Question

WITHOUT USING TECHNOLOGY OR TI-84

the author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. here are the observed frequencies for the outcomes of 1,2,3,4,5 and 6, respectively: 27,31,42,40,28,32. use a 0.05 significance level to test the claim that the outcomes are not equally likely.

Answer 1

Chi square test for Goodness of fit

Ho: outcomes are equally likely

H1: outcomes are not equally likely

expected proportions = 1/ 6 = 0.7

expected frequncy,E = expected proportions*total frequency
total frequency=   200

category	observed frequencey, O	expected proportion	expected frequency,E			(O-E)²/E
1	27	0.17	33.33			1.203
2	31	0.17	33.33			0.163
3	42	0.17	33.33			2.253
4	40	0.17	33.33			1.333
5	28	0.17	33.33			0.853
6	32	0.17	33.33			0.053

chi square test statistic,X² = Σ(O-E)²/E =   5.860              
                  
level of significance, α=   0.05              
Degree of freedom=k-1=   6   -   1   =   5
                  
P value >0.10     
Decision: P value >α , Do not reject Ho                  
                  
there is not enough evidence to conclude that the outcomes are not equally likely at α=0.05