Question: Tilting in online poker. In poker, making bad decisions due to negative emotions is known as tilting. A study in the Journal of Gambling Studies (March, 2014) investigated the factors that affect the severity of tilting for online poker players. A survey of 214 online poker players produced data on the dependent variable, severity of tilting (y), measured on a 30-point scale (where higher values indicate a higher severity of tilting). Two independent variables measured were poker experience (x₁, measured on a 30-point scale) and perceived effect of experience on tilting (x₂, measured on a 28-point scale). The researchers fit the interaction model, . The results are shown below (p-values in parentheses).

1. Evaluate the overall adequacy of the model using α = .01.

b. The researchers hypothesize that the rate of change of severity of tilting (y) with perceived effect of experience on tilting (x₂) depends on poker experience (x₁). Do you agree? Test using α = .01.

Step by step solution

01

Overall adequacy of the model

To test the overall adequacy of the model, F-test is conducted

$$\begin{gathered} {\mathrm{H}}_0:{\mathrm{\beta }}_1={\mathrm{\beta }}_2={\mathrm{\beta }}_3=0 \\ {\mathrm{H}}_{\mathrm{a}}:\mathrm{At}\mathrm{least}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{parameters}{\mathrm{\beta }}_1,{\mathrm{\beta }}_2,\mathrm{or}{\mathrm{\beta }}_3\mathrm{is}\mathrm{non}\mathrm{zero} \end{gathered}$$

Here, F test statistic = $\frac{SSE}{n-k+1}=31.98$

Value of F_0.05,213,213 is 2.605

H₀is rejected if F static >F_0.05,213,213 for $\alpha =0.05$,Since F >F_0.05,213,213

Sufficient evidence to reject H₀at 95% confidence interval

$\mathrm{Thus},\mathrm{atleast}oneoftheparameters{\mathrm{\beta }}_1,{\mathrm{\beta }}_2,or{\mathrm{\beta }}_{3}isnonzero$

02

 Step 2: Overall adequacy of the model

$$\begin{gathered} H_0:{\beta }_3=0 \\ {H}_0:{\beta }_3\ne 0 \end{gathered}$$

Here, t-test statistic = $\frac{\hat{\beta }s}{s\hat{\beta }s}$=-5.61

Value oft_0.05,213 is 1.96

H₀is rejected if t static > t_0.05,24,24 .$\alpha =0.05$ since t < t_0.05,24,24

Not sufficient evidence to reject H₀at 95% confidence interval

Therefore, ${\beta }_3=0$

Hence it can be concluded with enough evidence that x₁ and x₂ donot interact in the model

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