Question

In: Statistics and Probability

Big screw 7 Small screw 21 Big nut 12 Small nut 10 washer 6 5% level...

Big screw 7 Small screw 21 Big nut 12 Small nut 10 washer 6

5% level of significance.

A.Sample size:

B.Expected frequencies:

C.Test statistics show work:

F.P-value:

G.Decision:

H.Conclusion:

Solutions

Expert Solution

Chi square test for Goodness of fit  
  
expected frequncy,E = expected proportions*total frequency  
total frequency=   56

category observed frequencey, O expected proportion expected frequency,E (O-E)²/E
big 7 0.2000 11.2 1.575
small 21 0.2000 11.2 8.575
big nut 12 0.2000 11.2 0.057
small nut 10 0.2000 11.2 0.129
washer 6 0.2000 11.2 2.414

A) sample size=56

b)

category expected frequency,E
big 11.2
small 11.2
big nut 11.2
small nut 11.2
washer 11.2

c)

chi square test statistic,X² = Σ(O-E)²/E =   12.750

f)

Degree of freedom=k-1=   5   -   1   =   4
                  
P value =   0.0126 [ excel function: =chisq.dist.rt(test-stat,df) ]          

g)

Decision: P value < α=0.05, Reject Ho  

h)

conclusion: there is enough evidence that distribution is not equally likely


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