Chapter 2.4, Problem 71E

(a)

To determine

To find: The predicted count values.

Answer to Problem 71E

Solution: The predicted count values are: $528.1$, $378.7$, $229.3,$ and $79.9$.

Explanation of Solution

Calculation: The provided least-squares regression equation is:

$\text{count}=602.8-\left(74.7\times \text{time}\right)$

The time values are provided in the data as 1, 3, 5 and 7.

Substituting the values of time in the above linear regression equation, the following results are obtained:

For $\text{time}\left(t_1\right)\text{=1}$, the predicted count $\left({\widehat{c}}_1\right)$ is calculated as:

$\begin{array}{c}\text{count}=602.8-\left(74.7\times \text{time}\right)\\ =602.8-\left(74.7\times t_1\right)\\ =602.8-\left(74.7\times 1\right)\\ {\widehat{c}}_1=528.1\end{array}$

For $\text{time}\left(t_2\right)\text{=3}$, the predicted count $\left({\widehat{c}}_2\right)$ is calculated as:

$\begin{array}{c}\text{count}=602.8-\left(74.7\times \text{time}\right)\\ =602.8-\left(74.7\times t_2\right)\\ =602.8-\left(74.7\times 3\right)\\ {\widehat{c}}_2=378.7\end{array}$

For $\text{time}\left(t_3\right)\text{=5}$, the predicted count $\left({\widehat{c}}_3\right)$ is calculated as:

$\begin{array}{c}\text{count}=602.8-\left(74.7\times \text{time}\right)\\ =602.8-\left(74.7\times t_3\right)\\ =602.8-\left(74.7\times 5\right)\\ {\widehat{c}}_3=229.3\end{array}$

For $\text{time}\left(t_4\right)\text{=7}$, the predicted count $\left({\widehat{c}}_4\right)$ is calculated as:

$\begin{array}{c}\text{count}=602.8-\left(74.7\times \text{time}\right)\\ =602.8-\left(74.7\times t_4\right)\\ =602.8-\left(74.7\times 7\right)\\ {\widehat{c}}_4=79.9\end{array}$

Hence, the predicted count values obtained are $528.1$, $378.7$, $229.3,$ and $79.9$.

(b)

Section 1

To determine

To find: The difference between the observed and the predicted counts.

Answer to Problem 71E

Solution: The differences obtained are $49.9$, $-61.7$, $-26.3,$ and $38.1$.

Explanation of Solution

Calculation: The respective observed counts are provided in the exercise as:

$c_1=578$

$c_2=317$

$c_3=203$

$c_4=118$

for the times

$t_1\text{=1}$

$t_2\text{=3}$

$t_3\text{=5}$

$t_4\text{=7}$

The respective predicted counts for the above times are obtained from part (a) as:

${\widehat{c}}_1=528.1$

${\widehat{c}}_2=378.7$

${\widehat{c}}_3=229.3$

${\widehat{c}}_4=79.9$

The differences $\left(d_i\right)\text{where }i\text{=1,}\text{2,}\text{3, and 4}$ between the observed and the predicted counts are calculated as follows:

For $\text{time}\left(t_1\right)\text{ as 1}$, the difference $\left(d_1\right)$ is calculated as:

$\begin{array}{c}{d}_1=c_1-{\widehat{c}}_1\\ =578-528.1\\ =49.9\end{array}$

For $\text{time}\left(t_2\right)\text{ as 3}$, the difference $\left(d_2\right)$ is calculated as:

$\begin{array}{c}{d}_2=c_2-{\widehat{c}}_2\\ =317-378.7\\ =-61.7\end{array}$

For $\text{time}\left(t_3\right)\text{ as 5}$, the difference $\left(d_3\right)$ is calculated as:

$\begin{array}{c}{d}_3=c_3-{\widehat{c}}_3\\ =203-229.3\\ =-26.3\end{array}$

For $\text{time}\left(t_4\right)\text{ as 7}$, the difference $\left(d_4\right)$ is calculated as:

$\begin{array}{c}{d}_4=c_4-{\widehat{c}}_4\\ =118-79.9\\ =38.1\end{array}$

Hence, the differences obtained are $49.9$, $-61.7$, $-26.3,$ and $38.1$.

Section 2

To determine

The number of positive differences between the observed and the predicted counts.

Answer to Problem 71E

Solution: The two positive differences are $49.9$ and $38.1$.

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 1 of part (b) above as:

$d_1=49.9$

$d_2=-61.7$

$d_3=-26.3$

$d_4=38.1$

Clearly, the positive difference values are $d_1$, $d_4$ as $49.9$ and $38.1$. Hence, 2 differences $d_1$ and $d_4$ are positive.

To determine

The number of negative differences between the observed and the predicted counts.

Answer to Problem 71E

Solution: The two negative differences are $-61.7$ and $-26.3$.

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 1 of part (b) above as

$d_1=49.9$

$d_2=-61.7$

$d_3=-26.3$

$d_4=38.1$

Clearly, the negative difference values are $d_2$, $d_3$ $-61.7$ and $-26.3$. Hence, 2 differences $d_2$ and $d_3$ are negative.

(c)

Section 1

To determine

To find: Squares of the differences obtained in part(b).

Answer to Problem 71E

Solution: The squares of the differences are:

$d_1^2=2490.01$

$d_2^2=3806.89$

$d_3^2=691.69$

$d_4^2=1451.61$

Explanation of Solution

Calculation: The differences obtained in the part (b) above are:

$d_1=49.9$

$d_2=-61.7$

$d_3=-26.3$

$d_4=38.1$

The squares of the differences obtained are calculated as follows:

The square of the difference $\left(d_1\right)$ is calculated as:

$\begin{array}{c}{d}_1^2=d_1\times d_1\\ =49.9\times 49.9\\ =2490.01\end{array}$

The square of the difference $\left(d_2\right)$ is calculated as:

$\begin{array}{c}{d}_2^2=d_2\times d_2\\ =\left(-61.7\right)\times \left(-61.7\right)\\ =3806.89\end{array}$

The square of the difference $\left(d_3\right)$ is calculated as:

$\begin{array}{c}{d}_3^2=d_3\times d_3\\ =\left(-26.3\right)\times \left(-26.3\right)\\ =691.69\end{array}$

The square of the difference $\left(d_4\right)$ is calculated as:

$\begin{array}{c}{d}_4^2=d_4\times d_4\\ =38.1\times 38.1\\ =1451.61\end{array}$

Hence, the squares of the differences obtained are $2490.01$, $3806.89$, $691.69,$ and $1451.61$.

Section 2

To determine

To find: The sum of the squares of the differences obtained in Section 1 above.

Answer to Problem 71E

Solution: The sum of the differences obtained is $\underset{\_}{8440.2}$.

Explanation of Solution

Calculation: The squares of the differences are obtained in Section 1 above as:

$d_1^2=2490.01$

$d_2^2=3806.89$

$d_3^2=691.69$

$d_4^2=1451.61$

The sum of the differences $\left(\Sigma d_n^2\right)$ $\text{where }n\text{ is }1\text{,}\text{2,}\text{3, and 4}$ is calculated as follows:

$\begin{array}{c}\underset{n=1}{\overset{4}{\Sigma d_n^2}}=d_1^2+d_2^2+d_3^2+d_4^2\\ =\left(2490.01\right)+\left(3806.89\right)+\left(691.69\right)+\left(1451.61\right)\\ =8440.2\end{array}$

Hence, the sum of the differences obtained is $\underset{\_}{8440.2}$.

(d)

Section 1

To determine

To find: The predicted count values.

Answer to Problem 71E

Solution: The predicted count values obtained are:

${\widehat{c}}_5=400$

${\widehat{c}}_6=200$

${\widehat{c}}_7=0$

${\widehat{c}}_8=-200$

Explanation of Solution

Calculation: The provided least-squares regression equation is:

$\text{count}=500-\left(100\times \text{time}\right)$

The time values are provided in the data as 1, 3, 5 and 7.

Substituting the values of time in the above linear regression equation, the following results are obtained:

For $\text{time}\left(t_1\right)\text{=1}$, the predicted count $\left({\widehat{c}}_5\right)$ is calculated as:

$\begin{array}{c}\text{count}=500-\left(100\times \text{time}\right)\\ =500-\left(100\times t_1\right)\\ =500-\left(100\times 1\right)\\ {\widehat{c}}_5=400\end{array}$

For $\text{time}\left(t_2\right)\text{=3}$, the predicted count $\left({\widehat{c}}_6\right)$ is calculated as:

$\begin{array}{c}\text{count}=500-\left(100\times \text{time}\right)\\ =500-\left(100\times t_2\right)\\ =500-\left(100\times 3\right)\\ {\widehat{c}}_6=200\end{array}$

For $\text{time}\left(t_3\right)\text{=5}$, the predicted count $\left({\widehat{c}}_7\right)$ is calculated as:

$\begin{array}{c}\text{count}=500-\left(100\times \text{time}\right)\\ =500-\left(100\times t_3\right)\\ =500-\left(100\times 5\right)\\ {\widehat{c}}_7=0\end{array}$

For $\text{time}\left(t_4\right)\text{=7}$, the predicted count $\left({\widehat{c}}_8\right)$ is calculated as:

$\begin{array}{c}\text{count}=500-\left(100\times \text{time}\right)\\ =500-\left(100\times t_4\right)\\ =500-\left(100\times 7\right)\\ {\widehat{c}}_8=-200\end{array}$

Hence, the predicted count values obtained are:

${\widehat{c}}_5=400$

${\widehat{c}}_6=200$

${\widehat{c}}_7=0$

${\widehat{c}}_8=-200$

For the times 1, 3, 5 and 7 respectively.

Section 2:

To determine

To find: The difference between the observed and the predicted counts.

Answer to Problem 71E

Solution: The differences obtained are

$d_5=178$

$d_6=117$

$d_7=203$

$d_8=318$

Explanation of Solution

The observed counts are provided in the exercise as: $578$, $317$, $203$ and $118$ for the times as 1, 3, 5 and 7 respectively. The predicted counts obtained from section 1 above are: $400$, $200$, $0,$ and $-200$ for the times as 1, 3, 5 and 7 respectively.

The differences $\left(d_i\right)\text{where }i\text{is 5,}\text{6,}\text{7, and 8}$ between the observed and the predicted counts are calculated as follows:

For $\text{time}\left(t_1\right)\text{ as 1}$, the difference $\left(d_5\right)$ is calculated as:

$\begin{array}{c}{d}_5=c_1-{\widehat{c}}_5\\ =578-400\\ =178\end{array}$

For $\text{time}\left(t_2\right)\text{ as 3}$, the difference $\left(d_6\right)$ is calculated as:

$\begin{array}{c}{d}_6=c_2-{\widehat{c}}_6\\ =317-\left(200\right)\\ =117\end{array}$

For $\text{time}\left(t_3\right)\text{ as 5}$, the difference $\left(d_7\right)$ is calculated as:

$\begin{array}{c}{d}_7=c_3-{\widehat{c}}_7\\ =203-0\\ =203\end{array}$

For $\text{time}\left(t_4\right)\text{ as 7}$, the difference $\left(d_8\right)$ is calculated as:

$\begin{array}{c}{d}_8=c_4-{\widehat{c}}_8\\ =118-\left(-200\right)\\ =318\end{array}$

Hence, the differences obtained are $178$, $117$, $203,$ and $318$.

Section 3:

To determine

The number of positive differences between the observed and the predicted counts.

Answer to Problem 71E

Solution: The four positive differences are $178$, $117$, $203,$ and $318$.

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 2 above as,

$d_5=178$

$d_6=117$

$d_7=203$

$d_8=318$

Clearly, all the difference values are positive. Hence, all the 4 differences $d_5$, $d_6$, $d_7,$ and $d_8$ are positive.

To determine

The number of negative differences between the observed and the predicted counts.

Answer to Problem 71E

Solution: There are no negative differences.

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 1 above as,

$d_5=178$

$d_6=117$

$d_7=203$

$d_8=318$

Clearly, none of the values are negative. Hence, 0 out of 4 differences are negative.

Section 4:

To determine

To find: Squares of the differences obtained in section 2 of part (d).

Answer to Problem 71E

Solution: The squares of the differences are:

$d_5^2=31,684$

$d_6^2=13,689$

$d_7^2=41,209$

$d_8^2=101,124$

Explanation of Solution

Calculation: The differences obtained in the section 2 of part (d) above are:

$d_5=178$

$d_6=117$

$d_7=203$

$d_8=318$

The squares of the differences obtained are calculated as follows:

The square of the difference $\left(d_5\right)$ is calculated as:

$\begin{array}{c}{d}_5^2=d_5\times d_5\\ =178\times 178\\ =31,684\end{array}$

The square of the difference $\left(d_6\right)$ is calculated as:

$\begin{array}{c}{d}_6^2=d_6\times d_6\\ =117\times 117\\ =13,689\end{array}$

The square of the difference $\left(d_7\right)$ is calculated as:

$\begin{array}{c}{d}_7^2=d_7\times d_7\\ =203\times 203\\ =41,209\end{array}$

The square of the difference $\left(d_8\right)$ is calculated as:

$\begin{array}{c}{d}_8^2=d_8\times d_8\\ =318\times 318\\ =101,124\end{array}$

Hence, the squares of the differences obtained are $31,684$, $13,689$, $41,209,$ and $101,124$.

Section 5:

To determine

To find: The sum of the squares of the differences obtained in Section 3 above.

Answer to Problem 71E

Solution: The sum of the differences is $\underset{\_}{187,706}$.

Explanation of Solution

The squares of the differences are obtained in Section 4 above as:

$d_5^2=31,684$

$d_6^2=13,689$

$d_7^2=41,209$

$d_8^2=101,124$

The sum of the differences $\left(\Sigma d_n^2\right)$ $\text{where }n=\text{5,6,7, and 8}$ is calculated as follows:

$\begin{array}{c}\underset{n=5}{\overset{8}{\Sigma d_n^2}}=d_5^2+d_6^2+d_7^2+d_8^2\\ =\left(31,684\right)+\left(13,689\right)+\left(41,209\right)+\left(101,124\right)\\ =187,706\end{array}$

Hence, the sum of the differences is $\underset{\_}{187,706}$.

(e)

To determine

To explain: The least-squares inference based on the calculations performed.

Answer to Problem 71E

Solution: The following least-square regression is a better measure of the relationship between the count and the time:

$\text{count}=602.8-\left(74.7\times \text{time}\right)$

Explanation of Solution

In a linear least-square equation, the graph of the residuals must have an unbiased pattern, that is, the scatter plot should be scattered above and below the x-axis. The residual plots must be both positive and negative. In the calculations performed above, it is observed that for the regression line,

$\text{count}=602.8-\left(74.7\times \text{time}\right)$

The differences (residuals) are both positive and negative values, whereas, for the regression line,

$\text{count}=500-\left(100\times \text{time}\right)$

all the differences (residuals) are high and positive values. Thus, the residual plots for the first regression line will lie both below and above the x-axis. Also, its predicted regression line will lie much near to the observed regression line. Whereas, the residual plots for the second regression line will lie only above the x-axis. Also, with such high and positive differences, the predicted regression line will lie far to the observed regression line. Hence, the first line better depicts the relationship between the count and the time because it gives a better approximate value of the response variable.