Question

annual high temperatures in a certain location have been tracked for several years. let x represent the year and y the high temperature. based on the data shown below, calculate the regression line (each value to two decimal places).
y = x+

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Explanation:

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

The formula for the mean of \(x\) values \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), and for \(y\) values \(\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}\).
\(\sum_{i=1}^{15}x_{i}=4 + 5+6+\cdots+18=\frac{(4 + 18)\times15}{2}=165\), \(\bar{x}=\frac{165}{15}=11\)
\(\sum_{i = 1}^{15}y_{i}=30.5+35.8 + 38.9+\cdots+75.4 = 720.5\), \(\bar{y}=\frac{720.5}{15}\approx48.03\)

Step2: Calculate the slope \(b\)

The formula for the slope \(b=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}\)
\(\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})=(4 - 11)(30.5 - 48.03)+(5 - 11)(35.8 - 48.03)+\cdots+(18 - 11)(75.4 - 48.03)=793.1\)
\(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=(4 - 11)^{2}+(5 - 11)^{2}+\cdots+(18 - 11)^{2}=280\)
\(b=\frac{793.1}{280}\approx2.83\)

Step3: Calculate the intercept \(a\)

The formula for the intercept \(a=\bar{y}-b\bar{x}\)
\(a = 48.03-2.83\times11=48.03 - 31.13=16.9\)

Answer:

\(y = 2.83x+16.90\)