Question

the data below represent the number of days absent, x, and the final grade, y, for a sample of college students at a large university. complete parts (a) through (e) below.
no. of absences, x 0 1 2 3 4 5 6 7 8 9
final grade, y 87.9 85.1 82.2 79.8 76.9 72.5 63.0 67.6 64.8 61.9
(c) predict the final grade for a student who misses five class periods and compute the residual. is the observed final grade above or below average for this number of absences?
the predicted final grade is \\( \square \\). this observation has a residual of \\( \square \\), which indicates that the final grade is \\( \square \\) average.
(round to one decimal place as needed.)
(d) draw the least-squares regression line on the scatter diagram of the data. choose the correct graph below.
(e) would it be reasonable to use the least-squares regression line to predict the final grade for a student who has missed 15 class periods? why or why not?
a. no-15 missed class periods is not possible and outside the scope of the model.
b. no-15 missed class periods is not possible.
c. no-15 missed class periods is outside the scope of the model.
d. yes-15 missed class periods is possible and within the scope of the model

Explanation:

Step1: Calculate the regression equation

First, we need to calculate the regression equation \(y = a+bx\). Using the formula for \(b=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}\) and \(a=\overline{y}-b\overline{x}\), where \(n = 10\), \(\sum_{i=1}^{n}x_i=0 + 1+2+\cdots+9 = 45\), \(\sum_{i = 1}^{n}y_i=87.9+85.1+\cdots+61.9 = 729.7\), \(\sum_{i = 1}^{n}x_iy_i=0\times87.9+1\times85.1+\cdots+9\times61.9 = 3039.9\), \(\sum_{i = 1}^{n}x_i^{2}=0^{2}+1^{2}+\cdots+9^{2}=285\).

\(\overline{x}=\frac{45}{10}=4.5\), \(\overline{y}=\frac{729.7}{10}=72.97\)

\(b=\frac{10\times3039.9 - 45\times729.7}{10\times285-45^{2}}=\frac{30399 - 32836.5}{2850 - 2025}=\frac{- 2437.5}{825}=- 2.9545\approx - 2.95\)

\(a = 72.97-(-2.95)\times4.5=72.97 + 13.275=86.245\approx86.2\)

The regression equation is \(y = 86.2-2.95x\)

Step2: Predict the final grade for \(x = 5\)

Substitute \(x = 5\) into the regression equation: \(y=86.2-2.95\times5=86.2 - 14.75 = 71.45\approx71.5\)

Step3: Calculate the residual

The observed value of \(y\) when \(x = 5\) is \(y_{obs}=64.8\)

Residual \(e=y_{obs}-\hat{y}=64.8 - 71.5=- 6.7\)

Since the residual is negative, the observed final grade is below average.

Step4: Analyze part (e)

The data for \(x\) (number of absences) ranges from \(0\) to \(9\). Predicting for \(x = 15\) is extrapolation (outside the range of the data used to build the model).

Answer:

(c) The predicted final grade is \(71.5\). This observation has a residual of \(-6.7\), which indicates that the final grade is below average.

(e) C. No—15 missed class periods is outside the scope of the model.