eigenmath/86.tex
George Weigt c7b3ee7cc4 add
2006-05-01 15:36:13 -07:00

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\beginsection
1a) What is the independent variable?
Length of femur
\beginsection
1b) What is the dependent variable?
Height of person
\beginsection
1c) What data would you need to collect?
How do you want the independent variable to be distributed?
The data pairs would consist of a person's height
and femur length.
We expect $\rho>0$. In other words, we expect
taller people to have longer femurs.
\beginsection
1d) What is the necessary assumption you need to make?
A general belief that the linear
correlation technique is valid for this case?
\beginsection
2a) What is the value for the Regression's degree of freedom?
Regression has 1 degree of freedom.
\beginsection
2b) What is the value for the Residual's degree of freedom?
The residual error has $N-2=57$ degrees of freedom.
\beginsection
2c) What is the value for the Regression's Sum of Squares?
SSR = sum of squares regression\par
MSR = mean square of regression\par
We have SSR = MSR = 72.
\beginsection
2d) What is the value for Residual's Sum of Squares?
SSE = sum of squares (residual) error\par
SST = sum of squares total\par
${\rm SSE}={\rm SST}-{\rm SSR}=300-72=228$
\beginsection
2e) What is the estimate value for the variance of random error?
See the PDF (note 8) p. 9.
$$S^2={{\rm SSE}\over n-2}={228\over 57}=4$$
$$\sigma^2={2S^4\over n-2}={2\cdot4^2\over57}=0.5614$$
\beginsection
2f) What is the value for the $F$ statistic?
See p. 13 of note08.
$F^*=MSR/MSE=72/4=18$
\beginsection
2g) What is the value for Coefficient of Determination?
See slide number 8 of lecture082.
$$\eqalign{
\hbox{Explained Variation}&={\rm SSR}=\sum_{i=1}^n(\hat Y_i-\bar Y)^2=72\cr
\hbox{Unexplained Variation}&={\rm SSE}=\sum_{i=1}^n(Y_i-\hat Y_i)^2=228\cr
\hbox{Total Variation}&={\rm SST}=\sum_{i=1}^n(Y_i-\bar Y)^2=300\cr
}$$
$$\hbox{Coefficient of Determination}=R^2={{\rm SSR}\over{\rm SST}}={72\over300}=0.24$$
\beginsection
2h) What is the correlation coefficient?
See slide 12 of lecture082.
$$r^2=R^2$$
$$r=\sqrt{0.24}=0.49$$
\beginsection
2i) What is the conclusion?
Find the 95\% confidence interval for $\rho$.
$$n=59$$
$$r=0.49$$
$$Z'={1\over2}\ln((1+r)/(1-r))=0.5361$$
$$Z_{LB}=Z'-Z_{\alpha/2}/\sqrt{59-3}=0.2742$$
$$Z_{UB}=Z'+Z_{\alpha/2}/\sqrt{59-3}=0.7980$$
$${\exp(2Z_{LB})-1\over\exp(2Z_{LB}+1}={0.7305\over2.7305}=0.2675$$
$${\exp(2Z_{UB})-1\over\exp(2Z_{UB}+1}={3.9333\over5.9333}=0.6629$$
$$\hbox{confidence interval}=(0.2675,0.6629)$$
\end