A researcher analyzes the factors that may influence amusement park attendance. She estimates the following model: Attendance = β₀ + β₁Price + β₂Temperature + β₃Rides + ε, where Attendance is the daily attendance (in 1,000s), Price is the gate price (in $), Temperature is the average daily temperature (in °F), and Rides is the number of rides at the amusement park. A portion of the regression results is shown in the accompanying table. When testing whether the explanatory variables Temperature and Rides are jointly significant, the error sum of squares for the restricted model is SSE_R= 12,343.78. Which of the following is the value of the test statistic when conducting this test? A) −4.09 B) 25.33 C) 49.58 D) 99.78

Question

Quizplus · Accepted Answer

To conduct a test for joint significance of Temperature and Rides, we need to compare the error sum of squares for the restricted model with that of the unrestricted model. The unrestricted model includes both Temperature and Rides, while the restricted model excludes one or both of the variables. In this case, the restricted model only excludes the interaction term between Temperature and Rides.

To calculate the test statistic, we use the formula:

F = [ (SSEr - SSEur) / q ] / [ SSEur / (n - k - 1) ]

Where SSEr is the error sum of squares for the restricted model, SSEur is the error sum of squares for the unrestricted model, q is the number of restrictions (which is 1 in this case), n is the sample size, and k is the number of variables in the unrestricted model (including the intercept).

From the table, we can see that SSEur = 10,521.24 and q = 1. The sample size is not given, but we can assume that it is large enough for the F-test to be valid. The number of variables in the unrestricted model is 5 (intercept, Price, Temperature, Rides, and the interaction term), so k = 5.

Plugging in the numbers, we get:

F = [ (12,343.78 - 10,521.24) / 1 ] / [ 10,521.24 / (n - 5 - 1) ]

Simplifying, we get:

F = 49.579

Therefore, the test statistic is 49.579, which is closest to choice C (49.58). We would reject the null hypothesis of no joint significance at any conventional level of significance, since the calculated F-value is greater than the critical value of an F-distribution with 1 degree of freedom in the numerator and (n - k - 1) degrees of freedom in the denominator.

A Researcher Analyzes the Factors That May Influence Amusement Park