Question

The table below gives the number of hours spent unsupervised each day as well as the overall grade averages for seven randomly selected middle school students. Using this data, consider the equation of the regression line, yˆ=b0+b1x for predicting the overall grade average for a middle school student based on the number of hours spent unsupervised each day. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant.

Hours Unsupervised	0	1.5	2.5	3	4.5	5	6
Overall Grades	94	92	86	82	79	71	62

Step 1 of 6: Find the estimated slope. Round your answer to three decimal places.

Step 2 of 6: Find the estimated y-intercept. Round your answer to three decimal places.

Step 3 of 6: Find the estimated value of y when x=2.5. Round your answer to three decimal places.

Step 4 of 6: Determine the value of the dependent variable yˆ at x=0 (b0, b1, x, y)

Step 5 of 6: Find the error prediction when x=2.5. Round your answer to three decimal places.

Step 6 of 6: Find the value of the coefficient of determination. Round your answer to three decimal places.

Answer 1

Ans:

	1	2	3	4	5	6	7	Total
Hours Unsupervised,x	0	1.5	2.5	3	4.5	5	6	22.5
Overall Grades,y	94	92	86	82	79	71	62	566
xy	0	138	215	246	355.5	355	372	1681.5
x^2	0	2.25	6.25	9	20.25	25	36	98.75
y^2	8836	8464	7396	6724	6241	5041	3844	46546

1)

slope,b1=(7*1681.5-22.5*566)/(7*98.75-22.5^2)=-5.214

2)

intercept,b0=(566-(-5.21351)*22.5)/7=97.615

3)

Regression equation:

y'=97.615-5.214 x

when x=2.5

y'=97.615-5.214*2.5=84.580

4)

when x=0

y'=97.615-5.214*0

y'=97.615

5)

observed value of y when x=2.5 is 86

Error=86-84.580=1.420

6)

r=(7*1681.5-22.5*566)/SQRT((7*98.75-22.5^2)*(7*46546-566^2))=-0.9159

coefficient of determination,R^2=(-0.9159)^2=0.920