Attending Class The following data represent the number of days absent, x, and t

Question

Attending Class The following data represent the number of days absent, x, and the final grade, y, for a sample of college students in a general education course at a large state university. Find the least-squares regression line treating number of absences as the explanatory variable and final grade as the response variable. Interpret the slope and y-intercept, if appropriate. Predict the final grade for a student who misses five class periods and compute the residual. Is the final grade above or below average for this number of absences?

Explanation / Answer

Part a

The required regression output for the estimation of the dependent variable or response variable as final grade is given as below:

SUMMARY OUTPUT

Regression Statistics

Multiple R

0.947394997

R Square

0.897557281

Adjusted R Square

0.884751941

Standard Error

3.067320418

Observations

10

ANOVA

df

SS

MS

F

Significance F

Regression

1

659.4613636

659.4613636

70.0924217

3.14343E-05

Residual

8

75.26763636

9.408454545

Total

9

734.729

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Intercept

88.73272727

1.802829273

49.2185969

3.2139E-11

84.57539552

92.89005903

X

-2.827272727

0.337700864

-8.3721217

3.1434E-05

-3.606012315

-2.04853314

The least square regression line is given as below:

Y = 88.7327 – 2.8273*X

Final Grade = 88.7327 – 2.8273*No. Of absences

Part b

The slope for the above regression equation is given as -2.8273 which is negative in nature. This negative slope represents that there is a negative linear relationship or association exists between the dependent variable final grade and independent variable number of absences. The value of y-intercept is given as 88.7327 which is the value of final grade when there are no any absences.

Part c

Here, we have to find the value for final grade for X = 5.

Y = 88.7327 – 2.8273*X

Y = 88.7327 – 2.8273*5

Y = 74.5962

Final Grade = 74.5962

Residual = (Real value for x=5) – (estimated value for x =5)

Residual = 73.9 – 74.6 = -0.7

Final grade is below average for this number of absences.

SUMMARY OUTPUT

Regression Statistics

Multiple R

0.947394997

R Square

0.897557281

Adjusted R Square

0.884751941

Standard Error

3.067320418

Observations

10

ANOVA

df

SS

MS

F

Significance F

Regression

1

659.4613636

659.4613636

70.0924217

3.14343E-05

Residual

8

75.26763636

9.408454545

Total

9

734.729

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Intercept

88.73272727

1.802829273

49.2185969

3.2139E-11

84.57539552

92.89005903

X

-2.827272727

0.337700864

-8.3721217

3.1434E-05

-3.606012315

-2.04853314

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