a) Find the AD (Anderson-Darling) for both Springfield and Evansville (using Minitab). (round to 3 decimal places)
b) Find the p-value for this test (using Minitab). (round to 3 decimal places)
City | Price($) |
Springfield | 116480 |
Springfield | 114746 |
Springfield | 119697 |
Springfield | 116625 |
Springfield | 116877 |
Springfield | 110206 |
Springfield | 113552 |
Springfield | 115015 |
Springfield | 115058 |
Springfield | 117310 |
Springfield | 114553 |
Springfield | 114770 |
Springfield | 115249 |
Springfield | 113548 |
Evansville | 120070 |
Evansville | 113786 |
Evansville | 120177 |
Evansville | 117433 |
Evansville | 116873 |
Evansville | 112948 |
Evansville | 112260 |
Evansville | 117860 |
Evansville | 114275 |
Evansville | 115145 |
Evansville | 122196 |
Evansville | 112747 |
Evansville | 113422 |
Evansville | 116383 |
Evansville | 116277 |
Evansville | 119371 |
Evansville | 115739 |
Evansville | 116609 |
Evansville | 116977 |
Evansville | 115778 |
Evansville | 116861 |
Evansville | 109303 |
Evansville | 117424 |
Evansville | 117552 |
Evansville | 115460 |
Evansville | 115295 |
Evansville | 111973 |
a) H0: The data follows Normal distribution
H1: The data does not follows Normal distribution
Let the los be alpha = 5%
Test Statistic AD = 0.242
Test Statistic AD = 0.426
b) P-value of Price($)_Evansville is 0.746 which is > alpha 0.05 so we accept H0
Thus we conclude that Price_Evansville is follows normal distribution.
P-value of Price($)_Evansville is 0.270 which is > alpha 0.05 so we accept H0
Thus we conclude that Price_Evansville is follows normal distribution.
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