Question

A sample of 5 observations from the population indicated that sample variance s12 is 441. A second sample of 10 observations from the same population indicated that sample variance s22 is 196. 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. Reject H0 in favour of H1 if the computed value of the statistic is greater than or equal to FL. Reject H0 in favour of H1 if the computed value of the statistic is less than or equal to FL. Reject H0 in favour of H1 if the computed value of the statistic is less than FL. Reject H0 in favour of H1 if the computed value of the statistic is greater than FL. None of the above b) Compute the left F critical value. c) What is the value of the test statistic? d) What is your decision regarding H0? There is sufficient evidence, at the given significance level, to reject H0 and accept H1. There is insufficient evidence, at the given significance level, to reject H0 and so H0 will be accepted or at least not be rejected. There is insufficient evidence to make it clear as to whether we should reject or not reject the null hypothesis.

Answer 1

Test Statistic :-

f = 441 / 196
f = 2.25

Part a)

Reject null hypothesis if f < f(1 - α , n1-1 , n2-1 )

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, 4 , 9 ) = 0.1667 ( Critical value from F table )

Part c)

f = 2.25

Part d)

f > f(1 - 0.05 , n1-1 , n2-1 ) = 2.25 > 0.1667 , hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

There is insufficient evidence, at the given significance level, to reject H0 and so H0 will be accepted or at least not be rejected.