Question

Reality TV and cosmetic surgery. Refer to the Body Image: An International Journal of Research (March 2010) study of the impact of reality TV shows on a college student’s decision to undergo cosmetic surgery, Exercise 12.17 (p. 725). Recall that the data for the study (simulated based on statistics reported in the journal article) are saved in the file. Consider the interaction model, , where y = desire to have cosmetic surgery (25-point scale), = {1 if male, 0 if female}, and = impression of reality TV (7-point scale). The model was fit to the data and the resulting SPSS printout appears below.

a.Give the least squares prediction equation.

b.Find the predicted level of desire (y) for a male college student with an impression-of-reality-TV-scale score of 5.

c.Conduct a test of overall model adequacy. Use a= 0.10.

d.Give a practical interpretation of R²a.

e.Give a practical interpretation of s.

f.Conduct a test (at a = 0.10) to determine if gender (x₁) and impression of reality TV show (x₄) interact in the prediction of level of desire for cosmetic surgery (y).

Short answer

a.The least-square prediction equation is E(y) =11.779-1.972x₁+ 0.585x₄+0.553 x₁x₄

b. The predicted score for a male student with an impression-of-reality-TV score of 5 is 9.967.

c.At 95% confidence interval, it can be concluded that$\beta$₁$\ne \beta$₂$\ne \beta$₃$\ne 0$ .

d.The value of adjusted R² is 0.439 which indicates that the model is not a good fit for the data.

e. The value of s is 2.350 which is a lower value indicating that the data is close to the regression line plotted and that the data is not spr.

f.At 95% significance, $\beta$₃= 0 .ead Hence it can be concluded with enough evidence that x₁and x₂do not interact in the model.

Step-by-step solution

Least square prediction equation

The least-square prediction equation is formed by substituting model parameters in the estimated equation. Here, the least square prediction equation isE(y) =11.779-1.972x₁+ 0.585x₄+0.553 x₁x₄

Finding prediction level of desire

Since the impression-of-reality-TV-scale score is 5, the value becomes 1 (since we want to predict the level of desire for a male college student). Therefore, mathematically

E(y) = $\beta$₀+ $\beta$₁x₁+ $\beta$₂x₄ + $\beta$₃x₁x₄

E(y) = 11.779- 1.972(1) + 0.585(5) -0.533(1) (5)

E(y)= 9.967

The predicted score for a male student with an impression-of-reality-TV score of 5 is 9.967.

Significance of the model

H₀: $\beta$_{1 =}$\beta$_{2 = $\beta$3}= 0

H_a : at least one of the parameters localid="1649922695702" $\beta$_{1 ,}$\beta$$\beta$₂,$\beta$₃ is non zero

Here, F test statistic =SSE/n-(k-1) = 916.787/393-4 = 2.356

Value of F_0.05389389 is 1

H₀is rejected if F statistic > F_0.05389389

For a= 0.05, since F > F_0.05389389 Sufficient evidence to reject H₀ at 95% confidence interval.

Interpretation of R2 a

Adjusted R²denoted by R²a explains the variation in the variables which is explained by the model when additional independent variables are added in the model. The high value of R²a indicates that the model is a good fit for the data while a lower value denotes that the model is not a good fit for the data. Here the value of adjusted R² is 0.439 which indicates that the model is not a good fit for the data.

Analysis of s

The standard error of the regression (s) measures the distance of the data points from the regression line. It gives an estimate of the spread of the data points. Higher value denotes that the data is spread while lower value denotes the data points are closer to the regression line meaning the regression line is a good fit of the data. Here, the value of s is 2.350 which is a lower value indicating that the data is close to the regression line plotted and that the data is not spread.

Importance ofβ

H₀: $\beta$₃=0

H₀: $\beta$$\ne$0

Here, t-test statistic= 0.0536/0.276 = -0.2039

Value of t_0.05374 is 1.645 H₀is rejected if t is statistic >t_0.05374. For a=0.05 since t >t_0.05374. Not sufficient evidence to reject H₀at a 95% confidence interval.

Thus, $\beta$=0 .Hence it can be concluded with enough evidence that x₁and x₂ do not interact in the model.