solved: Were there any statistically significant findings? Explain why your…
Were there any statistically significant findings? Explain why your…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: 27Number of wait times higher than average from ride 1: 18Number of wait times lower than average from ride 2: 25Number of wait times higher than average from ride 2:20Total lower or higher than average wait times from ride 1:27+18= 45Total lower or higher than average wait times from ride 2: 25+20=45Total lower than average wait times: 52Total higher than average wait times: 38Grand total: 27+18+25+20=90CONTINGENCY TABLE RIDE1 SoaringRIDE 2 tsmmTOTALLOW272552HIGH182038TOTAL454590 Probabilities P(Ride 1) = 45 / 90=0.5P(Ride 2) = 45/ 90 =0.5P(Low) = 52/90 =0.576P(High) = 38/ 90=0.423P(Ride 1 È Low) = 70/ 90=0.778P(Ride 1 È High) = 65/ 90 =0.722P(Ride 2 È Low) = 72/ 90=0.8P(Ride 2 È High) = 63 / 90=0.7P(Ride 1 Ç Low) = 27/ 90 =0.3P(Ride 1 Ç High) = 18/ 90 =0.2P(Ride 2 Ç Low) = 25/ 90 =0.278P(Ride 2 Ç High) = 20/ 90 =0.222P(Ride 1 | Low) = 27/ 52 =0.519P(Ride 1 | High) = 18/ 38 =0.462P(Ride 2 | Low) = 25/ 52=0.481P(Ride 2 | High) = 20 / 38=0.526P(Low | Ride 1) = 27/ 45 =0.6P(High | Ride 1) = 18/ 45 =0.4P(Low | Ride 2) = 25/ 45=0.556P(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 the ride 1 sample 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__________________ ____________22.5________ =_______3.3541___________b ( x <27,27.5, 3.3541) = 0.9319814 Use the table above to determine the probability of the higher than average wait times for the ride 2 sample 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 2. n = ___________45_________ p = ____.50______________ q = _____.50_______________ x = 25____________________ approximated x value = _________25.5___________ ________22.5____________ =____3.3541_________________________b ( x < _______25______________; ____25.5_______________ , ___3.3541____________________________ ) = 0.8144533 MathStatistics and Probability MAT 243