4-13 Students in a management science class have just re- ceived their grades on the first test. The instructor has provided information about the first test grades in some previous classes as well as the final average for the same students. Some of these grades have been sampled and are as follows:STUDENT 1 2 3 4 5 6 7 8 9 1st test grade 98 77 88 80 96 61 66 95 69 Finalaverage 93 78 84 73 84 64 64 95 76 (a) Develop a regression model that could be used to predict the final average in the course based on the first test grade. (b) Predict the final average of a student who made an 83 on the first test. (c) Give the values of r and r2 for this model. Inter- pret the value of r2 in the context of this problem

Answer:a) y = 0.74x + 18.99; b) 80; c) r = 0.92, r² = 0.85; r² tells us that 85% of the variance in the dependent variable, the final average, is predictable from the independent variable, the first test score.Step-by-step explanation:For part a,We first plot the data using a graphing calculator.  We then run a linear regression on the data.In the form y = ax + b, we get an a value that rounds to 0.74 and a b value that rounds to 18.99.  This gives us the equationy = 0.74x + 18.99.For part b,To find the final average of a student who made an 83 on the first test, we substitute 83 in place of x in our regression equation:y = 0.74(83) + 18.99y = 61.42 + 18.99 = 80.41Rounded to the nearest percent, this is 80.For part c,The value of r is 0.92.  This tells us that the line is a 92% fit for the data.The value of r² is 0.85.  This is the coefficient of determination; it tells us how much of the dependent variable can be predicted from the independent variable.