Question

Role of retailer interest on shopping behavior. Retail interest is defined by marketers as the level of interest a consumer has in a given retail store. Marketing professors investigated the role of retailer interest in consumers’ shopping behavior (Journal of Retailing, Summer 2006). Using survey data collected for n = 375 consumers, the professors developed an interaction model for y = willingness of the consumer to shop at a retailer’s store in the future (called repatronage intentions) as a function of = consumer satisfaction and = retailer interest. The regression results are shown below.

(a) Is the overall model statistically useful for predicting y? Test using a=0.05

(b )Conduct a test for interaction at a= 0.05.

(c) Use the estimates to sketch the estimated relationship between repatronage intentions (y) and satisfaction when retailer interest is x₂=1 (a low value).

(d)Repeat part c when retailer interest is x₂= 7(a high value).

(e) Sketch the two lines, parts c and d, on the same graph to illustrate the nature of the interaction.

Short answer

(a) At 95% significance level, it can be concluded $\beta$_{1 $\ne \beta$2}$\ne$$\beta$₃$\ne 0$

(b) Since, $\beta \ne 0$. Hence it can be concluded with enough evidence that x₁and x₂ interact in the model. (c) Graph

(d)Graph

(e) Graph

Step-by-step solution

Overall goodness of the fit for the model

H₀= $\beta$₁= $\beta$₂= $\beta$₃= 0

H_a= At least one of the parameters is non zero

Here, F test statistic = SSE/ n-(k+1 ) = 226.35

Value of F_0.05,371,371 is 1

H₀ is rejected if F statistic > F_0.05,371,371 . For a= 0.05, since F > F_0.05,371,371

Sufficient evidence to reject at 95% confidence interval.

Therefore,$\beta$₁$\ne \beta$₂$\ne \beta$₃$\ne 0$

Significance ofβ3

H₀: $\beta$₃= 0

H_a: $\beta$₃$\ne 0$

Here, t-test statistic = -0.157/-3.09 =0.0508

Value of t_0.05374 is 1.645

H₀is rejected if t statistic > t_0.05374.

For a= 0.05, since t < t_0.05374.

Not sufficient evidence to reject at a 95% confidence interval.

Therefore, $\beta$₃ $\ne$0.$\beta$

Hence it can be concluded with enough evidence that x₁and x₂ interact in the model.

Graph

Given, E(y) = 0.4226x₁ + 0.044x₂ - 0.157x₁x₂for x₂=1 .

y= 0.4226x₁ + 0.044(1) - 0.157(1)x₁for x₂ =1 .

y = 0.044 + 0.269x₁

for Now to plot this equation, make a table

Y	0.044	0
X1	0	-0.1635

Graph

Given, E(y) = 0.4226x₁ + 0.044x₂ - 0.157x₁x₂for x₂=7 .

y= 0.4226x₁ + 0.044(7) - 0.157(7 )x₁for x₂ =7 .

y = 0.044 + 0.269x₁

Now to plot this equation, make a table

X	0.308	0
Y	0	0.457

Graph