Question

Consider the hypothesis test Ho : 67 = oz against H, : 67 +0. Suppose the sample sizes are ni = 16 and n2 = 21 and the sample standard deviations are si = 1.7 and $2 = 1.3. Use a = 0.05. a) Test the hypothesis. Find the P-value. Round your answer to three decimal places (e.g. 9.876). p-value = Ho b) Construct a 95% two-sided confidence interval of oʻrelations. Round your answers to two decimal places (e.g. 9.87).

Answer 1

Part a)

Test Statistic :-

f = 2.89 / 1.69
f = 1.7101

Test Criteria :-
Reject null hypothesis if f > f(α/2,n1-1,n2-1) OR f < f(1 - α/2), n1-1 , n2-1)
f(0.05/2 , 15 , 20 ) = 2.5731
f(1 - 0.05/2 ,15 , 20 ) = 0.3629
1.7101 lies between the value 2.5731 and 0.3629 , hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Decision based on P value
P value = 2 * P ( f > 1.7101 ) = 0.2604  
Reject null hypothesis if P value < α = 0.05
Since P value = 0.2604  > 0.05, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Part b)

$(S_{1}^{2} / S_{2}^{2}) * ( 1 / F_{\alpha/2, n1,n2}) < \sigma_{1}^{2} / \sigma_{2}^{2} < (S_{1}^{2} / S_{2}^{2}) * ( F_{\alpha/2, n2,n1})$
$( 2.89 / 1.69 ) * ( 1 / F_{0.05/2, 16,21}) < \sigma_{1}^{2} / \sigma_{2}^{2} < ( 2.89 / 1.69 ) * ( F_{0.05/2, 21,16})$
F(0.05/2, n1,n2) = F(0.05/2, 15,20) = 2.57
F(0.05/2, n1,n2) = F(0.05/2, 20,15) = 2.76
Lower Limit = ( 2.89 / 1.69 ) * ( 1 / F(0.05/2, 15,20)) = 0.6654 ≈ 0.67
Upper Limit = ( 2.89 / 1.69 ) * ( F(0.05/2, 20,15)) = 4.7198 ≈ 4.72

95% Confidence interval is ( 0.67 , 4.72 )

$( 0.67 < \sigma_{1}^{2} / \sigma_{2}^{2} < 4.72 )$