Statistics and Curve Fitting

Vipuil Kishore

Lab Lecture 2

Statistics and Curve Fitting

We will learn the following:

Mean

Standard Deviation and Standard Error

Gaussian or Normal Distribution

Confidence Intervals

Curve Fitting using DataFit

Significant Figures

Solving y = axb

Mean (Average)

Example:

Set of test scores: X1 = 80; X2 = 100; X3 = 60; X4 = 70; X5 = 90

𝑀𝑒𝑎𝑛 = 𝑋 =

𝑖=1

𝑁𝑋𝑖𝑁

𝑋 =80 + 100 + 60 + 70 + 90

5= 80

N – number of data points

𝑋 = 80

Standard Deviation (Spread of the data)

Example:

Set of test scores: X1 = 80; X2 = 100; X3 = 60; X4 = 70; X5 = 90

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑠 =

𝑖=1

𝑁

𝑠𝑞𝑟𝑡[(𝑋𝑖−𝑋)

2

𝑁 − 1]

𝑠 =

𝑖=1

𝑁

𝑠𝑞𝑟𝑡[(80 − 80)2+(80 − 100)2+(80 − 60)2+(80 − 70)2+(80 − 90)2

5 − 1]

𝑠 = 15.81

Gaussian or Normal Distribution

Example:

Set of test scores: X1 = 80; X2 = 100; X3 = 60; X4 = 70; X5 = 90

60 70 80 90 100

X (values)

0.2

Actual Distribution Normal Distribution

(for large sample size)

X (values)

X

s

Standard Error

Example:

Set of test scores: X1 = 80; X2 = 100; X3 = 60; X4 = 70; X5 = 90

Standard Error: Standard deviation of the sampling means

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐸𝑟𝑟𝑜𝑟 =𝑠

𝑁

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐸𝑟𝑟𝑜𝑟 =15.81

5

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐸𝑟𝑟𝑜𝑟 = 7.07

Confidence Intervals (CI)

Confidence interval is an estimated range of values within

which the true value is likely to be present

Two Assumptions: 1) large sample size and 2) all errors are

random

For most experimental cases, 95% CI is acceptable

𝑋𝑖 = 𝑋 ±𝑧(𝑠)

𝑁

𝐼 = −𝑧

+𝑧 1

2πexp−𝑧2

2𝑑𝑧

z I

0 0

1 0.6826

1.96 0.95

6 0.999997

X

-z +z

Confidence Intervals (CI)

Example:

Set of test scores: X1 = 80; X2 = 100; X3 = 60; X4 = 70; X5 = 90

𝑋𝑖 = 𝑋 ±𝑧(𝑠)

𝑁

For 95% CI, 80 ±1.96(15.81)

5

95% CI = 80 ± 13.85

Upper Interval: 80 + 13.85 = 93.85

Lower Interval: 80 − 13.85 = 67.15

Therefore, 95% CI is 67.15 to 93.85

Curve Fitting

DataFit: datafit6.zip (Install from CD or http://my.fit.edu/~vkishore/CHE3265/Spring%202014/)

Licensed to Florida Tech; License Key: BTOG-MTPI-HWFZ-LFEC

Curve Fitting: Process of finding a mathematical equation

which reproduces the experimental data

Attempts to fit the data with as few parameters as possible

𝑖=1

𝑁

(𝑦𝑖,𝑒𝑥𝑝 − 𝑦𝑖,𝑐𝑎𝑙𝑐)2

X

Y

Residual Sum of Squares

yi,calc = f(xi) = ax + b

Curve Fitting

Example: Fit the stress vs. strain data using Datafit and

calculate the slope and intercept (y = ax + b)

Stress (psi) Strain (in/in)

0 0.00000

2087 0.00015

4174 0.00035

6261 0.00055

8348 0.00075

10435 0.00090

12522 0.00105

14609 0.00130

16696 0.00155

18783 0.00170

20870 0.00190

22957 0.00215

25044 0.00235

27131 0.00260

29218 0.00285

31304 0.00310

33391 0.00335

35478 0.00370

37565 0.00465

38191 0.00560

95% Confidence IntervalsVariable Value 95% (+/-)a 9753608.341 383448.105b 1320.759293 772.5472262

Datafit Output:

Model Plot:

a is slope and b is intercept

Significant Figures

The only significant digit in the error is the leftmost non-zero

The last significant digit in the error is in the same decimal

place as the first significant digit in the number itself

Rounding: 5+, round up; 4-, round down

Example:

Slope (a): 9753608.341 ± 383448.105

9800000 ± 400000

Intercept (b): 1320.75923 ± 772.5472262

1300 ± 800

Linearizing an Exponent Equation

𝑦 = 𝑎𝑥𝑏 a, b – unknown constants

𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑏𝑙𝑜𝑔𝑥

Taking log on both sides,

𝑦 = 𝑐 +𝑚𝑥

log X

Slope = b

Intercept = loga