Problem
Listed below are systolic blood pressure measurements (in $\mathrm{mm} \mathrm{Hg}$ ) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor $(\mathrm{x})$ variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is $80 \mathrm{~mm}$ Hg. Use a significance level of 0.05 . \begin{tabular}{l|rrrrr} Right Arm & 100 & 99 & 91 & 75 & 75 \\ \hline Left Arm & 174 & 168 & 180 & 146 & 147 \end{tabular} Click the icon to view the critical values of the Pearson correlation coefficient $r$ The regression equation is $\hat{y}=67.1+1.1 x$. (Round to one decimal place as needed.) Given that the systolic blood pressure in the right arm is $80 \mathrm{~mm} \mathrm{Hg}$, the best predicted systolic blood pressure in the left arm is $\square \mathrm{mm} \mathrm{Hg}$. (Round to one decimal place as needed.)
Solution
Step 1 ：Given the regression equation \(\hat{y}=67.1+1.1x\), where \(\hat{y}\) is the predicted systolic blood pressure in the left arm and \(x\) is the systolic blood pressure in the right arm.
Step 2 ：Substitute \(x=80\) into the regression equation and solve for \(\hat{y}\).
Step 3 ：Calculate \(\hat{y}=67.1+1.1*80\).
Step 4 ：The best predicted systolic blood pressure in the left arm is \(\boxed{155.1 \mathrm{~mm} \mathrm{Hg}}\).
