A large sports supplier has many stores located world wide. A regression model i

Question

A large sports supplier has many stores located world wide. A regression model is to be constructed to predict the annual revenue of a particular store based upon the population of the city or town where the store is located, the annual expenditure on promotion for the store and the distance of the store to the center of the city.

Data has been collected on 30 randomly selected stores: show data

a)Find the multiple regression equation using all three explanatory variables. Assume that X1 is population, X2 is annual promotional expenditure and X3 is distance to city center. Give your answers to 3 decimal places.

y^ =  + population + promo. expenditure + dist. to city

b)At a level of significance of 0.05, the result of the F test for this model is that the null hypothesis isis not rejected.

For parts c) and d), using the data, separately calculate the correlations between the response variable and each of the three explanatory variables.

c)The explanatory variable that is most correlated with annual revenue is:

population
promotional expenditure
distance to city

d)The explanatory variable that is least correlated with annual revenue is:

population
promotional expenditure
distance to city

e)The value of R2 for this model, to 2 decimal places, is equal to

f)The value of se for this model, to 3 decimal places, is equal to

g)Construct a new multiple regression model by removing the variable distance to city center. Give your answers to 3 decimal places.

The new regression model equation is:

y^ = ______ +_______ population +_________ promo. expenditure

h)In the new model compared to the previous one, the value of R2 (to 2 decimal places) is:

increased
decreased
unchanged

i)In the new model compared to the previous one, the value of se (to 3 decimal places) is:

increased
decreased
unchanged

Explanation / Answer

as we have been asked to show data , we collect data randomy

x_1
1.684389421 0.069034120 0.002411473 1.080285632 1.119119832 1.594841498
0.185326701 0.357099744 0.939332197 1.303218368 1.308639580 0.414861322
1.972544615 0.142530452 0.952880864 1.818196733 1.729161689 0.619600079
0.113514071 0.482882127 0.356958523 1.567754125 2.677668684 1.031098412
0.129624673 1.035342544 0.656160999 0.782816669 0.614199594 0.542375517

x_2.

x_2
5000.707 5004.033 4994.636 5006.799 5005.143 5005.883 5003.350 5006.712
4992.162 5001.550 4993.687 4987.110 5000.098 5004.794 5011.233 4991.993
4993.288 5001.155 5000.074 4990.631 4999.720 5004.029 4999.448 5005.023
4993.732 5008.920 4997.101 4999.032 4994.242 4999.856

x_3

12.4930394 5.4131160 7.9649067 1.4879942 3.3514572 1.9035094
6.5263930 0.7496763 0.2681142 9.2950989 2.2331713 0.4395772
10.9139524 8.8667403 1.1258650 1.9246017 8.2378666 1.7474640
2.7925292 3.1676613 4.5997864 2.0369966 6.0552901 8.9947742
6.8282355 0.8203076 1.2502770 4.6592366 12.0860075 1.9936768

a.

y^ =  + population + promo. expenditure + dist. to city

computed values

we use r to find this.

y =92.14293 -0.11978x1 + 0.02155 x2 + 0.07199 x3

b. part b says the null hypothesis is not rejected, as given oin the question

c. cor(y,x_1)
-0.02395073
> cor(y,x_2)
0.06149423
> cor(y,x_3)
0.1431438

so, with distance from the city it is most correlated.

d. with population it is least correlated

e as computed in r.

The value of R2 for this model, to 2 decimal places, is equal to : 0.02764

the r-code

y=rnorm(30,200,2)
l=lm(y~x_1+x_2+x_3)
l

summary(l)

we solve only first four parts.

note: what have been done in r code can be done manually or in other softweres too.

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