Question

Zenith Computers, Texas would like to predict weekly Internet sales based on the number of orders. Data (over 15 weeks), relating the sales volume (in thousands of dollars) to the number of orders were available. Regression analysis was performed using Excel. Output related to the regression is given below.

Week	Orders	Sales ($1000)
1	265	15.3
2	150	18.4
3	131	11.6
.	.	.
.	.	.

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.795
R Square	0.632
Adjusted R Square	0.604
Standard Error	10.706
Observations	15.000
ANOVA
	df	SS	MS	F	Significance F
Regression	1	2564.2	2564.2	22.37	0.000
Residual	13	1490.0	114.6
Total	14	4054.2
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	5.521	5.690	0.970	0.350	-6.77	17.81
Orders	0.044	0.009	4.730	0.000	0.024	0.064

What is the slope of the least squares line?

a) 5.521

b) 10.706

c) 0.632

d) 0.044

About what percentage of the variation in Sales is explained by its regression on Orders?

a) 5.52

b) 63.2

c) 10.706

d) 4.4

What are the predicted sales in dollars for the third observation appearing in the data set?

a) 44

b) 10,706

c) 5521

d) 11,285

What is the average distance between the actual sales and the predicted sales in $1000s?

a) 60.4

b) 10.706

c) 4.4

d) 63.2

Answer 1

Here y=sales

X=oredes

From given output

1)What is the slope of the least squares line

slope =Coefficient of orders =0.044

Answer- d) 0.044

2)About what percentage of the variation in Sales is explained by its regression on Orders?

R square explain the percentage of variation in sales =63.2

Answer - b) 63.2

3) What are the predicted sales in dollars for the third observation appearing in the data set?

The regression line equation is

Sales =5.521+0.044*orders

Here Orders =5.521+131*0.044=11285 dollars

Answer - d) 11,285

4)What is the average distance between the actual sales and the predicted sales in $1000s?

Standard error gives the average distance between the actual sales and predicted sales

Answer - b) 10.706