Question

(1 point) Consider the following outputs for fitting a normal and triangular distributions to a data set with the Arena Input Analyzer. Which one is the better fit to the points? Explain you answer. 2. Distribution Summary Distribution Sumary Distribution: Normal Expression: Square Error: 0.000535 Distribution: Triangular Expression: TRIA (69, 153, 226) Square Error: 0.003363 NORM (149, 25.1) Chi Square Test Chi Square Test Mumber of intervals 31 Degreeธ of freedom 28 Test Statistic Corresponding p-value 0.249 Number of intervals37 Degrees of freedom 35 Test Statistic Corresponding p-value <0.005 32.7 - 269 Kolmogorov-Smirnov Test Kolmogorov-Smirnov Test Test Statistic Corresponding p-value - 0.011 0.15 Test Statistic Corresponding p-value < 0.01 0.0811 Data Summary Data Sumary umber of Data Points 2000 Min Data Value Max Data Value Sample Mean umber of Data Points2000 Min Data Value Max Data Value Sample Yean Sample Std Dev - 69.8 - 225 e 149 69.8 - 225 - 149 ample Std Dev 25.1 # 25.1 Bistogran Summary Histogram sumary Histogran Range Mumber of Intervala40 69 to 226 iscogzan Range Number of Intervals 40 69 to 226

Answer 1

Answer

Data for normal distribution

Chi square statistics = 32.7 with p value= 0.249

Kolmogorov–Smirnov test statistics is 0.011 with p value > 0.15

p value for chi square as well as Kolmogorov–Smirnov test are greater than 0.05 significance level, so result is not significant.

Data for triangular distribution

Chi square statistics = 269 with p value < 0.005

Kolmogorov–Smirnov test statistics is 0.0811 with p value < 0.01

p value for chi square as well as Kolmogorov–Smirnov test are less than 0.05 significance level, so result is significant.

Therefore, triangular distribution provides better fit to the points based on the given data table