Q. 4.63 - Elementary Statistics

Question

Study Time and Score. An instructor at Arizona State University asked a random sample of eight students to record their study times in a beginning calculus course. She then made a table for total hours studied (x) over 2 weeks and test score (y) at the end of the 2 weeks. Here are the results. For part (g), predict the score of a student who studies for 15 hours.

X	10	15	12	20	8	16	14	22
Y	92	81	84	74	85	80	84	80

• a. fond the regression equation for the data points.
• b. graph the regression equation and the data points.
• c. describe the apparent relationship between the two variables under consideration.
• d. interpret the slope of the regression line.
• e. identify the predictor and response variables.
• f. identify outliers and potential influential observations.
• g. predict the values of the response variable for the specified values of the predictor variable, and interpret your results.

Step-by-Step Solution

Answer

Ans: 

part (a): $\hat{y}=94.9-0.8x$

part (b): Graph representation 

part (c): The test score in calculus decreases as the study hours increase. 

part (d): The slope of the regression line is -0.8 

part (e): The predictor variable is study hours and the response variable is a test score. 

part (f): All the points are near the straight line so there are no outliers in the data given. 

Part (g): The predicted score of a student who studies for 15 hours is $82.2$ points.

Step 1. GIven information:

Given Data points: 

X	10	15	12	20	8	16	14	22
Y	92	81	84	74	85	80	84	80

Step 2. Solving part (a):

$\begin{array}{|llll|}\hline x& y& xy& x^2\\ 10& 92& 920& 100\\ 15& 81& 1215& 225\\ 12& 84& 1008& 144\\ 20& 74& 1480& 400\\ 8& 85& 680& 64\\ 16& 80& 1280& 256\\ 14& 84& 1176& 196\\ 22& 80& 1760& 484\\ \sum x=117& \sum y=660& \sum xy=9519& \sum x^2=1869\\ \hline\end{array}$

Step 3. Continue:

We first need to compute the $b_1$ and $b_0$ to find the regression equation. The slope of the regression line is,

$$\begin{gathered} b_1=\frac{S_{xy}}{S_{xx}} \\ =\frac{\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n}{\sum x_i^2-{\left(\sum x_i\right)}^2/n} \\ =\frac{9519-\left(117\right)\left(80\right)/8}{1869-(117{)}^2/8} \\ =\frac{-133.5}{157.875} \\ =-0.84561 \\ \approx -0.8 \end{gathered}$$

The y-intercept is,

$$\begin{gathered} b_0=\frac{1}{n}\left(\sum y_i-b_1\sum x_i\right) \\ =\frac{1}{8}(660-(-0.84561\left)\right(117\left)\right) \\ =94.86698 \\ \approx 94.9 \end{gathered}$$

So the regression equation is $\hat{y}=94.9-0.8x$

Step 4. Solving part (b):

The graph representation for the equation:

Step 5. Solving part (c):

c) From the above graph we can see that the test score in calculus decreases as the study hours increase.

Step 6. Solving part (d):

The slope of the regression line is -0.8, which means that the test score in calculus decreases an estimated 0.8 point for each increase in study time of 1 hour.

Step 7. Solving part (e):

From the regression equation, the predictor variable is study hours and the response variable is a test score.

Step 8. Solving part (f):

From the obtained regression equation we can see that all the points are near to the straight line so there are no outliers in the data given.

Step 9 . Solving part (g):

The predicted score of a student who studies for 15 hours,

We have the regression line as $\hat{y}=94.86698-0.84561x$

Substituting the value x=15, we get

$$\begin{gathered} \hat{y}=94.86698-0.84561x \\ =94.86698-0.84561\left(15\right) \\ =82.1829 \\ \approx 82.2\text{ points } \end{gathered}$$

Therefore, the predicted score of a student who studies for 15 hours is $82.2$ points.