Question: Were there any statistically significant findings? Explain why your data is/isn't statistically significant. Did you have any unusual values? Explain why your data points are/aren't
- Were there any statistically significant findings? Explain why your data is/isn't statistically significant.
- Did you have any unusual values? Explain why your data points are/aren't considered unusual.
- What are a few possible reasons your data may/may not have unusual/significant data points?
- this is the data Number of wait times lower than average from ride 1:27
Number of wait times higher than average from ride 1:18
Number of wait times lower than average from ride 2:25
Number of wait times higher than average from ride 2:20
Total lower or higher than average wait times from ride 1:27+18=45
Total lower or higher than average wait times from ride 2: 25+20=45
Total lower than average wait times:52
Total higher than average wait times:38
Grand total:27+18+25+20=90
CONTINGENCY TABLE
| RIDE1 Soaring | RIDE 2 tsmm | TOTAL | |
LOW | 27 | 25 | 52 |
HIGH | 18 | 20 | 38 |
TOTAL | 45 | 45 | 90 |
Probabilities
- P(Ride 1) = 45 / 90=0.5
- P(Ride 2) = 45/ 90 =0.5
- P(Low) = 52/90 =0.576
- P(High) = 38/ 90=0.423
- P(Ride 1 Low) = 70/ 90=0.778
- P(Ride 1 High) = 65/ 90 =0.722
- P(Ride 2 Low) = 72/ 90=0.8
- P(Ride 2 High) = 63 / 90=0.7
- P(Ride 1 Low) = 27/ 90 =0.3
- P(Ride 1 High) = 18/ 90 =0.2
- P(Ride 2 Low) = 25/ 90 =0.278
- P(Ride 2 High) = 20/ 90 =0.222
- P(Ride 1 | Low) = 27/ 52 =0.519
- P(Ride 1 | High) = 18/ 38 =0.462
- P(Ride 2 | Low) = 25/ 52=0.481
- P(Ride 2 | High) = 20 / 38=0.526
- P(Low | Ride 1) = 27/ 45 =0.6
- P(High | Ride 1) = 18/ 45 =0.4
- P(Low | Ride 2) = 25/ 45=0.556
- P(High | Ride 2) = 20/ 45=0.444
Approximating Binomial Probabilities
Use the table above to determine the probability of the higher-than-average wait times for theride 1sample data set happening by accident. you'll need to find the binomial approximation of the probability that a sample that has 45 wait times will have a probability that falls below the value you got in your contingency table assuming there's a 50% chance of a day not falling in the higher-than-average wait times group (high) for variable 1.
n = ________45____________ p = ____.50______________ q = ______1-.50=.50______________
x = ___________27________ approximated x value = __27.5__________________
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