Question 1 a) Independent variable is advertising. Dependent variable is sales.
The scatter plot shows that there is a positive relationship between amount spent on advertising and sales b) Sales= -4.9913+0.1227(advertising) c) For every additional dollar spent on advertising we can expect sales to increase by an average of 0.1227 dollars d) 0.87092
87% of variance in sales is predictable by amount spent on advertising. e) No. The histogram of residuals is evenly distributed around zero. Most error are at or close to zero f) Yes. Points in the residual plot is lying on a curve around zero. A pattern is seen. (Cheng-Few, John, Alice, 2013, p.800). This indicates that the model is not satisfactory. g) -4.9913+0.1227*90000
=$11038.0087
h) H0: p=0, H1: p ≠0 at α= 0.05 p-value= 5.68×10-14

If p-value is less than α reject H0

P-value < 0.05. This means that the probability that the observed results are due to random choice is low. So reject null hypothesis (H0)
Conclusion: At the 5% significance level there is evidence of linear association between advertising expenditure and sales. i) b± t*( sb ) (see appendix 1 )
=0.1227 ± (2.048* 0.0089)
= (0.104, 0.141)
It could be as low as 0.104 or as high as 0.141

A=8, B= 90, C=3, D=40, E=9.7327, F= 1.9759, G= 4.8247, H=-2.8247, I=6.2971 (See Appendix 2)
A=8, B= 90, C=3, D=40, E=9.7327, F= 1.9759, G= 4.8247, H=-2.8247, I=6.2971 (See Appendix 2)
j)

Question 2

a)

= -0.66061
b) There is a moderate negative correlation between the scores of judge 1 and the scores of judge 2. Scores of judge 2 decreases when scores of judge 1 increases.

Appendix
Appendix 1: Estimating slope of regression line b± t*( SEb ) where b= slope of least squares regression line i.e 0.1227 t*= the upper (1−C)/2 critical value from the t(n – 2) distribution (which means