In 1929 edwin hubble investigated the claim that distance (explanatory) and radial velocity (response) of extragalactic nebulae are positively linearly related. hubble's data is plotted below along with the relevant diagnostic plots. these are the plots and charts needed to analyze data are given below. (assume all observations are independent) reference: hubble, e. (1929) "a relationship between distance and radial velocity among extra-galactic nebulae," proceedings of the national academy of science, 168.

the explanatory variable is distance (in megaparsecs) and the response is radial velocity (velocity away or towards the earth). use 3 decimal places for the following questions.
(a) what is the regression equation?
(b) what is the estimated mean velocity of objects that are 1.9 megaparsecs from earth?

a) Regression equation which is y/x = -40.758 + 454.158 x

b) estimated mean velocity of objects that are 1.9 mega parsecs from Earth

basically  = 862.9

Step-by-step explanation:

a) Regression Variable

In this statement the independent variable is the distance form the earth and the dependent variable is the velocity of the Extra - Galactic Nebulae.

Construction of hypothesis

H₀ : b₁ = 0 and H₁ : b₁ < 0

H₀ is null hypothesis that is tested the distance of extra - galactic nebulae from Earth .It does not have any effect on the velocity of the extra - galactic nebulae

From the given statement

coefficient of the constant is -40.784 after calculations and standard deviation is 83.439 and t -statistic is -0.49 and p value is 0.6300

and coefficient of distance from earth is 454158 and standard error is -75 237 and t value is 6.04

from these calculation we calculate the regression equation which is

y/x = -40.758 + 454.158 x

B) estimated mean velocity of objects that are 1.9 mega parsecs from Earth

basically

coefficient of distance from earth is 454.158 * 1.9 from Earth

= 454.158 * 1.9 = 862.9

c. 16 units

step-by-step explanation:

mark brainliest and have a great day!

step-by-step explanation: