Question

The manager of an amusement park would like to be able to predict daily attendance in...

The manager of an amusement park would like to be able to predict daily attendance in order to develop more accurate plans about how much food to order and how many ride operators to hire. After some consideration, he decided that the following three factors are critical: Yesterday’s attendance Weekday or weekend Predicted weather He then took a random sample of 36 days. For each day, he recorded the attendance, the previous day’s attendance, day of the week, and weather forecast(mostly sunny, rain, cloudy). The first independent variable is interval, but the other two are nominal. a. Create the three indicator variables you need. b. Conduct a regression analysis. c. Is this model valid? Explain. d. Can we conclude that weather is a factor in determining attendance? e. Do these results provide sufficient evidence that weekend attendance is, on average, larger than weekday attendance? f. Do these results provide sufficient evidence that mostly sunny attendance is, on average, larger than cloudy attendance?

Attendance Yest Att day of the week weather forecast
7882 8876 2 1
6115 7203 2 3
5351 4370 2 3
8546 7192 1 1
6055 6835 2 3
7367 5469 2 1
7871 8207 2 1
5377 7026 2 3
5259 5592 2 1
4915 3190 2 3
6538 7012 2 3
6607 5434 2 3
5118 3764 2 3
6077 7575 2 3
4475 6047 2 3
3771 4430 2 3
6106 5697 2 3
7017 3928 1 2
5718 5552 2 3
5966 3142 1 2
8160 8648 1 2
4717 3397 2 3
7783 7655 2 3
5124 5920 2 3
7495 7831 1 2
5848 6355 2 3
5166 3529 2 3
4487 4220 2 3
7320 7526 2 1
6925 4083 1 1
8133 6382 1 1
7929 6459 2 3
7291 3432 1 2
5419 8077 2 3
3634 3353 2 3
6859 3803 1 2
1   weekend 1   mostly sunny
2   weekdays 2   rain
3   cloudy
0 0
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Answer #1

A.

Since "day of the week" can assume two values (1 and 2), we will need only one variable e.g.

A = 0 ; if weekend

A = 1 ; if weekday

Since "weather forecast" can assume three values (1, 2 and 3), we will need two variable e.g.

B1 = 1, if sunny; B1 = 0, otherwise.

B2 = 1, if rain ; B2 = 0, otherwise.

if both the variables gets 0 then it is cloudy.

B.

Summary of regression analysis:

SUMMARY OUTPUT
Regression Statistics
Multiple R 0.824568883
R Square 0.679913843
Adjusted R Square 0.638612404
Standard Error 790.6786272
Observations 36
ANOVA
df SS MS F Significance F
Regression 4 41166949.31 10291737.33 16.4622311 2.47624E-07
Residual 31 19380353.44 625172.6915
Total 35 60547302.75
Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept 4619.366527 695.8111859 6.638821882 2.02115E-07 3200.250257 6038.482797 3200.250257 6038.482797
Yest Att 0.383998499 0.080450863 4.773081158 4.11184E-05 0.219917882 0.548079116 0.219917882 0.548079116
A -1207.558126 586.0985572 -2.060332876 0.047844232 -2402.914014 -12.20223744 -2402.914014 -12.20223744
B1 988.54631 410.8190614 2.406281506 0.022266169 150.6753102 1826.41731 150.6753102 1826.41731
B2 541.7985097 685.57383 0.790284702 0.435365659 -856.438535 1940.035555 -856.438535 1940.035555

C.

Since the significance of F-statistics (i.e. the p-value) = 2.476*10-7 < 0.05 (if \alpha = 0.05)

Thus model is significant.

D.

The p-value for B1 = 0.022 < 0.05

Thus, the weather is a significant factor in determining attendance.

Since for B2, p-value = 0.435 > 0.05 i.e not significant. It means knowing if a day is sunny is more helpful than knowing if it is rainy.

E.

Regression Equation:

Attendance = 4619.366 + 0.384*Yest. Att. -1207.558*A + 988.546*B1 + 541.7985097*B2

For the weekend:

A = 0

Attendance = 4619.366 + 0.384*Yest. Att. + 988.546*B1 + 541.7985097*B2

For the weekday:

day of the week = 1

Attendance = 4619.366 + 0.384*Yest. Att. + 988.546*B1 + 541.7985097*B2 - 1207.558

Thus if all the other factors are constant, attendance on a weekend will be more as compared to a weekday on average.

F.

Attendance = 4619.366 + 0.384*Yest. Att. -1207.558*A + 988.546*B1 + 541.7985097*B2

For mostly sunny:

B1 = 1

B2 = 0

Attendance = 4619.366 + 0.384*Yest. Att. -1207.558*A + 988.546

For cloudy:

B1 and B2  = 0

Attendance = 4619.366 + 0.384*Yest. Att. -1207.558*A

Thus if all the other factors are constant, attendance on a sunny day will be more as compared to a cloudy day on average.

Please upvote if you have liked my answer, would be of great help. Thank you.

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