A researcher analyzes the factors that may influence amusement park attendance. She estimates the following model: Attendance = β_{0} + β_{1}Price + β_{2}Temperature + β_{3}Rides + ε, where Attendance is the daily attendance (in 1,000s), Price is the gate price (in $), Temperature is the average daily temperature (in ^{o}F), and Rides is the number of rides at the amusement park. A portion of the regression results is shown in the accompanying table.
df | SS | MS | F | Significance F | ||
Regression | 3 | 29,524.41 | 9,841.47 | 2.18E-14 | ||
Residual | 26 | 2,564.37 | 98.63 | |||
Total | 29 | 32,088.78 | ||||
Coefficients | Standard Error | t-stat | p-value | Lower 95% | Upper 95% | |
Intercept | 27.328 | 40.254 | 0.5032 | −55.415 | 110.071 | |
Price | −1.201 | 0.294 | 0.0004 | −1.805 | −0.598 | |
Temperature | 0.008 | 0.208 | 0.9693 | -0.419 | 0.435 | |
Rides | 3.621 | 0.364 | 2.32E-10 | 2.874 | 4.369 |
When testing whether Temperature is significant at the 5%
significance level, she ________.
Multiple Choice
rejects H_{0}:β_{2} = 0, and concludes that Temperature is significant
does not reject H_{0}:β_{2} = 0, and concludes that Temperature is significant
rejects H_{0}:β_{2} = 0, and cannot conclude that Temperature is significant
does not reject H_{0}:β_{2} = 0, and cannot conclude that Temperature is significant
Answer: does not reject H0:β2 = 0, and cannot conclude that Temperature is significant
Explanation:
Null Hypothesis, Ho: β2 = 0 i.e Temperature is not significant
Alternative Hypothesis, Ha: β2 0 i.e. Temperature is significant
Decision Rule: Reject the Null Hypothesis Ho if p-value <
Conclusion: Since p-value = 0.9693 > so we fail to Reject the Null Hypothesis and conclude that β2 = 0 i.e Temperature is not significant
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