Question

find the equation for the least - squares regression line of the data described below. herman and molly are judges for the lansberry ice skating federation. due to claims of malfeasance at recent competitions, a reporter is investigating the relationship between the scores awarded by the two judges. she has collected the scores awarded by herman, x, and molly, y, for each performance in the last competition.
hermans score mollys score
1.72 0.25
1.84 2.44
2.73 1.99
2.85 4.63
3.30 4.60
round your answers to the nearest thousandth.
y =

Explanation:

Step1: Calculate the means of x and y

Let \(x\) be Herman's scores and \(y\) be Molly's scores.
\(n = 5\)
\(\bar{x}=\frac{1.72 + 1.84+2.73+2.85+3.30}{5}=\frac{12.44}{5}=2.488\)
\(\bar{y}=\frac{0.25 + 2.44+1.99+4.63+4.60}{5}=\frac{14.91}{5}=2.982\)

Step2: Calculate the numerator and denominator for the slope \(b_1\)

\(\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=(1.72 - 2.488)(0.25-2.982)+(1.84 - 2.488)(2.44 - 2.982)+(2.73-2.488)(1.99 - 2.982)+(2.85 - 2.488)(4.63-2.982)+(3.30 - 2.488)(4.60 - 2.982)\)
\(=(- 0.768)(-2.732)+(-0.648)(-0.542)+(0.242)(-0.992)+(0.362)(1.648)+(0.812)(1.618)\)
\(=2.098176+0.351216-0.240064 + 0.596576+1.313816\)
\(=4.11972\)
\(\sum_{i=1}^{n}(x_i-\bar{x})^2=(1.72 - 2.488)^2+(1.84 - 2.488)^2+(2.73-2.488)^2+(2.85 - 2.488)^2+(3.30 - 2.488)^2\)
\(=(-0.768)^2+(-0.648)^2+(0.242)^2+(0.362)^2+(0.812)^2\)
\(=0.589824+0.419904+0.058564+0.131044+0.659344\)
\(=1.85868\)
\(b_1=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}=\frac{4.11972}{1.85868}\approx2.216\)

Step3: Calculate the y - intercept \(b_0\)

\(b_0=\bar{y}-b_1\bar{x}=2.982-2.216\times2.488\)
\(=2.982-5.513408\approx - 2.531\)

Answer:

\(y = 2.216x-2.531\)