Question

Forecasting movie revenues with Twitter. Refer to the IEEE International Conference on Web Intelligence and Intelligent Agent Technology (2010) study on using the volume of chatter on Twitter.com to forecast movie box office revenue, Exercise 12.10 (p. 723). The researchers modelled a movie’s opening weekend box office revenue (y) as a function of tweet rate (x₁ ) and ratio of positive to negative tweets (x₂) using a first-order model.

a) Write the equation of an interaction model for E(y) as a function of x₁ and x₂ .

b) In terms of the$\mathit{\beta }$ in the model, part a, what is the change in revenue (y) for every 1-tweet increase in the tweet rate (x₁ ) , holding PN-ratio (x₂)constant at a value of 2.5?

c) In terms of the in the model, part a, what is the change in revenue (y) for every 1-tweet increase in the tweet rate (x₁ ) , holding PN-ratio (x₂)constant at a value of 5.0?

d) In terms of the$\mathit{\beta }$ in the model, part a, what is the change in revenue (y) for every 1-unit increase in the PN-ratio (x₂) , holding tweet rate (x₁ )constant at a value of 100?

e) Give the null hypothesis for testing whether tweet rate (x₁ ) and PN-ratio (x₂) interact to affect revenue (y).

Short answer

a) The equation would be $E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x}_1{x}_2+\epsilon$.

b) The equation becomes, $E\left(y\right)=\left({\beta }_0+2.5{\beta }_2\right)+\left({\beta }_1+2.5{\beta }_3\right)x_1+\epsilon$. Hence, revenue (y) changes by $\left({\beta }_1+2.5{\beta }_3\right)$ for every 1-unit increase in the tweet rate (x₁) holding PN-ratio (x₂) constant.

c)The equation becomes,$E\left(y\right)=\left({\beta }_0+5{\beta }_2\right)+\left({\beta }_1+5{\beta }_3\right)x_1+\epsilon$ . Hence, revenue (y) changes by localid="1651172612533" $\left({\beta }_2+100{\beta }_3\right)$for every 1-unit increase in the tweet rate (x₁) holding PN-ratio (x₂) constant.

d) The equation becomes, $E\left(y\right)=\left({\beta }_0+100{\beta }_1\right)+\left({\beta }_2+100{\beta }_3\right)x_2+\epsilon$. Hence, revenue (y) changes by $\left({\beta }_2+100{\beta }_3\right)$ for every 1-unit increase in PN-ratio (x₂) holding the tweet rate (x₁ ) constant.

e) The null hypothesis and alternate hypothesis will be $H_0:{\beta }_3=0$ and $H_a:{\beta }_3\ne 0$

Step-by-step solution

Equation for the interaction model

The equation of an interaction model for E(y) as a function of x₁ and x₂ would have an additional variable (x₁x₂) to represent the interaction amongst the variable.

Therefore, the equation would be $\mathit{E}\left(y\right)\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}{\mathit{\beta }}_{\mathbf{3}}{\mathit{x}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}\mathit{\epsilon }$ .

Change in y for change in  x1

The interaction model equation is $E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x}_1{x}_2+\epsilon$.

Here, holding x₂ value constant at 2.5, the model becomes

$E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2\left(2.5\right)+{\beta }_3{x}_1\left(2.5\right)+\epsilon$

$E\left(y\right)=\left({\beta }_0+2.5{\beta }_2\right)+\left({\beta }_1+2.5{\beta }_3\right)x_1+\epsilon$

The equation becomes$\mathit{E}\left(y\right)\mathbf{=}\left({\beta }_0+2.5{\beta }_2\right)\mathbf{+}\left({\beta }_1+2.5{\beta }_3\right){\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathit{\epsilon }$ Hence, revenue (y) changes by localid="1651173294032" $\left({\beta }_1+2.5{\beta }_3\right)$ for every 1-unit increase in the tweet rate (x₁) holding PN-ratio (x₂) constant.

Convert in y for convert in x1 

The interaction model equation is $E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x}_1{x}_2+\epsilon$.

Here, holding x₂ value constant at 5, the model becomes

$E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2\left(5\right)+{\beta }_3{x}_1\left(5\right)+\epsilon$

$E\left(y\right)=\left({\beta }_0+5{\beta }_2\right)+\left({\beta }_1+5{\beta }_3\right)x_1+\epsilon$

The equation becomes,localid="1651173183555" $\mathit{E}\left(y\right)\mathbf{=}\left({\beta }_0+5{\beta }_2\right)\mathbf{+}\left({\beta }_1+5{\beta }_3\right){\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathit{\epsilon }$. Hence, revenue (y) changes bylocalid="1651173324259" ${\mathit{\beta }}_{\mathbf{2}}\mathbf{+}\mathbf{100}{\mathit{\beta }}_{\mathbf{3}}$ for every 1-unit increase in the tweet rate (x₁) holding PN-ratio (x₂) constant.

Switch in y for the switch in x2 

The interaction model equation is $E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x}_1{x}_2+\epsilon$.

Here, holding x₁ value constant at 100, the model becomes

$E\left(y\right)={\beta }_0+{\beta }_1\left(100\right)+{\beta }_2{x}_2+{\beta }_3\left(100\right)x_2+\epsilon$

$E\left(y\right)=\left({\beta }_0+100{\beta }_1\right)+\left({\beta }_2+100{\beta }_3\right)x_2+\epsilon$

The equation becomes, role="math" localid="1651173154939" $\mathit{E}\left(y\right)\mathbf{=}\left({\beta }_0+100{\beta }_1\right)\mathbf{+}\left({\beta }_2+100{\beta }_3\right){\mathit{x}}_{\mathbf{2}}\mathbf{+}\mathit{\epsilon }$. Hence, revenue (y) changes by ${\mathit{\beta }}_{\mathbf{2}}\mathbf{+}\mathbf{100}{\mathit{\beta }}_{\mathbf{3}}$ for every 1-unit increase in PN-ratio x₂ holding the tweet rate x₁ constant.

Significance of  β3

To test whether x₁ and x₂ variables interact in the model, the presence of the model parameter ${\beta }_3$ is tested.

Hence the null hypothesis would be the absence of the model parameter ${\beta }_3$and the alternate hypothesis would be the presence of ${\beta }_3$

Mathematically, ${\mathit{H}}_{\mathbf{0}}\mathbf{:}{\mathit{\beta }}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{;}{\mathit{H}}_{\mathbf{a}}\mathbf{:}{\mathit{\beta }}_{\mathbf{3}}\mathbf{\ne }\mathbf{0}$