Question

Right side Hypothesis Testing 03 (two independent samples 3. Two different formulations of ormulations of an Oxygenated motor fuel are being tested to study their road Varance of road octane number for formulation 1. 1.5 and for forma sample of size n = 15 and n = 20 are tested, and the mean road octane 12 = 92.5. Assume Normality. motor fuel are being tested to study their road octane numbers. The 1.5 and for formulation 2 It is of 1.2. Two random 0, and the mean road octane numbers observed are X, 89.6 and e down the null and alternative hypotheses to test if the mean octane number for formulation 2 IS Significantly greater than that for formulation 1. Define all the terms used. b. What is the type of statistical test that is appropriate? Explain. Test the hypotheses at 0.05, using the critical value approach. What is your decision on nur hypothesis (Ho)? Interpret the test results. d. What is the P Value for the test? Test the hypotheses using the P Value. Do you get the same answer part (c)? e Construct a 95% confidence interval for the mean difference of octane numbers. Test the hypoth using the confidence interval approach. Do you get the same answer as parts (c) and (d)?

Answer 1

a)

Ho :   µ1 - µ2 =   0
Ha :   µ1-µ2 <   0
      

b) Z test for two independent sample

c) difference in sample means = x̅1 - x̅2 =    89.6   -   92.5   =   -2.9
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.4000          
                  
Z-statistic = ((x̅1 - x̅2)-µd)/SE =    -2.9   /   0.4000   =   -7.2500
                  
Z-critical value , Z* =        -1.645   [excel function =NORMSINV(α)]      
                  
Desison: test stat < -1.645 , Reject null hypothesis              
Conclusion:   There is enough evidence to support the claim

d)

  p-value =        0.000 [excel function =NORMSDIST(z)]  
Desison:   p-value <α , Reject null hypothesis          

yes, we get same answer

e)

Level of Significance ,    α =    0.05          
z-critical value =    Z α/2 =    1.960   [excel function =normsinv(α/2) ]      
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.4000          
margin of error, E = Z*SE =    1.960   *   0.400   =   0.7840
                  
difference of means = x̅1 - x̅2 =    89.6   -   92.5   =   -2.900
confidence interval is                   
Interval Lower Limit= (x̅1 - x̅2) - E =    -2.900   -   0.784   =   -3.6840
Interval Upper Limit= (x̅1 - x̅2) + E =    -2.900   +   0.784   =   -2.1160

since, confidence interval do not contain 0, also, both ends of interval are negative.

it supports the alternative hypothesis

yes, we get same answer