C) y=0.16x+ 159.8
and r =0.956
1) Let's plot that trendline and scatterplot with the given data:
2) Now, let's find out the Regression Line equation by making another table for that:
Note that x is for Resting and y for the Maximum. So we can find the slope through this formula:
[tex]m=\frac{n\Sigma xy-\Sigma x\Sigma y}{n\Sigma x^{2}-(\Sigma x)^2}=\frac{5\cdot60010-350\cdot855}{5\cdot25500-(350)^2}=0.16[/tex]
We can now find the linear coefficient (b):
[tex]b=\frac{\Sigma y-m\cdot\Sigma x}{n}=\frac{855-0.16\cdot350}{5}=159.8[/tex]
Hence, the equation for this Regression Line is:
[tex]\hat{y}=0.16x+159.8[/tex]
Letter C.
For the Correlation Coefficient let's make use of another formula plugging into that the given values. The Summation is the last line on the table:
[tex]\begin{gathered} r=\frac{n\Sigma xy-\Sigma x\Sigma y}{\sqrt[]{\lbrack n\Sigma x^{2}-(\Sigma x)^2\rbrack\lbrack n\Sigma y^{2}-(\Sigma y)^2\rbrack}}=\frac{5\cdot60010-350\cdot855}{\sqrt[]{\lbrack5\cdot25500-(350)^2\rbrack\cdot\lbrack5\cdot146233-(855)^2}} \\ r=0.956 \end{gathered}[/tex]
2.3) The predicted maximum heart rate at a resting heart rate of 41 is found out when we plug it into the formula.
y = 016x +159.8 Plug into x
y= 0.16(41) +159.8
y= 166.36 Roundoff to the nearest integer
y=166
2.4) The predicted maximum heart rate at a resting heart rate of 56 is also found out by plugging into x, the value of 56
y = 016x +159.8
y= 0.16 (56) +159.8
y= 168.76 Round off to the nearest integer
y= 169
2.5) And finally, proceeding likewise we can find out the Maximum Heart Rate predicted
y = 016x +159.8
y= 0.16 *(77) +159.8
y = 172.12 Round off to the nearest integer
y= 172
3) Hence, the answers are:
y=0.16x+ 159.8
and r =0.956
Yes. It's a good fit for it reaches the maximum number of points.
Maximum heart rate of 166
Maximum heart rate of 169
Maximum heart rate of 172
