In: Statistics and Probability
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:
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