Question

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 squaw fish (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.
x=524.880,
y=413.975,sx=12.251,sy=12.300,
r=0.9562
options:
a) ^y=0.960xâ89.910
b) ^y=0.960x+89.910
c) ^y=â89.910x+0.960
d) ^y=0.952x+89.910
e) ^y=0.952xâ89.910
f) None of the above

Answer 1

the correct answer is not there. here it is:

^y = 0.960x - 89.910

Explanation:

we have regression line equation as:

^y = a+bx

we have the following details available

x = 524.88

y = 413.975

Sx = 12.251

Sy = 12.300

r = 0.9562

we have to find values of a and b

b = rSy/Sx

= 0.9562(12.3)/12.251

= 11.76126/12.251

= 0.960

a = y - intercept

= 413.975-0.96(524.88)

= 413.975-503.8848

= - 89.9098

a = -89.910

putting these values into the equation, we have:

^y = -89.910 + 0.9600x

which is also

^y = 0.960x - 89.910