101 lines
No EOL
2.5 KiB
TeX
101 lines
No EOL
2.5 KiB
TeX
\parindent=0pt
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\beginsection
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1a) What is the independent variable?
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Length of femur
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\beginsection
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1b) What is the dependent variable?
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Height of person
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\beginsection
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1c) What data would you need to collect?
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How do you want the independent variable to be distributed?
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The data pairs would consist of a person's height
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and femur length.
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We expect $\rho>0$. In other words, we expect
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taller people to have longer femurs.
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\beginsection
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1d) What is the necessary assumption you need to make?
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A general belief that the linear
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correlation technique is valid for this case?
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\beginsection
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2a) What is the value for the Regression's degree of freedom?
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Regression has 1 degree of freedom.
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\beginsection
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2b) What is the value for the Residual's degree of freedom?
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The residual error has $N-2=57$ degrees of freedom.
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\beginsection
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2c) What is the value for the Regression's Sum of Squares?
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SSR = sum of squares regression\par
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MSR = mean square of regression\par
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We have SSR = MSR = 72.
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\beginsection
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2d) What is the value for Residual's Sum of Squares?
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SSE = sum of squares (residual) error\par
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SST = sum of squares total\par
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${\rm SSE}={\rm SST}-{\rm SSR}=300-72=228$
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\beginsection
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2e) What is the estimate value for the variance of random error?
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See the PDF (note 8) p. 9.
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$$S^2={{\rm SSE}\over n-2}={228\over 57}=4$$
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$$\sigma^2={2S^4\over n-2}={2\cdot4^2\over57}=0.5614$$
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\beginsection
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2f) What is the value for the $F$ statistic?
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See p. 13 of note08.
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$F^*=MSR/MSE=72/4=18$
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\beginsection
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2g) What is the value for Coefficient of Determination?
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See slide number 8 of lecture082.
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$$\eqalign{
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\hbox{Explained Variation}&={\rm SSR}=\sum_{i=1}^n(\hat Y_i-\bar Y)^2=72\cr
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\hbox{Unexplained Variation}&={\rm SSE}=\sum_{i=1}^n(Y_i-\hat Y_i)^2=228\cr
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\hbox{Total Variation}&={\rm SST}=\sum_{i=1}^n(Y_i-\bar Y)^2=300\cr
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}$$
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$$\hbox{Coefficient of Determination}=R^2={{\rm SSR}\over{\rm SST}}={72\over300}=0.24$$
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\beginsection
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2h) What is the correlation coefficient?
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See slide 12 of lecture082.
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$$r^2=R^2$$
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$$r=\sqrt{0.24}=0.49$$
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\beginsection
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2i) What is the conclusion?
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Find the 95\% confidence interval for $\rho$.
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$$n=59$$
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$$r=0.49$$
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$$Z'={1\over2}\ln((1+r)/(1-r))=0.5361$$
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$$Z_{LB}=Z'-Z_{\alpha/2}/\sqrt{59-3}=0.2742$$
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$$Z_{UB}=Z'+Z_{\alpha/2}/\sqrt{59-3}=0.7980$$
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$${\exp(2Z_{LB})-1\over\exp(2Z_{LB}+1}={0.7305\over2.7305}=0.2675$$
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$${\exp(2Z_{UB})-1\over\exp(2Z_{UB}+1}={3.9333\over5.9333}=0.6629$$
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$$\hbox{confidence interval}=(0.2675,0.6629)$$
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\end |