Answer to Question #154454 in Statistics and Probability for Alexander

Question

Scores made by students in a statistics class in the mid-term and final examination are given here. Develop a regression equation which may be used to predict final examination scores from the midterm score. (Hint: The regression equation is given as 𝑌̂ = 𝑎 + 𝑏𝑥.)

Student 1 2 3 4 5 6 7 8 9 10

Mid Term (x) 94 60 96 94 81 40 70 50 72 70

Final (y)(y) 87 73 94 87 76 61 77 71 84 77

Answer 1

$n=10\\ \sum x_{i}=94+60+...+70=727\\ \sum y_{i}=87+73+...+77=787\\ \sum x_{i}^{2}=94^{2}+60^{2}+...+70^{2}=56133\\ \sum x_{i}\times \ y_{i} \\=94\times87+60\times73+...+70\times77=58734\\ \hat b= \frac{n\sum x_{i}\times \ y_{i}-\sum x_{i}\sum y_{i}} {n\sum x_{i}^{2}-( \sum x_{i})^{2}}\\ =\frac{10(58734)-727(787)}{10(56133)-727^{2}}\approx 0.4631\\ \bar x=\frac{ \sum x_{i}} {n} =\frac{ 727} {10}=72.7\\ \bar y=\frac{ \sum y_{i}} {n} =\frac{ 787} {10}=78.7\\ \hat a=\bar y-\hat b\bar x\\ =87.7-0.4631(72.7)\approx 45.033\\ \therefore \hat y_{i}=45.033+0.4631x_{i}$