Q. 4.149 - Elementary Statistics

Question

In below exercise, we repeat data from exercises in Section 4.2. For given exercise here.

a. obtain the linear correlation coefficient. 

b. interpret the value of r in terms of the linear relationship between the two variables in question. 

c. discuss the graphical interpretation of the value of r and verify that it is consistent with the graph you obtained in the corresponding exercise in Section 4.2.

d. square r and compare the result with the value of the coefficient of determination you obtained in the corresponding exercise in Section 4.3.

Study Time and Score. Following are the data on study time and score for calculus students from Exercises 4.63 and 4.103.

x	10	15	12	20	8	16	14	22
y	92	81	84	74	85	80	84	80

Step-by-Step Solution

Answer

Therefore the linear coefficient is -0.7749

From the two values we observe that both the values are equal.

Step 1. Given

x	10	15	12	20	8	16	14	22
y	92	81	84	74	85	80	84	80

Step 2. Table for obtaining the linear correlation coefficient.

x	y	xy	x2	y2
10	92	920	100	8584
15	81	1215	225	6561
12	84	1008	144	7056
20	74	1480	400	5476
8	85	680	64	7225
16	80	1280	256	6400
14	84	1176	196	7056
2	80	1760	484	6400
117	660	9516	1869	54638

$$\begin{gathered} r=\frac{\underset{}{\overset{}{\sum x_i{y}_{i }- {\displaystyle \frac{\left(\underset{}{\overset{}{\sum x_i}}\right) \left(\underset{}{\overset{}{\sum y_i}}\right)}{n}}}} }{\sqrt{\left[\sum {x^2}_i-{\displaystyle \frac{{\left(\sum x_i\right)}^2}{n}}\right]\left[\sum {y^2}_i-\frac{{\left(\sum y_i\right)}^2}{n}\right]}} \\ =\frac{9519-(117*660)/8}{\sqrt{\left[ 1869-{\displaystyle \frac{{\left(117\right)}^2}{8}}\right]\left[54638-{\displaystyle \frac{{\left(660\right)}^2}{8}}\right]}} \\ = -0.7749 \end{gathered}$$

Therefore the linear coefficient is -0.7749

Step 3. Solution b

The value of the correlation coefficient r suggests that there is a negative linear relationship between study time and student school.

Step 4. Solution c

The correlation coefficient, r = 00.7789, suggests a negative correlation between study time and school of calculus students.

In particular, it shows that as study time progresses there is a tendency for points to drop.

The graph from Exercise 14.93 is as follows

Step 5. Finding square of r.

Square r is given by 

$$\begin{gathered} r^2={(-0.7749)}^2 =0.60 \\ \mathrm{The} \mathrm{average} \mathrm{of} \mathrm{tax} \mathrm{efficinecy} \mathrm{is} \\ \overline{y}=\frac{{\displaystyle \sum _{}}y}{n} \\ =\frac{660}{8} \\ =82.5 \end{gathered}$$

$$\begin{gathered} SSR=\sum _{}^{}{(\overbrace{y_i}-\overline{y})}^2=112.887 \\ SSR=\sum _{}^{}{(\overbrace{y_i}-\overline{y})}^2 \\ =188 \\ \mathrm{Coefficient} \mathrm{of} \mathrm{determination} \\ {r}^2=\frac{SSR}{SST} \\ =\frac{112.887}{188} \\ =0.60 \\ \mathrm{The} \mathrm{value} \mathrm{of} {\mathrm{r}}^2 \mathrm{is} \mathrm{same} \mathrm{in} \mathrm{both} \mathrm{methods} \end{gathered}$$