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HW 6 Due in recitation 2-24-
Go to a computer lab before recitation. Launch stat
2-10-09 found on www.stt.msu.edu/~lepage (be sure to
launch the 2-10-09 edition near the end of the file list). Math-
ematica will launch. Follow the instructions on Lecture Out-
line 2-20-09 and do the following:
- Enter the following matrix to Mathematica :
myx= {{1, 2.3, 3.6}, {1, 2.4, 3.5}, {1, 2.0, 3.1}, {1, 2.4, 3.7},
In[51]:= myx^ =^881 ,^ 2.3,^ 3.6<,
Out[51]= 881 ,^ 2.3,^ 3.6<,
- Enter y = the last five digits of your student number. For
example, if your student number ends in 47680 you enter:
myy = {4, 7, 6, 8, 0}
In[52]:= myy^ =^84 ,^7 ,^6 ,^8 ,^0 <
Out[52]= 84 ,^7 ,^6 ,^8 ,^0 <
- Compute the coefficients b
`
0 ,^ b
`
1 ,^ b
`
2 of^ a^ least^ squares^ fit^ of
the model y = b 0 , + b 1 x 1 + b 2 x 2 for the n = 5 data values.
In[53]:= mybetahats^ =^ betahat@myx,^ myyD
Out[53]= 8 7.73563,^ - 16.092,^ 9.88506<
- Compute the multiple correlation R.
In[54]:= R@myx,^ myyD
Out[54]= 0.
- Determine the fraction of s y
2 explained by regression on the
columns of myx.
In[55]:= 0.472217^ ^^2
Out[55]= 0.
- Determine a 95% CI for b 1 that would apply if n were large
(here it is only 5) and specified assumptions on the "errors in
regression" were satisfied.
In[57]:= MatrixForm@betahatCOV@myx,^ myyDD
Out[57]//MatrixForm=
893.385 110.746 - 327.
110.746 491.214 - 357.
In[58]:= - 16.092^ +^8 -^1 ,^1 <^ 1.96^ [email protected]
Out[58]= 8 - 59.5322,^ 27.3482<
The role of n = 5 is concealed in the above calculation of CI.
Had n been large we'd have seen a narrower (more informa-
tive) CI.
- Calculate the predicted value y
`
for independent variable val-
ues
It is the value 1 b
`
0 +^ 2.4^ b
`
1 +^ 3.0^ b
`
2 and^ is^ simply^ calculated
using the "dot product" below.
In[111]:= normalprobabilityplot@resid@myx,^ myyD,^ .02D
Out[111]=
zpercentile
1
2
3
orderstat
Submit a one page printout of your results
(as above) in recitation.
Here are some additional exercises for you to work through.
You will be quizzed on these ideas in recitation but don't hand
them in. I can respond to questions in lecture.
1-4. A model for the strength of a concrete mixture is
strength = b 0 + b 1 agg + b 2 add + b 3 temp + b 4 cure
where
agg is a measure of aggregate in the mix
add is the amount of an additive to the mix
temp is a measure of the temperature during curing
cure is the time allowed to cure before strength testing
- What is the dependent variable? List the independent vari-
ables (including constant term).
- The coefficients obtained from least squares (i.e. regression)
are b
`
0 =^ 28.2,^ b
`
1 =^ 1.22,^ b
`
2 =^ 2.31,^ b
`
3 =^ 0.26,^ b
`
4 =^ 0.36.
Determine the estimated strength for a mix
agg = .3 add = 6.3 temp = 47 cure = 12.
where
agg is a measure of aggregate in the mix
add is the amount of an additive to the mix
temp is a measure of the temperature during curing
cure is the time allowed to cure before strength testing
- What is the dependent variable? List the independent vari-
ables (including constant term).
- The coefficients obtained from least squares (i.e. regression)
are b
`
0 =^ 28.2,^ b
`
1 =^ 1.22,^ b
`
2 =^ 2.31,^ b
`
3 =^ 0.26,^ b
`
4 =^ 0.36.
Determine the estimated strength for a mix
agg = .3 add = 6.3 temp = 47 cure = 12.
- For R = 0.8 give the fraction of s y
2 explained by regression on
the independent variables.
- Suppose sy = 34 and R = 0.8. For an elliptical plot, give the
distribution of the y values in the vertical cylinder (not strip) for
agg = .3 add = 6.3 temp = 47 cure = 12.
Give the mean, sd, and form of the distribution.
- Suppose the residuals are
Here is a normal probability plot of these residuals (required
computer).
normalprobabilityplot@ 8 3.7125, 1.7125, 0.7125,
- 1.3875, - 2.3875, - 0.9875, - 1.9875, 0.6125<, .02D
HW 2-24-09.nb 5
