Question

The following table displays the mathematics test scores for a random sample of, along with their final SY16C grades.

a.Fit the regression line y=a+bx to the data

and interpret the results.

b.Use the regression equation to determine the SY16C grade for a college student who scored 60 on their achievement test. What would their SY16C grade be?

Mathematics (x)SY16C grade (y)139652437832152464825579264789728738759893456

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:} & X & Y & XY & X^2 & Y^2 \\ \hline & 39 & 65 & 2535 & 1521 & 4225 \\ \hdashline & 43 & 78 & 3354 & 1849 & 6084 \\ \hdashline & 21 & 52 & 1092 & 441 & 2704 \\ \hdashline & 64 & 82 & 5248 & 4096 & 6724 \\ \hdashline & 57 & 92 & 5244 & 3249 & 8464 \\ \hdashline & 47 & 89 & 4183 & 2209 & 7921 \\ \hdashline & 28 & 73 & 2044 & 784 & 5329 \\ \hdashline & 75 & 98 & 7350 & 5625 & 9604 \\ \hdashline & 34 & 56 & 1904 & 1156 & 3136 \\ \hdashline Sum= & 408 & 685 & 32954 & 20930 & 54191 \\ \hdashline \end{array}$$

$$\bar{X}=\dfrac{1}{n}\sum_iX_i=\dfrac{408}{9}\approx45.333333$$

$$\bar{Y}=\dfrac{1}{n}\sum_iY_i=\dfrac{685}{9}\approx76.111111$$

$$SS_{XX}=\sum_iX_i^2-\dfrac{1}{n}(\sum_iX_i)^2$$

$$=20930-\dfrac{408^2}{9}=2434$$

$$SS_{YY}=\sum_iY_i^2-\dfrac{1}{n}(\sum_iY_i)^2$$

$$=54191-\dfrac{685^2}{9}=2054.888889$$

$$SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum_iX_i)(\sum_iY_i)$$

$$=32954-\dfrac{408(685)}{9}=1900.666667$$

$$slope=m=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{1900.666667}{2434}$$

$$=0.7809$$

$$n=\bar{Y}-m\bar{X}=76.111111-0.7809(45.333333)$$

$$=40.7111$$

a.The regression equation is:

$$Y=40.7111+0.7809X$$

$$r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}}\sqrt{SS_{YY}}}$$

$$=\dfrac{1900.666667}{\sqrt{2434}\sqrt{2054.888889}}\approx0.849868$$

Strong positive correlation.

The SY16C grades increases from $40.7111.$ The slope is $0.7809.$

b.

$$Y=40.7111+0.7809(60)$$

$$SY16C=88$$