Question

1. using the linear best - fit model for the data, what is the predicted walking heart rate of a girl whose running heart rate is 130 bpm? (round your answer to the hundredths place.) type a response

Explanation:

Step1: Find Linear Regression Equation

First, we identify the data points. Let \( x \) be Walking Heart Rate (WHR) and \( y \) be Running Heart Rate (RHR). The data points are: \((66, 128)\), \((72, 136)\), \((74, 134)\), \((78, 138)\), \((80, 142)\), \((84, 146)\), \((86, 148)\), \((88, 152)\).

We calculate the mean of \( x \) (\(\bar{x}\)) and mean of \( y \) (\(\bar{y}\)):

\(\bar{x}=\frac{66 + 72+74 + 78+80 + 84+86 + 88}{8}=\frac{628}{8} = 78.5\)

\(\bar{y}=\frac{128+136 + 134+138+142+146+148+152}{8}=\frac{1124}{8}=140.5\)

Next, calculate the slope \( m \) using the formula \( m=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}\)

Calculate \((x_i-\bar{x})(y_i-\bar{y})\) and \((x_i-\bar{x})^2\) for each point:

• For \((66,128)\): \((66 - 78.5)(128 - 140.5)=(- 12.5)(-12.5)=156.25\); \((-12.5)^2 = 156.25\)
• For \((72,136)\): \((72 - 78.5)(136 - 140.5)=(-6.5)(-4.5)=29.25\); \((-6.5)^2=42.25\)
• For \((74,134)\): \((74 - 78.5)(134 - 140.5)=(-4.5)(-6.5)=29.25\); \((-4.5)^2 = 20.25\)
• For \((78,138)\): \((78 - 78.5)(138 - 140.5)=(-0.5)(-2.5)=1.25\); \((-0.5)^2=0.25\)
• For \((80,142)\): \((80 - 78.5)(142 - 140.5)=(1.5)(1.5)=2.25\); \((1.5)^2 = 2.25\)
• For \((84,146)\): \((84 - 78.5)(146 - 140.5)=(5.5)(5.5)=30.25\); \((5.5)^2=30.25\)
• For \((86,148)\): \((86 - 78.5)(148 - 140.5)=(7.5)(7.5)=56.25\); \((7.5)^2 = 56.25\)
• For \((88,152)\): \((88 - 78.5)(152 - 140.5)=(9.5)(11.5)=109.25\); \((9.5)^2=90.25\)

Sum of \((x_i-\bar{x})(y_i-\bar{y})\): \(156.25+29.25 + 29.25+1.25+2.25+30.25+56.25+109.25 = 414\)

Sum of \((x_i-\bar{x})^2\): \(156.25+42.25+20.25 + 0.25+2.25+30.25+56.25+90.25=398\)

So, \( m=\frac{414}{398}\approx1.0399\)

Then, the y - intercept \( b=\bar{y}-m\bar{x}=140.5-1.0399\times78.5\)

\(1.0399\times78.5\approx81.632\)

\(b = 140.5 - 81.632=58.868\)

The linear equation is \( y = 1.0399x+58.868\) (here \( y \) is Running Heart Rate, \( x \) is Walking Heart Rate. We need to find \( x \) when \( y = 130\))

Step2: Solve for \( x \) when \( y = 130\)

We have the equation \(130=1.0399x + 58.868\)

Subtract \(58.868\) from both sides: \(130 - 58.868=1.0399x\)

\(71.132 = 1.0399x\)

Then \(x=\frac{71.132}{1.0399}\approx68.39\)

Answer:

\(68.39\)