Question

find the equation for the least squares regression line of the data described below.
ruben wants to figure out how long it usually takes to get through a supermarket
checkout line. for several weeks, he observed the checkout lines he waited in.
ruben counted how many people were ahead of him in each line he joined, x,
and how many minutes it took him to get to the front of that line, y.

round your answers to the nearest thousandth.
y = x +

Explanation:

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

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\) and \(\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}\).
For \(x\) - values: \(x_1 = 1,x_2=1,x_3 = 3,x_4=5,x_5 = 10,x_6=10\). \(\sum_{i=1}^{6}x_{i}=1 + 1+3 + 5+10+10=30\), \(n = 6\), so \(\bar{x}=\frac{30}{6}=5\).
For \(y\) - values: \(y_1 = 8,y_2=8,y_3 = 4,y_4=9,y_5 = 12,y_6=6\). \(\sum_{i=1}^{6}y_{i}=8 + 8+4 + 9+12+6=47\), so \(\bar{y}=\frac{47}{6}\approx7.833\).

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

The formula for \(b_1=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}\)
\((x_1-\bar{x})(y_1-\bar{y})=(1 - 5)(8 - 7.833)=(-4)\times0.167=- 0.668\)
\((x_2-\bar{x})(y_2-\bar{y})=(1 - 5)(8 - 7.833)=(-4)\times0.167=- 0.668\)
\((x_3-\bar{x})(y_3-\bar{y})=(3 - 5)(4 - 7.833)=(-2)\times(-3.833)=7.666\)
\((x_4-\bar{x})(y_4-\bar{y})=(5 - 5)(9 - 7.833)=0\times1.167 = 0\)
\((x_5-\bar{x})(y_5-\bar{y})=(10 - 5)(12 - 7.833)=5\times4.167 = 20.835\)
\((x_6-\bar{x})(y_6-\bar{y})=(10 - 5)(6 - 7.833)=5\times(-1.833)=-9.165\)
\(\sum_{i = 1}^{6}(x_{i}-\bar{x})(y_{i}-\bar{y})=-0.668-0.668 + 7.666+0+20.835-9.165=18\)

\((x_1-\bar{x})^{2}=(1 - 5)^{2}=16\)
\((x_2-\bar{x})^{2}=(1 - 5)^{2}=16\)
\((x_3-\bar{x})^{2}=(3 - 5)^{2}=4\)
\((x_4-\bar{x})^{2}=(5 - 5)^{2}=0\)
\((x_5-\bar{x})^{2}=(10 - 5)^{2}=25\)
\((x_6-\bar{x})^{2}=(10 - 5)^{2}=25\)
\(\sum_{i = 1}^{6}(x_{i}-\bar{x})^{2}=16+16 + 4+0+25+25=86\)

So \(b_1=\frac{18}{86}\approx0.209\)

Step3: Calculate the intercept \(b_0\)

The formula for \(b_0=\bar{y}-b_1\bar{x}\)
\(b_0 = 7.833-0.209\times5=7.833 - 1.045=6.788\)

Answer:

\(y = 0.209x+6.788\)