find the equation for the least squares regression line of the data described below. a professor wants to…

find the equation for the least squares regression line of the data described below. a professor wants to know if a students first exam score can be used to accurately predict his or her final exam score. he looked at how some of his previous students did on their first exam, x. he also looked at their final exam scores, y. both exams were graded out of 100.\nfirst exam score final exam score\n70 61\n71 50\n75 51\n85 64\n89 92\nround your answers to the nearest thousandth.\ny = x +

Answer

Explanation:

Step1: Calculate means of x and y

Let (x_i) be first - exam scores and (y_i) be final - exam scores. (n = 5) (\bar{x}=\frac{70 + 71+75+85+89}{5}=\frac{390}{5}=78) (\bar{y}=\frac{61 + 50+51+64+92}{5}=\frac{318}{5}=63.6)

Step2: Calculate numerator and denominator for slope (b1)

(\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=(70 - 78)(61 - 63.6)+(71 - 78)(50 - 63.6)+(75 - 78)(51 - 63.6)+(85 - 78)(64 - 63.6)+(89 - 78)(92 - 63.6)) (=(- 8)\times(-2.6)+(-7)\times(-13.6)+(-3)\times(-12.6)+7\times0.4 + 11\times28.4) (=20.8+95.2 + 37.8+2.8+312.4=469) (\sum_{i = 1}^{n}(x_i-\bar{x})^2=(70 - 78)^2+(71 - 78)^2+(75 - 78)^2+(85 - 78)^2+(89 - 78)^2) (=(-8)^2+(-7)^2+(-3)^2+7^2+11^2) (=64 + 49+9+49+121=292) (b_1=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}=\frac{469}{292}\approx1.606)

Step3: Calculate y - intercept (b0)

(b_0=\bar{y}-b_1\bar{x}=63.6-1.606\times78) (=63.6 - 125.268=-61.668)

Answer:

(y = 1.606x-61.668)