Chapter 3, Problem 6CRE

(a)

To determine

To find: The equation for the appropriate least-squares regression line.

Answer to Problem 6CRE

The equation of least-squares regression line isY=30.2+0.16X

Explanation of Solution

Given:

r = 0.6 ,

The Pre −exam:

mean 280 and standard deviation 30,

The final-exam:

mean 75 and standard deviation 8.

Calculation:

The slope of the least-square regression line of Final exam scores based on pre-exam scores is calculated as follows:

  $b=0.6\times \frac{8}{30}$

=0.16

Also, the Y-intercept is calculated as follows:

  $a=\overline{Y}-b\overline{X}$

=75-0.15 x 280

=30.2

Hence, the equation of the least-squares line is:

Y=30.2+0.16X

Here, the slope b of above regression line (0.16) is the rate at which the predicted response Y changes among the line as the explanatory variable X changes. Specifically, b is the predicted change in Y when X increases by 1 unit.

Here b is positive; this shows of pre-exam totals increases by 1 mark then final examination score increases by 0.16 marks.

Conclusion:

Therefore, th equation of least-squares regression line is Y=30.2+0.16X

(b)

To determine

To predict: J’s final-exam score by using the regression line

Answer to Problem 6CRE

J’s final-exam scoreis 78.2.

Explanation of Solution

Calculation:

It is given that Julie’s pre-exam total was 300.

Then, the predicted final examination score is calculated as follows:

  $\widehat{Y}=30.2+0.16X$

  $=30.2+0.15\times 300$

  $=78.2$

Julie’s predicted final examination score is 78.2.

Conclusion:

Hence, J’s final-exam score is 78.2.

(c)

To determine

the value of $r^2$ .

Answer to Problem 6CRE

The value is R-square = 36%

Explanation of Solution

Calculation:

The correlation between the pre-exam totals and final examination score is given as 0.6.

That is, r=0.6

Then $r^2={(0.6)}^2$

=0.36

That is, R-square=36%

Based on the value of R-square, we can say that 36% of the variation in final examination score is accounted for by the least-squares regression line of final examination score to pre-exam total.

That implies, 64% of the variation in final examination scores is unaccounted for by the linear relationship for these data. Professor Friedman’s prediction of Julie’s final examination score may not be very accurate.

Conclusion:

Hence,the value is R-square = 36%.