Question

15-E6. Below is some output from a regression model of the amount spent on pay TV optional extras using the number of optional extras purchased in the previous year as the predictor. There were 83 individuals in the database. The mean expenditure was $20.02 with a standard deviation of $14.06. The mean number of optional purchases last year was 2.783, with a standard deviation of 1.828.

Results of multiple regression for Expend
Summary Measures				Predicted Value		12.261
Adj RSquare		13.1%		Pred Error		13.36
Model Error		13.19		Trend Error		2.75
p-value		0.0008
Regression coefficients
	Coefficient	Std Err	t-value	p-value	Lower	Upper	Partial
Constant	12.261	2.648	4.63	0.000	6.992	17.529
PP	2.787	0.797	3.50	0.001	1.202	4.372	0.363

1. Is the number of previous options purchased a significant predictor of expenditure this year? How sure are you?
2. How would you explain the regression equation in simple words?
3. How accurate is the regression equation either in terms of typical prediction error or in terms of variation in options explained?
4. Predict the expenditure of a customer who purchased five optional extras last year.
5. Suppose I did not give you an figures in the Summary Stats section. How could you calculate the correlation of Expenditure with Number last year?

Answer 1

a) The number of previous options purchased is a significant predictor of expenditure as the p-value associated with the t-statistic (t = 3.50) is 0.001. Hence, there is only a 0.001 probability of a type 1 error.

b) We can say that if there is increase in one previous options purchased, the expected increase in the expenditure is $2.787 on average. If there are no options purchased, on an average the expenditure is $12.261.

c) The % variation explained in the dependent variable is 13.1%.

d) The regression equation is:

y = 12.261 + 2.787*PP
y = 12.261 + 2.787*5
y = $26.196

e) The correlation coefficient can be calculated by the following formula:

r = b1*(Sx/Sy)
r = 2.787*(1.828/14.06) = 0.36
r-square = 0.131