Question

run a regression analysis on the following data set with y as the response variable.

x	y
56.9	73.1
89.3	81.7
86.4	79.8
76	77.5
79.3	80.2
81.7	79.2
67.2	78.2

what is the predicted value of the response variable when the explanatory variable has value 69.6?
y =
round to 1 decimal place.

Explanation:

Step1: Calculate sums

Let \(n = 8\).
\(\sum_{i = 1}^{n}x_{i}=62 + 56.9+89.3+86.4+76+79.3+81.7+67.2 = 608\)
\(\sum_{i = 1}^{n}y_{i}=77.3+73.1+81.7+79.8+77.5+80.2+79.2+78.2 = 627\)
\(\sum_{i = 1}^{n}x_{i}^{2}=62^{2}+56.9^{2}+89.3^{2}+86.4^{2}+76^{2}+79.3^{2}+81.7^{2}+67.2^{2}\)
\(=3844 + 3237.61+7974.49+7464.96+5776+6288.49+6674.89+4515.84 = 45776.37\)
\(\sum_{i = 1}^{n}x_{i}y_{i}=62\times77.3+56.9\times73.1+89.3\times81.7+86.4\times79.8+76\times77.5+79.3\times80.2+81.7\times79.2+67.2\times78.2\)
\(=4792.6+4159.39+7295.81+6894.72+5890+6359.86+6470.64+5255.04 = 46018.06\)

Step2: Calculate slope \(b_1\)

The formula for the slope \(b_1\) of the regression - line \(y = b_0 + b_1x\) is \(b_1=\frac{n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}}{n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}}\)
\(n\sum_{i = 1}^{n}x_{i}y_{i}- \sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}=8\times46018.06-608\times627\)
\(=368144.48 - 381216=-13071.52\)
\(n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}=8\times45776.37-608^{2}\)
\(=366210.96 - 369664=-3453.04\)
\(b_1=\frac{-13071.52}{-3453.04}\approx3.786\)

Step3: Calculate intercept \(b_0\)

The formula for the intercept \(b_0\) is \(b_0=\bar{y}-b_1\bar{x}\), where \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\) and \(\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}\)
\(\bar{x}=\frac{608}{8}=76\)
\(\bar{y}=\frac{627}{8}=78.375\)
\(b_0 = 78.375-3.786\times76\)
\(=78.375 - 287.736=-209.361\)

The regression equation is \(y=-209.361 + 3.786x\)

Step4: Predict \(y\) value

When \(x = 69.6\), \(y=-209.361+3.786\times69.6\)
\(y=-209.361 + 263.5056\)
\(y = 54.1446\approx54.1\)

Answer:

\(54.1\)