Question

4) A multiple regression model is developed to predict Innovative Index, to check for the possibility of collinearity among iust the predictor variables. Data were collected on the following variables: innovative index (higher scores indicate a more innovative and creative organizational culture), job growth (in % ) and number of employees. Based on the results shown below, a regression model was run to predict innovative index based on job growth and number of employees. The regression equation is: Innovative Index 40.9 +3.99 Job Growth 0.00612 Employees P-value 0.000 0.000 Predictor Coef SE Coef 8,162 t Constant 5.01 2 Job Growth No. of Employees 3.9863 -0.00612 0.8912 4.47 0.009296 -0.66 0.518 s13.1511 -52.5% R-Sq Find the adjusted R2 What is the predicted innovative index for a company with 500 employees and a job growth of 5% ? Test a hypothesis to determine if the number of employees is a significant predictor in the presence of job growth. Use a-0.05. What is the variance inflation factor (VIF) for the variable Innovative Index? (3) i) (3) ii) (3) iii) (3) iv)

Answer 1

i)

We need the value of n (number of observations) and k (number of predictor variables)

k = 2

Since n is not given, the degree of freedom for residual is 17 (based on 2 * P(t - 0.66) = 0.518))

Thus,

n - k - 1 = 17

=> n = 17 + 2 + 1 = 20

Adjusted R² = 1 - (1- R²) * (n - 1) / (n - k - 1)

= 1 - (1 - 0.525) * (20 - 1) / (20 - 2 - 1)

= 0.469

= 46.9%

(b)

The regression equation is,

Innovative Index = 40.9 + 3.99 Job Growth - 0.00612 Employees

For Employees = 500, Job Growth = 5

Innovative Index = 40.9 + 3.99 * 5 - 0.00612 * 500 = 57.79

(c)

H0:

Emplyees = 0

H1:

Emplyees 0

P-value of the test statistic for No. of employees is 0.518

Since, p-value is greater than 0.05 significance level, we fail to reject null hypothesis H0 and conclude that there is no strong evidence that and there is no strong evidence that number of employees is a significant predictor in presence of job growth.

Emplyees 0

(d)

VIF = 1 / (1 - R²)

= 1 / (1 - 0.525)

= 2.105263