the following results were obtained from a simple regression

the following results were obtained from a simple regression analysis:

=37.2895-1.2024X

r² = .6744 Sb= .2934

1) for each unit change in X (independent variable), what is the estimated change in Y (dependent variable)?

2) when X (independent variable) is equal to zero, what is the estimated value of Y (dependent variable)?

3) what is the proportion of the variation explained by the simple linear regression model?

4) if the result was obtained from a sample with 12 observations. determine if there is a significant negative relationship between the independent and dependent variables at =.05. give the test statistic and the resulting conclusion.

Solution

The following results were obtained from a simple regression analysis:

Y^ = 37.2895 - 1.2024X

r² = .6744

Sb= .2934

1) for each unit change in X (independent variable), what is the estimated change in Y (dependent variable)?

The coefficient indicates that for every X you can expect Y to decrease by 1.2024

2) when X (independent variable) is equal to zero, what is the estimated value of Y (dependent variable)?

When X = 0, then

Y^ = 37.2895 - (1.2024 * 0)

Y^ = 37.2895

3) what is the proportion of the variation explained by the simple linear regression model?

proportion of the variation explained by the simple linear regression model is denoted by R².

and R² = 0.6744 given in the question.

R² = 0.6744 * 100 = 67.44%

that is 67.44% of the variation of y-values around the mean are explained by the x-values.

4) if the result was obtained from a sample with 12 observations. determine if there is a significant negative relationship between the independent and dependent variables at =.05. give the test statistic and the resulting conclusion.

n = number of observations = 12

Here we have to test the hypothesis that,

H0 : rho = 0 Vs H1 : rho < 0

alpha = 0.05

The test statistic is,

t = r*sqrt(n-2) / sqrt(1-r² )

where r id the correlation coefficient.

r = sqrt(R²) = sqrt(0.6744) = 0.8212

n = 12

t = 0.8212 * sqrt(12-2) / sqrt(1-0.8212² ) = 4.5512

We calculate P-value for taking decision.

EXCEL syntax for P-value :

=TDIST(x, d.f., tails)

x is the test statistic value

d.f. = n - 2 = 12 - 2 = 10

tails = 1

P-value = 0.000528

P-value < alpha (0.05)

Reject H0 at 5% level of significance.

There is a significant negative relationship between the independent and dependent variables