On a typical night in a large city, about 25,000 people attend a theatrical event, paying an average cost of over $110 per ticket. The results of a multiple regression of weekly data for the receipts in millions of dollars, the paid attendance in thousands, the number of shows, and the average ticket price are found below.
Write the regression model.
Receipts = −17.0253 + 0.1116 Attendance + 0.0071 # Shows + 0.2016 Price
1.What does the coefficient of Attendance mean in the context of this regression model?
2.In a week in which the paid attendance was 200,000 customers attending 30 shows at an average ticket price of $105, what would you estimate the receipts would be?
About $_____________Million (Round to two decimal places as needed.)
3.Is this likely to be a good prediction? Why do you think that?
Multiple Regression Model
Dependent variable is: Receipts |
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R squared = 99.9% R squared (adjusted) = 99.9% |
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S = 0.0966 with 74 degrees of freedom |
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Source |
Sum of Squares |
df |
Mean Square |
F-ratio |
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Regression |
487.733 |
33 |
162.5777 |
17481 |
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Residual |
0.691 |
74 |
0.0093 |
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Variable |
Coeff |
SE(Coeff) |
t-ratio |
P-value |
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Intercept |
−17.0253 |
0.2943 |
−57.85 |
< 0.0001 |
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Attendance |
0.1116 |
0.0009 |
124.04 |
< 0.0001 |
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# Shows |
0.0071 |
0.0046 |
1.55 |
0.1254 |
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Price |
0.2016 |
0.0033 |
61.09 |
< 0.0001 |
1)
option A)
Compared to a week with a similar number of shows and average
ticket price, an increase in attendance is associated with an
increase in receipts proportional to the coefficient of
Attendance.
2)
Receipts = −17.0253 + 0.1116 Attendance + 0.0071 # Shows + 0.2016
Price
= -17.0253 + 0.1116 * 200 + 0.0071 * 30 + 0.2016 * 105
=26.6757
3)
R^2 = 0.999
option C)
Yes, because the high R2 value indicates that the model accounts
for almost all of the variation in the receipts.
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