Chapter 2.4, Problem 72E

(a)

To determine

To find: The predicted count values.

Answer to Problem 72E

Solution: The predicted count values obtained are

${\widehat{c}}_1=6.3324$

${\widehat{c}}_2=5.8112$

${\widehat{c}}_3=5.29$

${\widehat{c}}_4=4.7688$

Explanation of Solution

Calculation: The provided least-squares regression equation is

$\text{log count}=6.593-\left(0.2606\times \text{time}\right)$

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 log count $\left({\widehat{c}}_1\right)$ is calculated as

$\begin{array}{c}\text{log count}=6.593-\left(0.2606\times \text{time}\right)\\ =6.593-\left(0.2606\times t_1\right)\\ =6.593-\left(0.2606\times 1\right)\\ {\widehat{c}}_1=6.3324\end{array}$

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

$\begin{array}{c}\text{log count}=6.593-\left(0.2606\times \text{time}\right)\\ =6.593-\left(0.2606\times t_2\right)\\ =6.593-\left(0.2606\times 3\right)\\ {\widehat{c}}_2=5.8112\end{array}$

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

$\begin{array}{c}\text{log count}=6.593-\left(0.2606\times \text{time}\right)\\ =6.593-\left(0.2606\times t_3\right)\\ =6.593-\left(0.2606\times 5\right)\\ {\widehat{c}}_3=5.29\end{array}$

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

$\begin{array}{c}\text{log count}=6.593-\left(0.2606\times \text{time}\right)\\ =6.593-\left(0.2606\times t_4\right)\\ =6.593-\left(0.2606\times 7\right)\\ {\widehat{c}}_4=4.7688\end{array}$

Hence, the predicted count values obtained are $6.3324$, $5.8112$, $5.29,$ and $4.7688$.

(b)

Section 1

To determine

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

Answer to Problem 72E

Solution: The differences obtained are

$d_1=0.02717$

$d_2=-0.0523$

$d_3=0.02321$

$d_4=0.00188$

Explanation of Solution

Calculation: The observed counts are provided in the exercise as

$c_1=6.35957$

$c_2=5.75890$

$c_3=5.31321$

$c_4=4.77068$

The predicted counts obtained from part (a) are

${\widehat{c}}_1=6.3324$

${\widehat{c}}_2=5.8112$

${\widehat{c}}_3=5.29$

${\widehat{c}}_4=4.7688$

All of the above values are for the times 1, 3, 5, and 7.

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

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

$\begin{array}{c}{d}_1=c_1-{\widehat{c}}_1\\ =6.35957-6.3324\\ =0.02717\end{array}$

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

$\begin{array}{c}{d}_2=c_2-{\widehat{c}}_2\\ =5.75890-5.8112\\ =-0.0523\end{array}$

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

$\begin{array}{c}{d}_3=c_3-{\widehat{c}}_3\\ =5.31321-5.29\\ =0.02321\end{array}$

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

$\begin{array}{c}{d}_4=c_4-{\widehat{c}}_4\\ =4.77068-4.7688\\ =0.00188\end{array}$

Hence, the differences obtained are $0.02717$, $-0.0523$, $0.02321,$ and $0.00188$.

Section 2

To determine

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

Answer to Problem 72E

Solution: There are three difference values that are positive as $0.02717$, $0.02321,$ and $0.00188$.

Explanation of Solution

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

$d_1=0.02717$

$d_2=-0.0523$

$d_3=0.02321$

$d_4=0.00188$

Clearly, the positive difference values are $0.02717$, $0.02321,$ and $0.00188$.

Hence, three differences $d_1$, $d_3,$ and $d_4$ are positive.

To determine

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

Answer to Problem 72E

Solution: The negative difference is

$d_2=-0.0523$

Explanation of Solution

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

$d_1=0.02717$

$d_2=-0.0523$

$d_3=0.02321$

$d_4=0.00188$

Clearly, the negative difference value is $-0.0523$. Hence, the difference $d_2$ is negative.

(c)

Section 1

To determine

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

Answer to Problem 72E

Solution: The squares of the differences are

$d_1^2=0.000738$

$d_2^2=0.002735$

$d_3^2=0.000539$

$d_4^2=3.61\times {10}^{-6}\left(\text{approximately }0\right)$

Explanation of Solution

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

$d_1=0.02717$

$d_2=-0.0523$

$d_3=0.02321$

$d_4=0.00188$

The squares of the differences obtained are calculated as follows:

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

$\begin{array}{c}{d}_1^2=d_1\times d_1\\ =0.02717\times 0.02717\\ =0.000738\end{array}$

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

$\begin{array}{c}{d}_2^2=d_2\times d_2\\ =\left(-0.0523\right)\times \left(-0.0523\right)\\ =0.002735\end{array}$

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

$\begin{array}{c}{d}_3^2=d_3\times d_3\\ =0.02321\times 0.02321\\ =0.000539\end{array}$

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

$\begin{array}{c}{d}_4^2=d_4\times d_4\\ =0.0019\times 0.0019\\ =3.61\times {10}^{-6}\\ \simeq 0\end{array}$

Hence, the squares of the differences obtained are $0.000738$, $0.002735$, $0.000539$ and $3.61\times {10}^{-6}\left(\text{approximately equal to }0\right)$.

Section 2

To determine

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

Answer to Problem 72E

Solution: The sum of the squares of the differences obtained is $\underset{\_}{0.004011}$.

Explanation of Solution

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

$d_1^2=0.000738$

$d_2^2=0.002735$

$d_3^2=0.000538$

$d_4^2=3.61\times {10}^{-6}\left(\text{approximately }0\right)$.

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

$\begin{array}{c}\underset{n=1}{\overset{4}{\Sigma d_n^2}}=d_1^2+d_2^2+d_3^2+d_4^2\\ =\left(0.000738\right)+\left(0.002735\right)+\left(0.000538\right)+\left(0\right)\\ =0.004011\end{array}$

Hence, the sum of the differences obtained is $0.004011$.

(d)

Section 1

To determine

To find: The predicted count values for the new regression equation.

Answer to Problem 72E

Solution: The predicted count values obtained are:

${\widehat{c}}_1=6.8$

${\widehat{c}}_2=6.4$

${\widehat{c}}_3=6$

${\widehat{c}}_4=5.6$

Explanation of Solution

Calculation: The provided least-squares regression equation is

$\text{log count}=7-\left(0.2\times \text{time}\right)$

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

$\begin{array}{c}\text{log count}=7-\left(0.2\times \text{time}\right)\\ =7-\left(0.2\times t_5\right)\\ =7-\left(0.2\times \text{1}\right)\\ {\widehat{c}}_5=6.8\end{array}$

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

$\begin{array}{c}\text{log count}=7-\left(0.2\times \text{time}\right)\\ =7-\left(0.2\times t_6\right)\\ =7-\left(0.2\times 3\right)\\ {\widehat{c}}_6=6.4\end{array}$

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

$\begin{array}{c}\text{log count}=7-\left(0.2\times \text{time}\right)\\ =7-\left(0.2\times t_7\right)\\ =7-\left(0.2\times 5\right)\\ {\widehat{c}}_7=6\end{array}$

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

$\begin{array}{c}\text{log count}=7-\left(0.2\times \text{time}\right)\\ =7-\left(0.2\times t_8\right)\\ =7-\left(0.2\times 7\right)\\ {\widehat{c}}_8=5.6\end{array}$

Hence, the predicted count values obtained are $6.8$, $6.4$, $6,$ and $5.6$ 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 72E

Solution: The differences obtained are $-0.44043$, $-0.6411$, $-0.68679,$ and $-0.82932$.

Explanation of Solution

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

$c_1=6.35957$

$c_2=5.75890$

$c_3=5.31321$

$c_4=4.77068$

The predicted counts obtained from section 1 above are:

${\widehat{c}}_5=6.8$

${\widehat{c}}_6=6.4$

${\widehat{c}}_7=6$

${\widehat{c}}_8=5.6$

In both the cases, the time is 1, 3, 5 and 7 respectively.

The differences $\left(d_i\text{s where }i\text{ are 5, 6, 7, and 8}\right)$ 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

$\begin{array}{c}{d}_5=c_1-{\widehat{c}}_5\\ =6.35957-6.8\\ =-0.44043\end{array}$

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

$\begin{array}{c}{d}_6=c_2-{\widehat{c}}_6\\ =5.75890-6.4\\ =-0.6411\end{array}$

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

$\begin{array}{c}{d}_7=c_3-{\widehat{c}}_7\\ =5.31321-6\\ =-0.68679\end{array}$

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

$\begin{array}{c}{d}_8=c_4-{\widehat{c}}_8\\ =4.77068-5.6\\ =-0.82932\end{array}$

Hence, the differences obtained are $-0.44043$, $-0.6411$, $-0.68679,$ and $-0.82932$.

Section 3

To determine

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

Answer to Problem 72E

Solution: None of the differences are positive.

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 2 above as $-0.44043$, $-0.6411$, $-0.68679,$ and $-0.82932$.

Clearly, none of the difference values are positive. Hence, zero difference values are positive.

To determine

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

Answer to Problem 72E

Solution: There are four negative differences as: $-0.44043$, $-0.6411$, $-0.68679,$ and $-0.82932$.

Explanation of Solution

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

$d_5=-0.44043$

$d_6=-0.6411$

$d_7=-0.68679$

$d_8=-0.82932$

Clearly, all of the values are negative. Hence, the differences $d_5$, $d_6$, $d_7,$ and $d_8$ are negative.

Section 4

To determine

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

Answer to Problem 72E

Solution: The squares of the differences are $0.19397859$, $0.41100921$, $0.47168050,$ and $0.68777166$.

Explanation of Solution

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

$d_5=-0.44043$

$d_6=-0.6411$

$d_7=-0.68679$

$d_8=-0.82932$

The squares of the differences obtained are calculated as follows:

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

$\begin{array}{c}{d}_5^2=d_5\times d_5\\ =\left(-0.44043\right)\times \left(-0.44043\right)\\ =0.19397859\end{array}$

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

$\begin{array}{c}{d}_6^2=d_6\times d_6\\ =\left(-0.6411\right)\times \left(-0.6411\right)\\ =0.41100921\end{array}$

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

$\begin{array}{c}{d}_7^2=d_7\times d_7\\ =\left(-0.68679\right)\times \left(-0.68679\right)\\ =0.47168050\end{array}$

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

$\begin{array}{c}{d}_8^2=d_8\times d_8\\ =\left(-0.82932\right)\times \left(-0.82932\right)\\ =0.68777166\end{array}$

Hence, the squares of the differences obtained are $0.19397859$, $0.41100921$, $0.47168050,$ and $0.68777166$.

Section 5

To determine

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

Answer to Problem 72E

Solution: The sum of the differences is $\underset{\_}{1.76443996}$.

Explanation of Solution

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

$d_5^2=0.19397859$

$d_6^2=0.41100921$

$d_7^2=0.47168050$

$d_8^2=0.68777166$

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

$\begin{array}{c}\underset{n=5}{\overset{8}{\Sigma d_n^2}}=d_5^2+d_6^2+d_7^2+d_8^2\\ =\left(0.19397859\right)+\left(0.41100921\right)+\left(0.47168050\right)+\left(0.68777166\right)\\ =1.76443996\end{array}$

Hence, the sum of the differences is $1.76443996$.

(e)

To determine

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

Answer to Problem 72E

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

$\text{log count}=6.593-\left(0.2606\times \text{time}\right)$

Explanation of Solution

In a linear least-squares 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

$\text{log count}=6.593-\left(0.2606\times \text{time}\right)$

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

$\text{log count}=7-\left(0.2\times \text{time}\right)$

all the differences (residuals) are extremely low negative 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 below the x-axis, indicating it to be not a much accurate measure. Also, with such low negative differences, the predicted regression line will lie far from the observed regression line. Hence, the first line better predicts the relationship between the variables log count and the time because it gives a better approximate value of the response variable.