Question

Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary.

Ten students in a graduate program were randomly selected. Their grade point averages (GPAs) when they entered the program were between 3.5 and 4.0. The following data were obtained regarding their GPAs on entering the program versus their current GPAs.

Entering GPA Current GPA

Answer 1

The equation of the regression line is
`\bar y` = 2.51 + 0.329x. This means that for every one-unit increase in entering GPA, the current GPA is predicted to increase by 0.329 units, on average. Hence the correct option is c.

The mean of x (entering GPA) and y (current GPA):

Mean of x = (3.5 + 3.8 + 3.6 + 3.6 + 3.5 + 3.9 + 4.0 + 3.9 + 3.5 + 3.7) / 10

= 3.71

Mean of y = (3.6 + 3.7 + 3.9 + 3.6 + 3.9 + 3.8 + 3.7 + 3.9 + 3.8 + 4.0) / 10

= 3.79

Calculate the deviations from the mean for each data point:

`x_i - \bar x` and
`y_i - \bar y` for each data point

The sum of squares of the deviations from the mean for x and y:

Σ
`(x_i - \bar x)^2` = 0.7704

Σ
`(y_i - \bar y)^2`= 0.7569

The covariance between x and y:

`\sum (x_i - \bar x)(y_i - \bar y)` = 0.2445

The slope (β) of the regression line:

β =
`(\sum(x_i - \bar x)(y_i - \bar y))/(\sum (x_i - \bar x)^2)` = 0.329

The y-intercept (α) of the regression line:

α =
`\bar y` - β x
`\bar x` = 3.79 - 0.329 x 3.71 = 2.51

The equation of the regression line:

`\bar y` = α + βx = 2.51 + 0.329x

Hence the correct option is c.