Chapter 4, Problem 12RE

Commute times: Every morning, Tania leaves for work a few minutes after 7:00 A.M. For eight days, she keeps track of the time she leaves (the number of minutes after 7:00) and the number of minutes it takes her to get to work. Following are the results.

1. Construct a scatterplot of the length of commute (y) versus the time leaving (x).
2. Compute the least-squares regression line for predicting the length of commute from the time leaving.
3. Compute the coefficient of determination.
4. Which point is an outlier?
5. Remove the outlier and compute the least-squares regression line for predicting the length of commute from the time leaving.
6. Is the outlier influential? Explain.
7. Compute the coefficient of determination for the data set with the outlier removed. Is the relationship stronger. weaker; or about equally strong without the outlier?

To determine

To Graph:a scatter plot using the length of commute time $y$ versus time leaving $x$

Explanation of Solution

Given information: T leaves her home every day few minutes after $7.00$ A.M. The time she leaves the home (The number of minutes after $7.00$ A.M) and the number of minutes taken to office has been recorded as below.

$x$	$13$	$14$	$16$	$30$	$20$	$12$	$9$	$17$	$16$	$10$	$16$
$y$	$27$	$20$	$23$	$45$	$20$	$21$	$20$	$28$	$27$	$23$	$30$

Graph:The scatter plot shows the number of minutes after $7.00$ A.M on the $x$ -axis and the number of minutes taken to office on the $y$ -axis.

  

Interpretation:Each of the data in the table contributes an ordered pair of the form (number of minutes after $7.00$ A.M, number of minutes taken to office).So the ordered pairs to be plotted are

  $(13,27),(14,20),(16,23),(30,45),(20,20),(12,21),(9,20),(17,28),(16,27),(10,23),(16,30)$ .

We use a scatter plot for this example because to understand the relationship between the two variables as ordered pairs it is useful. The points tend to cluster around the straight line. Therefore, we conclude that the variable on the $x$ -axis and variable on the $y$ -axis have a linear relationship.

Now consider the three points $(9,20),(14,20),(20,20)$ . These show that the number of minutes after $7.00$ A.M on the $x$ -axis does not change with the number of minutes taken to an office on the $y$ -axis. There are few other points in the table to reflect the same idea.

Therefore, we conclude that the variable on the $x$ -axis and variable on the $y$ -axis have a linear relationship, but it is difficult to say whether it is negative or positive.

Because for positive linear relationship the large value of data associates with large values of data in the plot, while for negative linear relationship the large value of data associates with small values of data in the pot. And in this case, it is difficult to say that because the large values of data associate with both small and large and small values of data associate with both small and large at the same time.

Therefore it is good to measure how strong the linear relationship is, to know this we can calculate the correlation coefficient.

To determine

To Calculate: the least square regression line

Answer to Problem 12RE

When two variables have a linear relationship, the points on a scatter plot tend to cluster around a straight line called the least square regression line. It is simplified to be $\stackrel{\wedge }{y}=26+4\times {10}^{-31}x$ .

Explanation of Solution

Given information: T leaves her home every day few minutes after $7.00$ A.M. The time she leaves the home (The number of minutes after $7.00$ A.M) and the number of minutes taken to office has been recorded as below.

$x$	$13$	$14$	$16$	$30$	$20$	$12$	$9$	$17$	$16$	$10$	$16$
$y$	$27$	$20$	$23$	$45$	$20$	$21$	$20$	$28$	$27$	$23$	$30$

Formulas Used:

Sample mean: $\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}$

Sample variance: $s^2=\frac{{\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}}{n-1}$

Correlation Coefficient: $r=\frac{1}{(n-1)}.\frac{{\displaystyle \sum _{i=1}^n(x_i-\overline{x})}}{s_x}.\frac{{\displaystyle \sum _{i=1}^n(y_i-\overline{y})}}{s_y}$

The least-square regression line:

  $\begin{array}{l}\stackrel{\wedge }{y}=b_0+b_{1}x\\ b_1=r\frac{s_y}{s_x}\\ b_0=\overline{y}-b_1\overline{x}\end{array}$

Calculation: Using the below table for calculation.

  $\begin{array}{llllllll}x\hfill & y\hfill & \overline{x}\hfill & \overline{y}\hfill & {(x-\overline{x})}^2\hfill & {(y-\overline{y})}^2\hfill & x-\overline{x}\hfill & y-\overline{y}\hfill \\ 13\hfill & 27\hfill & 15.72727273\hfill & 25.81818\hfill & 7.438017\hfill & 1.396694\hfill & -2.727272727\hfill & 1.181818182\hfill \\ 14\hfill & 20\hfill & \hfill & \hfill & 2.983471\hfill & 33.85124\hfill & -1.727272727\hfill & -5.81818182\hfill \\ 16\hfill & 23\hfill & \hfill & \hfill & 0.07438\hfill & 7.942149\hfill & 0.272727273\hfill & -2.81818182\hfill \\ 30\hfill & 45\hfill & \hfill & \hfill & 203.7107\hfill & 367.9421\hfill & 14.27272727\hfill & 19.18181818\hfill \\ 20\hfill & 20\hfill & \hfill & \hfill & 18.2562\hfill & 33.85124\hfill & 4.272727273\hfill & -5.81818182\hfill \\ 12\hfill & 21\hfill & \hfill & \hfill & 13.89256\hfill & 23.21488\hfill & -3.727272727\hfill & -4.81818182\hfill \\ 9\hfill & 20\hfill & \hfill & \hfill & 45.2562\hfill & 33.85124\hfill & -6.727272727\hfill & -5.81818182\hfill \\ 17\hfill & 28\hfill & \hfill & \hfill & 1.619835\hfill & 4.760331\hfill & 1.272727273\hfill & 2.181818182\hfill \\ 16\hfill & 27\hfill & \hfill & \hfill & 0.07438\hfill & 1.396694\hfill & 0.272727273\hfill & 1.181818182\hfill \\ 10\hfill & 23\hfill & \hfill & \hfill & 32.80165\hfill & 7.942149\hfill & -5.727272727\hfill & -2.81818182\hfill \\ 16\hfill & 30\hfill & \hfill & \hfill & 0.07438\hfill & 17.4876\hfill & 0.272727273\hfill & 4.181818182\hfill \end{array}$

The sample means and the sample variances can be calculated as shown.

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ \overline{x}=\frac{13+14+16+30+20+12+9+17+16+10+16}{11}\\ \overline{x}=\frac{173}{11}\\ \overline{x}=15.72727\end{array}$

  $\begin{array}{l}{s}^2=\frac{{\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}}{n-1}\\ s_x{}^2=\frac{{(13-15.72727)}^2+\mathrm{..............}+{(16-15.72727)}^2}{10}\\ s_x{}^2=\frac{326.1818}{10}\\ s_x{}^2=32.61818\\ s_x=5.71123279161\end{array}$

  $\begin{array}{l}\overline{y}=\frac{{\displaystyle \sum _{i=1}^n{y}_i}}{n}\\ \overline{y}=\frac{27+20+23+45+20+21+20+28+27+23+30}{11}\\ \overline{y}=\frac{284}{11}\\ \overline{y}=25.81818\end{array}$

  $\begin{array}{l}{s}^2=\frac{{\displaystyle \sum _{i=1}^n{(y_i-\overline{y})}^2}}{n-1}\\ s_y{}^2=\frac{{(27-25.81818)}^2+\mathrm{..............}+{(30-25.81818)}^2}{10}\\ s_y{}^2=\frac{533.6364}{10}\\ s_y{}^2=53.36364\\ s_y=7.30504209433\end{array}$

Now, one can use these to calculate the correlation coefficient as shown.

  $\begin{array}{l}r=\frac{1}{(n-1)}.\frac{{\displaystyle \sum _{i=1}^n(x_i-\overline{x})}}{s_x}.\frac{{\displaystyle \sum _{i=1}^n(y_i-\overline{y})}}{s_y}\\ r=\frac{1}{(11-1)}.\frac{7.1054274\times {10}^{-15}}{7.30504209433}.\frac{1.77636\times {10}^{-14}}{5.71123279161}\\ r=\frac{1}{10}(0.972674395\times {10}^{-15})(0.311029171\times {10}^{-14})\\ r=0.030253011\times {10}^{-29}\\ r=3.0253011\times {10}^{-31}\end{array}$

Finally, to calculate the least square regression line as shown, theser can be used.

  $\stackrel{\wedge }{y}=b_0+b_{1}x$

Where $b_0$ (intercept) and $b_1$ (slope) can be calculated as shown.

  $\begin{array}{l}{b}_1=r\frac{s_y}{s_x}\\ =(3.0253011\times {10}^{-31})\frac{7.30504209433}{5.71123279161}\\ =3.869559\times {10}^{-31}\\ =4.0\times {10}^{-31}\\ b_0=\overline{y}-b_1\overline{x}\\ =25.81818-(3.869559\times {10}^{-31})(15.72727273)\\ =25.81818-0\\ =26\end{array}$

To determine

To Calculate: the coefficient of determination

Answer to Problem 12RE

The coefficient of determination is $r^2=$

  $9.1524467\times {10}^{-62}$

Explanation of Solution

Given information: T leaves her home every day few minutes after $7.00$ A.M. The time she leaves the home (The number of minutes after $7.00$ A.M) and the number of minutes taken to office has been recorded as below.

$x$	$13$	$14$	$16$	$30$	$20$	$12$	$9$	$17$	$16$	$10$	$16$
$y$	$27$	$20$	$23$	$45$	$20$	$21$	$20$	$28$	$27$	$23$	$30$

Formulas Used:

Sample mean: $\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}$

Sample variance: $s^2=\frac{{\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}}{n-1}$

Correlation Coefficient: $r=\frac{1}{(n-1)}.\frac{{\displaystyle \sum _{i=1}^n(x_i-\overline{x})}}{s_x}.\frac{{\displaystyle \sum _{i=1}^n(y_i-\overline{y})}}{s_y}$

Calculation: the correlation coefficient can be calculated as shown by using the formula.

  $\begin{array}{l}r=\frac{1}{(n-1)}.\frac{{\displaystyle \sum _{i=1}^n(x_i-\overline{x})}}{s_x}.\frac{{\displaystyle \sum _{i=1}^n(y_i-\overline{y})}}{s_y}\\ r=\frac{1}{(11-1)}.\frac{7.1054274\times {10}^{-15}}{7.30504209433}.\frac{1.77636\times {10}^{-14}}{5.71123279161}\\ r=\frac{1}{10}(0.972674395\times {10}^{-15})(0.311029171\times {10}^{-14})\\ r=0.030253011\times {10}^{-29}\\ r=3.0253011\times {10}^{-31}\end{array}$

The correlation coefficient $r$ measures the strength of a linear relationship. Positive values of $r$

indicates a positive linear association. Here the value of the correlation coefficient close to zero.

Therefore, one can conclude that the positive linear relationship is weak.

Also,

To calculate the coefficient of determination we need to square the correlation coefficient.

Therefore, the coefficient of determination is $r^2=$

  $9.1524467\times {10}^{-62}$

To determine

To Find: the outlier point

Explanation of Solution

Given information: Tania leaves her home every day few minutes after $7.00$ A.M. The time she leaves the home (The number of minutes after $7.00$ A.M) and the number of minutes taken to office has been recorded as below.

$x$	$13$	$14$	$16$	$30$	$20$	$12$	$9$	$17$	$16$	$10$	$16$
$y$	$27$	$20$	$23$	$45$	$20$	$21$	$20$	$28$	$27$	$23$	$30$

Graph: The scatter plot shows the number of minutes after $7.00$ A.M on the $x$ -axis and the number of minutes taken to an office on the $y$ -axis.

  

Interpretation: An outlier is a value that considerably larger or considerably smaller than most of the values in a data set. It may be resulting from an error in the process of sampling.

So in the given data set, an outlier point can be detected in the ordered pair $(30,45)$ .

Because it is much larger than the other ordered pairs.

To determine

To Calculate: the least square regression line without the outlier point.

Answer to Problem 12RE

The least square regression line without theoutlier point is $\stackrel{\wedge }{y}=24+{10}^{-30}x$

Explanation of Solution

Given information: T leaves her home every day few minutes after $7.00$ A.M. The time she leaves the home (The number of minutes after $7.00$ A.M) and the number of minutes taken to office has been recorded as below.

$x$	$13$	$14$	$16$	$30$	$20$	$12$	$9$	$17$	$16$	$10$	$16$
$y$	$27$	$20$	$23$	$45$	$20$	$21$	$20$	$28$	$27$	$23$	$30$

Formulas Used:

Sample mean: $\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}$

Sample variance: $s^2=\frac{{\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}}{n-1}$

Correlation Coefficient: $r=\frac{1}{(n-1)}.\frac{{\displaystyle \sum _{i=1}^n(x_i-\overline{x})}}{s_x}.\frac{{\displaystyle \sum _{i=1}^n(y_i-\overline{y})}}{s_y}$

The least square regression line:

  $\begin{array}{l}\stackrel{\wedge }{y}=b_0+b_{1}x\\ b_1=r\frac{s_y}{s_x}\\ b_0=\overline{y}-b_1\overline{x}\end{array}$

Calculation: the sample means and the sample variances can be calculated as shown without the outlier point.

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ \overline{x}=\frac{13+14+1+20+12+9+17+16+10+16}{10}\\ \overline{x}=\frac{143}{10}\\ \overline{x}=14.3\end{array}$

  $\begin{array}{l}\overline{y}=\frac{{\displaystyle \sum _{i=1}^n{y}_i}}{n}\\ \overline{y}=\frac{27+20+23+20+21+20+28+27+23+30}{10}\\ \overline{y}=\frac{239}{10}\\ \overline{y}=23.9\end{array}$

  $\begin{array}{l}{s}^2=\frac{{\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}}{n-1}\\ s_x{}^2=\frac{{(13-14.3)}^2+\mathrm{..............}+{(16-14.3)}^2}{9}\\ s_x{}^2=\frac{102.1}{9}\\ s_x{}^2=11.3444444\\ s_x=3.36815148115\end{array}$

  $\begin{array}{l}{s}^2=\frac{{\displaystyle \sum _{i=1}^n{(y_i-\overline{y})}^2}}{n-1}\\ s_y{}^2=\frac{{(27-23.9)}^2+\mathrm{..............}+{(30-23.9)}^2}{9}\\ s_y{}^2=\frac{128.9}{9}\\ s_y{}^2=14.3222222\\ s_y=3.78447\end{array}$

Now, one can use these to calculate the correlation coefficient without the outlier as shown.

  $\begin{array}{l}r=\frac{1}{(n-1)}.\frac{{\displaystyle \sum _{i=1}^n(x_i-\overline{x})}}{s_x}.\frac{{\displaystyle \sum _{i=1}^n(y_i-\overline{y})}}{s_y}\\ r=\frac{1}{(10-1)}.\frac{-7.1054\times {10}^{-15}}{3.36815148115}.\frac{1.42109\times {10}^{-14}}{3.78447119159}\\ r=\frac{1}{9}(-2.1096\times {10}^{-15})(3.75504\times {10}^{-15})\\ r=8.80179146\times {10}^{-31}\end{array}$

Finally, one can use these to calculate the least square regression line as shown.

  $\stackrel{\wedge }{y}=b_0+b_{1}x$

Where $b_0$ (intercept) and $b_1$ (slope) can be calculated as shown.

  $\begin{array}{l}{b}_1=r\frac{s_y}{s_x}\\ =(8.80179146\times {10}^{-31})\frac{3.78447119}{3.36815148115}\\ =9.889735182\times {10}^{-31}\\ b_0=\overline{y}-b_1\overline{x}\\ =23.9-(9.889735182\times {10}^{-31})(14.3)\\ =23.9-141.4232131\times {10}^{-31}\\ =23.9-0\\ =23.9\end{array}$

To determine

To Show: the outlier is influential

Answer to Problem 12RE

No, the outlier is not that much influential.

Explanation of Solution

Given information: T leaves her home every day few minutes after $7.00$ A.M. The time she leaves the home (The number of minutes after $7.00$ A.M) and the number of minutes taken to office has been recorded as below.

$x$	$13$	$14$	$16$	$30$	$20$	$12$	$9$	$17$	$16$	$10$	$16$
$y$	$27$	$20$	$23$	$45$	$20$	$21$	$20$	$28$	$27$	$23$	$30$

The least-square regression line with the outlier is $\stackrel{\wedge }{y}=26+4\times {10}^{-31}x$ .

The least-square regression line without the outlier is $\stackrel{\wedge }{y}=24+{10}^{-30}x$ .

The two least square regression lines are so much close and have no huge difference.

Therefore we can conclude that the outlier is not that much influential. Here, the outlier cannot be a result of an error. It is just random data measured along with the other data in the sampling process.

To determine

To Find: the coefficient of determination without the outlier and discuss its strength.

Answer to Problem 12RE

Coefficient of determination is $r^2=$

  $77.4715329053$

  $\times {10}^{-62}$.

Explanation of Solution

Given information: T leaves her home every day few minutes after $7.00$ A.M. The time she leaves the home (The number of minutes after $7.00$ A.M) and the number of minutes taken to office has been recorded as below.

$x$	$13$	$14$	$16$	$30$	$20$	$12$	$9$	$17$	$16$	$10$	$16$
$y$	$27$	$20$	$23$	$45$	$20$	$21$	$20$	$28$	$27$	$23$	$30$

Formula Used:Sample mean: $\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}$

Sample variance: $s^2=\frac{{\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}}{n-1}$

Correlation Coefficient: $r=\frac{1}{(n-1)}.\frac{{\displaystyle \sum _{i=1}^n(x_i-\overline{x})}}{s_x}.\frac{{\displaystyle \sum _{i=1}^n(y_i-\overline{y})}}{s_y}$

Calculation: to calculate the correlation coefficient without the outlier as shown.

  $\begin{array}{l}r=\frac{1}{(n-1)}.\frac{{\displaystyle \sum _{i=1}^n(x_i-\overline{x})}}{s_x}.\frac{{\displaystyle \sum _{i=1}^n(y_i-\overline{y})}}{s_y}\\ r=\frac{1}{(10-1)}.\frac{-7.1054\times {10}^{-15}}{3.36815148115}.\frac{1.42109\times {10}^{-14}}{3.78447119159}\\ r=\frac{1}{9}(-2.1096\times {10}^{-15})(3.75504\times {10}^{-15})\\ r=8.80179146\times {10}^{-31}\end{array}$

The correlation coefficient $r$ measures the strength of a linear relationship. Positive values of $r$

indicates a positive linear association. It is $r=8.8\times {10}^{-31}$ without the outlier.

The value is so close to zero .because of the ten to the power is minus thirty-one.

Therefore, one can conclude that the positive linear relationship is very weak without the outlier.

Also,

One can calculate the coefficient of determination by squaring the correlation coefficient.

Therefore, the coefficient of determination is $r^2=$$77.4715329053$$\times {10}^{-62}$.