91影视

A regression line is a straight line that shows how an explanatory variable $x$ affects a response variable $y$ You can use a regression line to forecast the value of $y$ for any value of $x$ by plugging this $x$ into the equation of the line.

From the above figure we can clearly see that $$\begin{gathered} slope\left(b\right)=鈭�2.9935 \\ Y鈭�intercept\left(a\right)=157.68 \end{gathered}$$

As a result, the least-squares regression line's equation is $\stackrel{铃�}{y}=鈭�2.9935x+157.68$

The percentage of males (banded for identification) who returned for the next breeding is recorded in the study. The % return is the response variable, and this is the study's outcome. A second variable is now breeding pairings. The entire goal of the study is to see if breeding pairings have any effect on the percent return. Explanatory variables include breeding pairs. To predict % return, we shall employ breeding pairs (explanatory variable or predictor) (response or predicted variable). The dependent variable is percent return, while the independent variable is breeding pairs. The equation for forecasting the % of men who return from the number of breeding couples using the least-squares regression line is.

Percent return $=157.68鈭�2.9935$(Breeding pairs)

After a season with 30 breeding pairs, the percentage of returning males is,

(Predicted) Percent return $$\begin{gathered} =157.68鈭�2.9935脳30 \\ =67.875 \end{gathered}$$

There are $67.875\%$of males that return after a season with$30$ breeding pairs.

We can observe from the preceding graph that $r^2=63.1\%$ is a positive number. We interpret this to mean that the straight-line association with a number of breeding pairs the previous year explains $63.1$ percent of the year-to-year fluctuation in the percent of returning males. Therefore, the percentage of variation explained by the regression line is$r^2=63.1\%$

From the above figure we can clearly see that $slope\left(b\right)=鈭�2.9935$

This demonstrates a negative relationship between variables.

$$\begin{gathered} R鈭�sq\left(R^2\right)=63.1\% \\ {r}^2=0.631 \\ correlation\left(r\right)=\sqrt{r^2} \\ =\sqrt{0.631} \\ =鈭�0.794 \end{gathered}$$

The percentage of returning males and the number of breeding couples are negatively associated, with a correlation coefficient of $0.794$

The residuals' standard deviation for these data is: $s=9.46334$

When utilizing the regression line, the standard deviation of the residuals s reflects the average size of the prediction errors (residuals), and s = 9.46334 indicates that there is too much prediction error.

Therefore, the prediction of the percent returning is varied by about $9.46334\%$ from the actual percent returning.