Question

use the given data to find the equation of the regression line. round the final values to three significant digits, if necessary. ten students in a graduate program were randomly selected. their grade point averages (gpas) when they entered the program were between 3.5 and 4.0. the following data were obtained regarding their gpas on entering the program versus their current gpas. entering gpa current gpa 3.5 3.6 3.8 3.7 3.6 3.9 3.6 3.6 3.5 3.9 3.9 3.8 4.0 3.7 3.9 3.9 3.5 3.8 3.7 4.0 \\( \hat { y } = 5.81 + 0.497 x \\) \\( \hat { y } = 3.67 + 0.0313 x \\) \\( \hat { y } = 2.51 + 0.329 x \\) \\( \hat { y } = 4.91 + 0.0212 x \\)

Explanation:

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

Let \(x\) values be \(x_1 = 3.5,x_2 = 3.8,x_3 = 3.6,x_4 = 3.6,x_5 = 3.5,x_6 = 3.9,x_7 = 4.0,x_8 = 3.9,x_9 = 3.5,x_{10}=3.7\)
Let \(y\) values be \(y_1 = 3.6,y_2 = 3.7,y_3 = 3.9,y_4 = 3.6,y_5 = 3.9,y_6 = 3.8,y_7 = 3.7,y_8 = 3.9,y_9 = 3.8,y_{10}=4.0\)
\(\bar{x}=\frac{\sum_{i = 1}^{10}x_i}{10}=\frac{3.5 + 3.8+3.6+3.6+3.5+3.9+4.0+3.9+3.5+3.7}{10}=\frac{37}{10}=3.7\)
\(\bar{y}=\frac{\sum_{i = 1}^{10}y_i}{10}=\frac{3.6 + 3.7+3.9+3.6+3.9+3.8+3.7+3.9+3.8+4.0}{10}=\frac{37.9}{10}=3.79\)

Step2: Calculate \(S_{xy}=\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})\) and \(S_{xx}=\sum_{i = 1}^{n}(x_i-\bar{x})^2\)

\(S_{xy}=(3.5 - 3.7)(3.6-3.79)+(3.8 - 3.7)(3.7 - 3.79)+(3.6-3.7)(3.9 - 3.79)+(3.6-3.7)(3.6 - 3.79)+(3.5 - 3.7)(3.9 - 3.79)+(3.9 - 3.7)(3.8 - 3.79)+(4.0 - 3.7)(3.7 - 3.79)+(3.9 - 3.7)(3.9 - 3.79)+(3.5 - 3.7)(3.8 - 3.79)+(3.7 - 3.7)(4.0 - 3.79)\)
\(S_{xy}=(- 0.2)(-0.19)+(0.1)(-0.09)+(-0.1)(0.11)+(-0.1)(-0.19)+(-0.2)(0.11)+(0.2)(0.01)+(0.3)(-0.09)+(0.2)(0.11)+(-0.2)(0.01)+(0)(0.21)\)
\(S_{xy}=0.038-0.009 - 0.011 + 0.019-0.022 + 0.002-0.027+0.022-0.002+0=0.01\)

\(S_{xx}=(3.5 - 3.7)^2+(3.8 - 3.7)^2+(3.6-3.7)^2+(3.6-3.7)^2+(3.5 - 3.7)^2+(3.9 - 3.7)^2+(4.0 - 3.7)^2+(3.9 - 3.7)^2+(3.5 - 3.7)^2+(3.7 - 3.7)^2\)
\(S_{xx}=(-0.2)^2+(0.1)^2+(-0.1)^2+(-0.1)^2+(-0.2)^2+(0.2)^2+(0.3)^2+(0.2)^2+(-0.2)^2+(0)^2\)
\(S_{xx}=0.04 + 0.01+0.01+0.01+0.04+0.04+0.09+0.04+0.04+0=0.32\)

Step3: Calculate the slope \(b_1=\frac{S_{xy}}{S_{xx}}\) and the intercept \(b_0=\bar{y}-b_1\bar{x}\)

\(b_1=\frac{S_{xy}}{S_{xx}}=\frac{0.01}{0.32}=0.03125\approx0.0313\)
\(b_0=\bar{y}-b_1\bar{x}=3.79-0.0313\times3.7 = 3.79-0.11581=3.67419\approx3.67\)

Answer:

\(\hat{y}=3.67 + 0.0313x\)