91影视

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

Let x represent the overhead width.

Let y represent the weight

Themean value of xis given as,

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{7.2 + 7.4 + .... + 8.4}}{6}\\ = 8.5\end{array}\)

Therefore, the mean value of x is 8.5.

Themean value of yis given as,

\(\begin{array}{c}\bar y = \frac{{\sum\limits_{i = 1}^n {{y_i}} }}{n}\\ = \frac{{116 + 154 + .... + 191}}{6}\\ = 184.667\end{array}\)

Therefore, the mean value of y is 184.67.

The standard deviation of xis given as,

\(\begin{array}{c}{s_x} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {7.2 - 8.5} \right)}^2} + {{\left( {7.4 - 8.5} \right)}^2} + ..... + {{\left( {8.4 - 8.5} \right)}^2}}}{{6 - 1}}} \\ = 1.049\end{array}\)

Therefore, the standard deviation of x is 1.05.

The standard deviation of y is given as,

\(\begin{array}{c}{s_y} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({y_i} - \bar y)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {116 - 184.667} \right)}^2} + {{\left( {154 - 184.667} \right)}^2} + ..... + {{\left( {191 - 184.667} \right)}^2}}}{{6 - 1}}} \\ = 44.433\end{array}\)

Therefore, the standard deviation of y is 44.43.

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.

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.

The interceptis computed as,

\(\begin{array}{c}{b_0} = \bar y - {b_1}\bar x\\ = 184.667 - \left( {40.176 \times 8.5} \right)\\ = - 156.829\\ \approx - 157\end{array}\)

Therefore, the value of intercept is -157.

Theregression equationis given as,

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = - 157 + 40.2x\end{array}\)

Thus, the regression equation is \(\hat y = - 157 + 40.2x\).

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.

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.

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.