Two different formulas of an oxygenated motor fuel are being tested to study the

Question

Two different formulas of an oxygenated motor fuel are being tested to study their road octane numbers. The variance of road octane number for formula 1 is sigma^2_1 = 1.5, and for formula 2 it is sigma^2_2 = 1.2. Two random samples of size n_1 = 15 and n_2 = 20 are tested, and the mean octane numbers observed are x_1 = 89.0 fluid ounces and x_2 = 93.5 fluid ounces. Assume normality. Round your answers to three decimal places (e.g. 98.765). Test the hypothesis that the formulations are equal versus the hypothesis that formulation 2 produces a higher mean road octane number than formulation 1. Calculate a 95% two-sided confidence interval on the mean difference road octane number. (Calculate using the following order: x_1 - x_2) Is there sufficient evidence to show that formulation 2 produces a higher mean road octane number than formulation 1?

Explanation / Answer

A)

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   >=   0  
Ha:   u1 - u2   <   0  
At level of significance =    0.05          
As we can see, this is a    left   tailed test.      
Calculating the means of each group,              
              
X1 =    89          
X2 =    93.5          
              
Calculating the standard deviations of each group,              
              
s1 =    1.224744871          
s2 =    1.095445115          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    15          
n2 = sample size of group 2 =    20          

Also, sD =    0.4          
              
Thus, the Z statistic will be              
              
z0 = [X1 - X2 - uD]/sD =    11.25 [ANSWER]
              
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b)      
              
For the   0.95   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.025          
z(alpha/2) =    1.959963985          
              
lower bound = [X1 - X2] - z(alpha/2) * sD =    -5.283985594          
upper bound = [X1 - X2] + z(alpha/2) * sD =    -3.716014406          
              
Thus, the confidence interval is              
              
(   -5.283985594   ,   -3.716014406   ) [ANSWER]

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c)

As the interval is totally negative, then YES, THERE IS SUFFICIENT EVIDENCE. [ANSWER]

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