A sample of 8 observations from the population indicated that sample variance s12 is 784. A second sample of 8 observations from the same population indicated that sample variance s22 is 144. Conduct the following test of hypothesis using a 0.05 significance level.
H0: σ12 =
σ22
H1: σ12 <
σ22
You should use the tables in the book for obtaining the
F values.
For full marks your answer should be accurate to at least two
decimal places.
a) | State the decision rule.
|
b) Compute the left F critical value.
FL critical value: 0
c) What is the value of the test statistic?
F-Test statistic: 0
d) | What is your decision regarding H0?
|
Test Statistic :-
f = 784 / 144
f = 5.4444
Test Criteria :-
Reject null hypothesis if f <= f(1 - α , n1-1 , n2-1 )
f(1 - 0.05, 7 , 7 ) = 0.2641
f > f(1 - 0.05 , n1-1 , n2-1 ) = 5.4444 > 0.2641 , hence we
fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0
Decision based on P value
P value = P ( f > 5.4444 )
P value = 0.9801
Reject null hypothesis if P value < α = 0.05
Since P value = 0.9801 > 0.05, hence we fail to reject the null
hypothesis
Conclusion :- We Fail to Reject H0
Part a)
Reject H0 in favour of H1 if the computed value of the statistic is less than or equal to FL.
Part b)
f(1 - 0.05, 7 , 7 ) = 0.2641
Part c)
f = 5.4444
Part d)
There is insufficient evidence, at the given significance level, to reject H0 and so H0 will be accepted or at least not be rejected.
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