Answer:
(a) The regression line is: y = 6.343 + 1.523 x
(b) For a gestation period of 10 months, the life span is 22 years.
Step-by-step explanation:
The regression line is of the form: y = a + b x.
Here,
y = dependent variable = life span of animals
x = independent variable = gestation periods
a = intercept
[tex]=\frac{\sum Y.\sum X^{2}-\sum X\sum XY}{n.\sum X^{2}-(\sum X)^{2}}[/tex]
b = slope
[tex]=\frac{n.\sum XY-\sum X\sum Y}{n.\sum X^{2}-(\sum X)^{2}}[/tex]
(a)
Compute the value of intercept as follows:
[tex]a=\frac{\sum Y.\sum X^{2}-\sum X\sum XY}{n.\sum X^{2}-(\sum X)^{2}}=\frac{(138\times836.37)-(57.3\times1637.1)}{(8\times836.37)-(57.3)^{2}} =6.343[/tex]
The intercept is 6.343.
Compute the value of slope as follows:
[tex]b=\frac{n.\sum XY-\sum X\sum Y}{n.\sum X^{2}-(\sum X)^{2}}=\frac{(138\times1637.1)-(57.3\times138)}{(8\times836.37)-(57.3)^{2}} =1.523[/tex]
The slope is 1.523.
The regression line is:
y = 6.343 + 1.523 x
(b)
Foe x = 10 compute the value of y as follows:
[tex]y = 6.343 + 1.523 x\\=6.343+(1.523\times10)\\=21.573\\\approx22[/tex]
Thus, for a gestation period of 10 months, the life span is 22 years.
