Page 1
Exam 4 Introductory Statistics for Engineers
Spring 2012
/150
Name ________________________
1. (60 points)
A study of the effects of 2D and 3D weaving processes on the tensile strengths of synthetic fibers
was carried out at the University of Delaware’s Center for Composite Materials. One part of the
study was an investigation of the curvature of the weave as a source for the degradation of the
mechanical properties of the fibers. A tow is an untwisted bundle of continuous filaments and
Vectran is a Registered Trademark for a commercial multifilament yarn spun from a liquid crystal
polymer. The data presented below gives the tensile strength of a Vectran tow calculated as the
failure load (the mass required to break the fiber) in grams divided by the average linear density
of the tow Each set of samples of tow were of a specific radius of curvature, which is stated at the
top of the table.
Tensile Strength for Different Radii
Radius = 1 mm
Radius = 2 mm
Radius = 3 mm
Radius = 5 mm
25.58
24.25
26.76
24.75
23.58
22.52
23.55
25.46
25.09
25.13
24.97
23.26
23.20
25.78
26.86
25.87
26.76
25.66
25.65
26.07
26.33
24.01
24.31
23.62
25.72
25.83
26.03
24.70
Test at a 5% level of significance whether there are any differences in fiber tensile strength
associated with the different radii of curvature. State and explain your conclusion. Explain how
you reached your conclusion. Compute and report the P- Value of your observed statistic.
Now determine by a Bonferroni procedure which, if any, of the radii are significantly different
from each other in their effect on fiber tensile strength.
What assumptions are necessary for the above Single Factor ANOVA to be valid?
Do these assumptions seem to be satisfied by this particular set of data? Explain your answer.
Hand in your spreadsheet with the exam.
Page 2
Exam 4 Introductory Statistics for Engineers
Spring 2012
2. (90 points)
Below a material’s elastic limit the relationship between stress (force per unit area) and strain
Stress
(fractional elongation) follows Hook’s Law. This is expressed by the equation: Strain =
,
E
where E, the Young’s Modulus, is an intrinsic constant of the material. The elastic limit for nickel
is about 20 MPa (the unit MPa is a Mega Pascal and is equal to 106 N/m 2 ). The following data
was obtained at room temperature for a uniform thickness strand of nickel.
Stress in MPa
Strain ( Fraction of Elongation )
1.50
6.391E-06
2.00
1.153E-05
2.50
1.275E-05
3.00
1.412E-05
3.50
1.736E-05
4.00
2.059E-05
4.50
2.273E-05
5.00
2.491E-05
5.50
2.811E-05
6.00
2.966E-05
6.50
3.276E-05
7.00
3.656E-05
7.50
3.779E-05
8.00
4.095E-05
8.50
4.350E-05
9.00
4.543E-05
The error in the applied stress is negligible, but the elongation measurements are subject to error.
In Excel perform a linear regression analysis on this data. Your spreadsheet should contain two
scatter plots. The first should display the data and the predictions of the linear regression
model with the latter drawn as a continuous line. The second should display the residuals plotted
versus the control variable.
Calculate the lower and upper limits for the 95% confidence intervals requested below.
95% Confidence Interval for
Population Slope β1
Population Intercept β 0
The mean strain at a stress of 10 MPa
A single measurement of strain at a stress of 10 MPa
correlation coefficient =
coefficient of determination =
Lower Limit
Upper Limit
Page 3
Exam 4 Introductory Statistics for Engineers
Fill in the following Regression ANOVA table as shown below.
Degrees of
Source
Sum of Squares
Freedom
Spring 2012
Mean Square
Linear Model
Error
Total
Observed F score
____________________
At a 5% level of significance, what conclusion do you draw from this ANOVA Table?
Comment on how well the linear model fits this data.
What does the pattern of residuals indicate about the theoretical model?
At a 5% level of significance is it true that the regression intercept is different than 0?
From this data construct a 95% confidence interval for the Young’s Modulus of Nickel.