1. A company has yearly data for its sales volumes, advertising expenditure (Adv
ID: 1110577 • Letter: 1
Question
1. A company has yearly data for its sales volumes, advertising expenditure (Advertising) and the number of sales people (People), for ten years. It thinks that the second two variables explain sales volumes and runs a regre results ssion with the following Coefficient Standard Deviation 7.0174 8.6233 0.0858 Predictor Constant Advertising People 5.3146 2.3968 0.1845 Each part is worth 5 marks. (a) Test the significance of the coefficient for Advertising, using a 5% significance level and the critical value approach. critical value approach. with explanations explaining how you got your answers) (b) Test the significance of the coefficient for People, using a 5% significance level and the (c) Provide the missing information for the following ANOVA table (reproduce it in your answer Sum of Degrees of Mean SquareF Squares Freedom p-value Source of Variation Regression Error Total 321.11 63.39 384.50 (d) At the 1% level of significance, does the fitted equation represent a significant relationship between the dependent variable and the independent variables. (e) If unadjusted R2 is 0.8351, what is adjusted R2? (f) Now assume a dummy variable is added to the regression. The dummy variable is equal to 1 if only TV advertising is used and 0 otherwise. The following ANOWA table results: Sum of Squares Freedom 360.59 3 23.91 6 384.5 9 Source of Variation Degrees of Mean Square 20.197 3.985 Regression Error Total Use this information to determine if the new variable contributes significantly to the model. Let = 0.05 30.16Explanation / Answer
a) Check the null hypothesis
H0: 1= 0
Against 1 0
We will calculate t = 1(hat)/SE(1(hat))
t= 8.6233/2.3968 = 3.60
Since the t calculated above is sufficiently large so we reject the null hypothesis
b) Check the null hypothesis
H0: 2= 0
Against 2 0
We will calculate t = 2(hat)/SE(2(hat))
t= 0.0858/0.1845 = 0.465
Since the t calculated above is very small so we do not reject the null hypothesis
c) Degrees of freedom
Regression = k-1 = 3-1 =2
Error = n-k = 10 – 3 =7
Total = n-1 = 9
Mean Square
Regression = 321.11/2 = 160.55
Error = 63.39/7 = 9.05
Total = 384.50/9 = 42.72
d) F-value = 160.55/42.72 = 3.75
We will reject the null at 1% level.
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