Question

find the equation for the least squares regression line of the data described below. jenna keeps detailed information about her model airplane collection. the data includes not only the characteristics of each model, but also the time it took to assemble it. she is thinking of buying a new model airplane and decided to look at this information to understand how long assembly might take. from her data, jenna found the number of pieces in each model, x, and how many minutes each took to assemble, y. pieces assembly time 40 38 49 48 54 55 69 51 93 55

Explanation:

Step1: Calculate the means of \(x\) and \(y\)

Let \(x_1 = 40,x_2=49,x_3 = 54,x_4=69,x_5 = 93\) and \(y_1 = 38,y_2=48,y_3 = 55,y_4=51,y_5 = 55\).
The mean of \(x\), \(\bar{x}=\frac{40 + 49+54+69+93}{5}=\frac{305}{5}=61\).
The mean of \(y\), \(\bar{y}=\frac{38 + 48+55+51+55}{5}=\frac{247}{5}=49.4\).

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

The numerator \(\sum_{i = 1}^{5}(x_i-\bar{x})(y_i - \bar{y})\):
\((40 - 61)(38-49.4)+(49 - 61)(48 - 49.4)+(54-61)(55 - 49.4)+(69-61)(51 - 49.4)+(93-61)(55 - 49.4)\)
\(=(- 21)(-11.4)+(-12)(-1.4)+(-7)(5.6)+(8)(1.6)+(32)(5.6)\)
\(=239.4 + 16.8-39.2+12.8+179.2\)
\(=399\)
The denominator \(\sum_{i=1}^{5}(x_i-\bar{x})^2\):
\((40 - 61)^2+(49 - 61)^2+(54 - 61)^2+(69 - 61)^2+(93 - 61)^2\)
\(=(-21)^2+(-12)^2+(-7)^2+(8)^2+(32)^2\)
\(=441+144 + 49+64+1024\)
\(=1722\)
The slope \(b_1=\frac{\sum_{i = 1}^{5}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{5}(x_i-\bar{x})^2}=\frac{399}{1722}\approx0.232\)

Step3: Calculate the intercept \(b_0\)

The intercept \(b_0=\bar{y}-b_1\bar{x}=49.4-0.232\times61\)
\(=49.4 - 14.152=35.248\)
The equation of the least - squares regression line is \(y=b_0 + b_1x\), so \(y = 35.248+0.232x\)

Answer:

\(y=35.248 + 0.232x\)