Question

Exercises 13–28 use the same data sets as Exercises 13–28

in Section 10-1. In each case, find the regression equation, letting the first variable be the predictor (x) variable. Find the indicated predicted value by following the prediction procedure summarized in Figure 10-5 on page 493.

Using the listed width/weight data, find the best predicted weight of a seal if the overhead width measured from a photograph is 2 cm. Can the prediction be correct? If not, what is wrong?

Overhead Width	7.2	7.4	9.8	9.4	8.8	8.4
Weight	116	154	245	202	200	191

Short answer

The regression equation is$\hat{y}=-157+40.2x$.

The best predicted weight of a seal if the overhead width measured from a photograph is 2 cm is -76.6.

The predicted weight is negative which is not possible in general. Also, the prediction is exptrapolated.

Step-by-step solution

Given informationa

Values are given on two variables namely, overhead width and weight.

Calculate the mean values

Let x represent the overhead width.

Let y represent the weight

Themean value of xis given as,

$\begin{array}{c}\overline{x}=\frac{\sum _{i=1}^n{x}_i}{n}\\ =\frac{7.2+7.4+....+8.4}{6}\\ =8.5\end{array}$

Therefore, the mean value of x is 8.5.

Themean value of yis given as,

$\begin{array}{c}\overline{y}=\frac{\sum _{i=1}^n{y}_i}{n}\\ =\frac{116+154+....+191}{6}\\ =184.667\end{array}$

Therefore, the mean value of y is 184.67.

Calculate the standard deviation of x and y

The standard deviation of xis given as,

$\begin{array}{c}{s}_x=\sqrt{\frac{\sum _{i=1}^n{(x_i-\overline{x})}^2}{n-1}}\\ =\sqrt{\frac{{\left(7.2-8.5\right)}^2+{\left(7.4-8.5\right)}^2+.....+{\left(8.4-8.5\right)}^2}{6-1}}\\ =1.049\end{array}$

Therefore, the standard deviation of x is 1.05.

The standard deviation of y is given as,

$\begin{array}{c}{s}_y=\sqrt{\frac{\sum _{i=1}^n{(y_i-\overline{y})}^2}{n-1}}\\ =\sqrt{\frac{{\left(116-184.667\right)}^2+{\left(154-184.667\right)}^2+.....+{\left(191-184.667\right)}^2}{6-1}}\\ =44.433\end{array}$

Therefore, the standard deviation of y is 44.43.

Calculate the correlation coefficient

The correlation coefficient is given as,

$r=\frac{n\left(\sum xy\right)-\left(\sum x\right)\left(\sum y\right)}{\sqrt{\left(\left(n\sum x^2\right)-{\left(\sum x\right)}^2\right)\left(\left(n\sum y^2\right)-{\left(\sum y\right)}^2\right)}}$

The calculations required to compute the correlation coefficient are as follows:

The correlation coefficient is given as,

$\begin{array}{l}r=\frac{n\left(\sum xy\right)-\left(\sum x\right)\left(\sum y\right)}{\sqrt{\left(\left(n\sum x^2\right)-{\left(\sum x\right)}^2\right)\left(\left(n\sum y^2\right)-{\left(\sum y\right)}^2\right)}}\\ =\frac{6\left(9639\right)-\left(51\right)\left(1108\right)}{\sqrt{\left(\left(6\times 439\right)-{\left(51\right)}^2\right)\left(\left(6\times 214482\right)-{\left(1108\right)}^2\right)}}\\ =0.9485\end{array}$

Therefore, the correlation coefficient is 0.9485.

Calculate the slope of the regression line

The slopeof the regression line is given as,

$\begin{array}{c}{b}_1=r\frac{s_Y}{s_X}\\ =0.9485\times \frac{44.433}{1.049}\\ =40.2\end{array}$

Therefore, the value of slope is 40.2.

Calculate the intercept of the regression line

The interceptis computed as,

$\begin{array}{c}{b}_0=\overline{y}-b_1\overline{x}\\ =184.667-\left(40.176\times 8.5\right)\\ =-156.829\\ \approx -157\end{array}$

Therefore, the value of intercept is -157.

Form a regression equation

Theregression equationis given as,

$\begin{array}{c}\hat{y}=b_0+b_{1}x\\ =-157+40.2x\end{array}$

Thus, the regression equation is $\hat{y}=-157+40.2x$.

Analyze the model

Referring to exercise 23 of section 10-1,

1)The scatter plot shows a linear relationship between the variables.

2)The P-value is 0.004.

As the P-value is less than the level of significance (0.05), this implies the null hypothesis is rejected.

Therefore, the correlation is significant.

Referring to figure 10-5, the criteria for a good regression model are satisfied.

Therefore, the regression equation can be used to predict the value of y.

Predict the value

The best predicted weight of a seal if the overhead width measured from a photograph is 2 cm is computed as,

$\begin{array}{c}\hat{y}=-157+\left(40.2\times 2\right)\\ =-76.6\end{array}$

Therefore, the best predicted weight of a seal if the overhead width measured from a photograph is 2 cm is -76.6 kg.

State if the prediction is correct

Thepredicted weight is negativewhich is not possible in general.

Also, the measure 2 cm is extreme as compared to observations of the sampled responses. Thus, the prediction is not suggested to be made as it would lead to extrapolation resulting in inaccurate prediction.

Therefore, the prediction is not correct.