmetrical,
0 < Z < 1.5 Area = 0.433
(See Table 4-1 in text)
0.3
Frequency
0 < Z < ∞ Area = 0.5
0.35
0.25
0.2
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
0
1
2
3
4
X or Z value
Note: non-normal distributions can be converted to normal distributions
as follows: Z = (x - μ)/σ
5
Chapter 4 – Gaussian Distributions
Now for limit problems – example 1 – population statistics:
A lake is stocked with trout. A biologist is able to randomly
sample 42 fish in the lake (and we can assume that 42
fish are enough for proper – Z-based statistics).
Each fish is weighed and the average and standard deviation of the
weight are 2.7 kg and 1.1 kg, respectively.
If a fisherman knows that the minimum weight for keeping the fish is
2.0 kg, what percent of the time will he have to throw fish back?
(assuming catching is not size-dependent)
1st part: convert limit (2.0 kg) to normalized (Z) value:
Z = (x – )/
2nd part: use Z area to get percent
Chapter 4 – Gaussian Distributions
Limit problem – example 2 – measurement statistics:
A man wants to get life insurance. If his measured
cholesterol level is over 240 mg/dL (2,400 mg/L), his
premium will be 25% higher. His level is measured and
found to be 249 mg/dL. His uncle, a biochemist who
developed the test, tells him that a typical standard
deviation on the measurement is 25 mg/dL. What is the
chance that a second measurement (with no crash diet
or extra exercise) will result in a value under 240 mg/dL
(e.g. beat the test)?
Graphical view of examples
Equivalent Area
Frequency
Normal Distribution
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
Table area
Desired area
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z value
240
249
X-axis
Chapter 4 – Calculation of Confidence
Interval
1.
2.
x
n
Z depends on area or desired
probability
At Area = 0.45 (90% both sides),
Z = 1.65
At Area = 0.475 (95% both sides), Z =
1.96 => larger confidence interval
Normal Distribution
Frequency
Confidence Interval = x + uncertainty
Calculation of uncertainty depends on
whether σ is “well known”
3.
When  is not well known (covered
later)
4.
When  is well known (not in text)
Value + uncertainty =
Z
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-3
-2
-1
0
Z value
1
2
3
Chapter 4 – Calculation of
Uncertainty
Example:
The concentration of NO3- in a sample is measured 2 times
and found to give 18.6 and 19.0 ppm. The method is
known to have a constant relative standard deviation of
2.0% (from past work). Determine the concentration
and 95% confidence interval.
Chapter 4 –
Calculation of Confidence Interval with  Not Known
Value + uncertainty =
tS
x
n
t = Student’s t value
t depends on:
- the number of samples (more samples => smaller t)
- the probability of including the true value (larger
probability => larger t)
Chapter 4 –
Calculation of Uncertainties Example
• Measurement of lead in drinking water
sample:
– values = 12.3, 9.8, 11.4, and 13.0 ppb
• What is the 95% confidence interval?