# statistics for biomedical engineering labs

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## Content

Laboratory in Biomedical Engineering (1)
0555.4180

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List of Experiments

  

Exp. 1: Bone Mechanics Exp. 2: Soft Tissue Mechanics Exp. 3: Pressure Pulse Propagation in Arteries Exp. 4: Pressure Wave Velocity in Arteries Exp. 5: Mechanical Properties of Bioresorbable Polymers

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Requirements
Take midterm and final exams.  Attend all 5 laboratory sessions & submit all “background & Protocol” and final reports (use predefined formats).  Sign a safety form.

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Final grade: 60% Experiments 20% Midterm Exam 20% Final Exam Each experiment: 70% Final Lab Reports 20% Initial “Background & Protocol” lab reports 10% Oral Quiz, “Background & Protocol”
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Basic Statistical Tools

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Why Statistics?
 

Easy way to sum up data. Enables you to compare data from your

experiments.

Quantitatively and qualitatively assesses relationships in your data.

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What is Statistics?

Allows quantitative analysis and calculation of
the uncertainty magnitude. Estimation of the measurements‟ uncertainty after they are made. Design experiments in an efficient process.

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Sources of Uncertainty

 

Systematic uncertainty - fixed error
Occurs every time a measurement is made under identical conditions.
Limitations and accuracy of the equipment.

  

Random error
Environmental variations. Limitations of human sensing ability (unavoidable). “Noise” in the process, random unstableness in the experimental system.

 

Illegitimate error (unacceptable human mistakes)
Sloppy experimental technique. Erroneous calculation.
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Common Statistics

Mean (the average) – a measure of where the center of your distribution lies. It is simply the sum of all observations divided by the number of observations.

Median (2nd quartile or 50th percentile) – the middle value. 50% of values are above and 50% below the median. If N is an odd number, then the median is the middle value. if N is an even number, then the median is half way between the two middle values.
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Common Statistics – Cont.

Standard Deviation (SD) – a measure of how far the observations in a sample deviate from the mean. It is analogous to an average distance (independent of direction) from the mean. Variance – the square value of SD.

Var  SD 
2

( xi  x) 2  n 1
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Example
Measured data: 1, 4, 4, 4, 5, 18, 20

1  4  4  4  5  18  20 8 Mean= 7
(1  8) 2  3  (4  8) 2  (5  8) 2  (18  8) 2  (20  8) 2  7.63762 7 1

SD =

Median = 4

Variance = SD2=58.33333

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Normal Distribution

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Normal Distribution & the SD
Mean Variance

± 1 SD from the mean is ~68.2% Interval ± 2 SD from the mean is ~95.4% Interval ±1.96 SD from the mean is 95% Interval

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Degrees of Freedom (DF)

The number of independent data points available for calculation.
For example, if we have ten data points (n=10), after calculating the mean, only nine will still be independent. Therefore, there will be only nine (n-1) degrees of freedom for variance estimation.
Var  SD 2  ( xi  x ) 2  n 1

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Standard Error of the Mean
(SE Mean)

SE Mean is an estimation of the dispersion that you would observe in the distribution of sample means, if you continued to take samples of the same size from the population.

SEMean  SD / N
(N – number of data points)

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Confidence Intervals for The Mean

A confidence interval for a mean specifies a range of

values within which the unknown population parameter,
in this case the mean, may lie.

We interpret an interval calculated at a 95% level, as we are 95% confident that the interval contains the true population mean. We could also say that 95% of all confidence intervals formed in this manner (from different samples of the population) will include the true population mean.
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Confidence Intervals for The Mean

In general, you compute the 95% confidence interval for the mean with the following formula:
Lower limit = M - Z.95σm Upper limit = M + Z.95σm

Z.95 is the number of standard deviations extending
from the mean of a normal distribution required to

contain 0.95 of the area.

σm is the standard error of the mean.

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Confidence Intervals - Example
Assume that the weights of 10-year old children are
normally distributed with a mean of 90 pounds and a standard deviation of 36 pounds.

What is the sampling distribution of the mean for a
sample size of 9 (including the 95% confidence interval for the mean)?
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Example – Cont.

The sampling distribution of the mean has a mean of 90 and a standard deviation of 36/3 = 12.
95% Interval = ±1.96 SD

90 - (1.96)(12) = 90-23.52=66.48
90 + (1.96)(12) = 90+23.52= 113.52
The middle 95% of the distribution is shaded

If we compute the mean (M) from a sample, and create an interval ranging

from M - 23.52 to M + 23.52, this interval has a 0.95 probability of
containing 90 (the true population mean).
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Confidence Interval – t distribution

CI- confidence interval around the mean

CI  x 

ts n

DF 2
3 4

0.95 4.303
3.182 2.776 2.571 2.306 2.228

0.99 9.925
5.841 4.604 4.032 3.355 3.169

CL = Confidence Limit

5 8 10

t is the „confidence‟ that the user wants to develop in the estimation.

Abbreviated t table 20

Example: Confidence Interval
Excel: TINV=(p,DF)
The set of data for Young‟s modulus of a bone specimen tested in tension is:

xi=(9.75, 10.2, 11.62, 9.43, 10.56, 10.4, 9.98,
11.02, 10.19, 10.35) Gpa.

What is the 95% confidence interval of the mean?
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Example – Cont.
 The probability for 95% confidence is p=0.05  DF=n-1, n=10, DF=9  t = TINV(p,DF)=2.262159  Mean = 10.35  SD = 0.623592  CL = 0.446

CI=10.35±0.45

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Test of Significance

How can we determine whether the data agrees with the theoretical value or not? t-test – test of significance

t test  x  x

*

n/s

For this type of analysis to be valid, both samples we are comparing should be drawn from the same population with the same distribution of variance.
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Example

Does the mean value of the example agrees with the theoretical value 10.2 GPa? (10.35-10.2)100.5/0.632=0.075 < 2.262

 ttest=

=> There‟s no significant difference.
|t| < tcr no statistical difference
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t-Test

The t-test assesses whether the means of two groups are statistically different from each other.

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t-Test Cont.

Paired t-test is employed when measuring
two different quantities from the same specimen, repeating this measurement across different samples.

Unpaired t-test is employed when
measuring two different quantities from different specimens. The sample size of the two samples may or may not be equal.
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Hypothesis Testing

The null hypothesis for the test is that all population means (level means) are the same. Ho: µ1 - µ2 = 0
The alternative hypothesis is that one or more population means differ from the others. H1: µ1 - µ2 ≠ 0
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How to test the difference between group means for significance?

First step: specify the null hypothesis and an alternative hypothesis.

Second step: choose a significance level. Usually a 0.05 level is chosen.
Third step: compute the t-value.

Fourth step: determine the probability value for the computed t-value
using a t table.

Fifth step: compare the probability value to the significance level. If it is
less than the significance level (0.05), then the effect is significant. If the effect is significant, the null hypothesis is rejected and vice versa.
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Excel: =TTEST()

2 1,3

p<0.05 statistical difference p>0.05 no statistical difference

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Excel:

Tools/ Data analysis/ t-test: paired two sample for means

0

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Regression

Regression analysis (curve fitting)- the objective technique for evaluating the best fit of data. Linear regression with linear relationship Y=a+bX.

The regression line is obtained by minimizing the sum of the actual data squared deviations from the best-fit line. The aim is to find coefficients a & b, that minimize the expression.

 (Y
i 1

n

i

 Y )   (Yi  (a  bX i ))
2 i 1

n

2
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R2-value

Regression correlation coefficients are shown as either R or R2. For R2:
0.8 – 1:  0.6 – 0.8:  0.4 – 0.6:  0.2 – 0.4:  0.0 – 0.2:

Very strong Strong Moderate Weak Very Weak
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Excel:

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Excel: Tools/ Data analysis/ Regression

a b

R Square- closer to 1, better the fit Standard Error- closer to 0, better the fit
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Correlation Example

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Experimental Design

For experimental design, we want to have a way to estimate the uncertainty inherent in the measurements, due to the nature of the equipment or method used:

Y=A X1X2X3

Y-dependent variable, X‟s –measured independent variables, A -constant. The change in Y due to a change in any of the X‟s will be:
 

DY/Y= DX1/X1 + DX2/X2 + DX3/X3 (DY/Y)max = S |DXi/Xi|
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Example: Experimental Design

We want to measure a distance. The data: 5 sec, 10 m/sec.

The stopwatch we are using allows us to time the

process to the nearest 0.05 sec and the speedometer
allows us to monitor velocity to the nearest 0.1 m/sec.
What is the maximum expected fractional distance uncertainty and the absolute uncertainty value?
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Experimental Design – Cont.

In this example Y is distance, X1 is velocity, & X2 is time. Accordingly, the fractional distance uncertainty is: (DY/50)max = 0.1/10 + 0.05/5 = 0.02 = 2%

(DY)max = 1 m

 The maximum expected fractional distance
uncertainty is 2%.

 The absolute uncertainty value is 1 m.

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Example: experimental design & t Test

Background:
 

Unique geometrical foot structure. Stress concentration.

Hypothesis:
The internal compression stress in the bone-soft tissue interface is higher than the superficial compression stress in the shoe-heel interface.
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Example: experimental design & t Test – Cont.

Method:

The two kinds of compression stresses were measured and computed during gait in the calcaneus

region.

For the purpose of comparison, the peak compression stress of each gait cycle was collected.

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Example: experimental design & t Test – Cont.

Results:

Analysis: Paired, 2 tailed t-test P = 1.45*10-13 < 0.05 hypothesis supported
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Some References…
 http://davidmlane.com/hyperstat
 http://www.statsoft.com/textbook/stathome.html  http://helios.bto.ed.ac.uk/bto/statistics/tress4a.html

 http://www.stat.yale.edu/Courses/1997-98/101/confint.htm
 http://www.phys.selu.edu/rhett/plab193/labinfo/Error_Analysis /05_Random_vs_Systematic.html

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