Question

Each year, students in an elementary school take a standardized math test at the end of the school year. For a class of fourth-graders, the average score was 55.1 with a standard deviation of 12.3. In the third grade, these same students had an average score of 61.7 with a standard deviation of 14.0. The correlation between the two sets of scores is r = 0.95. Calculate the equation of the least-squares regression line for predicting a fourth-grade score from a third-grade score.

a. y^=3.58+0.835x

b. y^=15.69+0.835x

c. y^=2.19+1.08x

d. y^=−11.54+1.08x

Short answer

The correct option is (a)$\stackrel{⏜}{y}=3.58+0.835x$

Step-by-step solution

Given information

$\stackrel{⏜}{y}=55.1,s_y=12.3,\overline{x}=61.7,s_x=14,r=0.95$

Explanation

The slope could be calculated as:

$$\begin{gathered} b=r\times \frac{s_y}{s_x} \\ =0.95(\frac{12.3}{14.0}) \\ =0.835 \end{gathered}$$

The intercept could be calculated as:

$$\begin{gathered} a=\overline{y}-\overline{b_x} \\ =55.1-0.835(61.7) \\ =3.60 \end{gathered}$$

Thus, the equation of the regression line is:

$$\begin{gathered} \stackrel{⏜}{y}=a+bx \\ =3.60+0.835x \end{gathered}$$

Hence, the correct option is (a).