Question

question number 6. the decline of salmon fisheries along the columbia river in oregon has caused great concern among commercial and recreational fishermen. the paper feeding of predaceous fishes on out - migrating juvenile salmonids in john day reservoir, columbia river (trans. amer. fisheries soc. (1991): 405 - 420) gave the accompanying data on y = maximum size of salmonids consumed by a northern squawfish (the most abundant salmonid predator) and x = squawfish length, both in mm. use the following statistics to give the equation of the least squares regression line. $\bar{x}=524.990$, $\bar{y}=492.689$, $s_x = 10.602$, $s_y = 12.100$, $r = 0.9726$ $hat{y}=1.110x + 90.050$ $hat{y}=1.110x - 90.050$ $hat{y}=0.852x - 90.050$ $hat{y}=-90.050x + 1.110$ $hat{y}=0.852x + 90.050$ none of the above

Explanation:

Step1: Recall the formula for the least - squares regression line

The equation of the least - squares regression line is $\hat{y}=b_0 + b_1x$, where $b_1=r\frac{s_y}{s_x}$ and $b_0=\bar{y}-b_1\bar{x}$.

Step2: Calculate the slope $b_1$

Given $r = 0.9726$, $s_x=10.602$, $s_y = 12.100$. Then $b_1=r\frac{s_y}{s_x}=0.9726\times\frac{12.100}{10.602}\approx1.110$.

Step3: Calculate the intercept $b_0$

Given $\bar{x}=524.990$, $\bar{y}=492.689$, $b_1\approx1.110$. Then $b_0=\bar{y}-b_1\bar{x}=492.689-1.110\times524.990=492.689 - 582.7389=- 90.0499\approx - 90.050$.

Answer:

$\hat{y}=1.110x - 90.050$