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The data in the chart shows the maximum and corresponding resting heart rate of a healthy 20-year old man. Fit a regression line to the data.

The data in the chart shows the maximum and corresponding resting heart rate of a-example-1
User Ian Mackinnon
by
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1 Answer

18 votes
18 votes

1) y = 0.19X + 157.3 (option C)

2) Yes, it is a good fit

3) i) 165

ii) 168

iii) 172

Step-by-step explanation:

1)2) Regression equation is given as:


\begin{gathered} ŷ=bX+a \\ \text{where b = slope} \\ a\text{ = y-intercept} \end{gathered}

To determine the regression equation, we will use a regression calculator:


\begin{gathered} \text{The regression equation of the table is given as:} \\ ŷ=0.19X+157.3\text{ (option C)} \end{gathered}

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


\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}

To get the maximum heart rate for a resting heart, we will substitute X with the values given

For resting heart rate of 42:


\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}

For resting heart rate of 57:


\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}

For resting heart rate of 78:


\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}

User J B
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