1) y = 0.19X + 157.3 (option C)
2) Yes, it is a good fit
3) i) 165
ii) 168
iii) 172
Explanation:1)2) Regression equation is given as:
[tex]\begin{gathered} ŷ=bX+a \\ \text{where b = slope} \\ a\text{ = y-intercept} \end{gathered}[/tex]To determine the regression equation, we will use a regression calculator:
[tex]\begin{gathered} \text{The regression equation of the table is given as:} \\ ŷ=0.19X+157.3\text{ (option C)} \end{gathered}[/tex]2) The correlation coefficient = r = 0.98510411
A correlation coefficient that is close to 1 is a strong correlation and it makes it a good fit.
Yes, it is a good fit
[tex]\begin{gathered} 3)\text{ The regression equation:} \\ ŷ=0.19X+157.3\text{ } \\ X\text{ = resting heart rate} \\ ŷ=max\text{ imum heart rate} \end{gathered}[/tex]To get the maximum heart rate for a resting heart, we will substitute X with the values given
For resting heart rate of 42:
[tex]\begin{gathered} X\text{ = 42} \\ ŷ=0.19(42)+157.3\text{ = }165.28 \\ Maximum\text{ heart rate for 42 = }165\text{ (nearest integer)} \end{gathered}[/tex]For resting heart rate of 57:
[tex]\begin{gathered} X\text{ = 57} \\ ŷ=0.19(57)+157.3\text{ = }168.13 \\ Maximum\text{ heart rate for 57 = }168\text{ (nearest integer)} \end{gathered}[/tex]For resting heart rate of 78:
[tex]\begin{gathered} X\text{ = 78} \\ ŷ=0.19(78)+157.3\text{ = }172.12 \\ Maximum\text{ heart rate for 78 = }172\text{ (nearest integer)} \end{gathered}[/tex]