A fueleconomy study was conducted for two Japanese automobil

A fuel-economy study was conducted for two Japanese automobiles, Honda and Toyota. One vehicle of brand was selected, and the mileage performance was observed for 11 and 9 tanks of fuel in each car respectively. The data are as follows (in MPG). Assume that MPGs have normal distributions.

**For the questions below, please specify your hypotheses.
a) In terms of the mean MPG, does Honda have significantly higher MPG than Toyota? (?=0.05)
b) What is the p-value for the test in a?
c) Please calculate the 99% confidence interval for the difference of mean MPGs of Honda and
Toyota.

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Solution

a)

Let

u1 = mean of Honda
u2 = mean of Toyota

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    right   tailed test.      
Calculating the means of each group,              
              
X1 =    32.57272727          
X2 =    31.84444444          
              
Calculating the standard deviations of each group,              
              
s1 =    3.271113238          
s2 =    2.793345266          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    11          
n2 = sample size of group 2 =    9          
Thus, df = n1 + n2 - 2 =    18          
Also, sD =    1.356362455          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    0.53693821          
              
where uD = hypothesized difference =    0          
              
Now, the critical value for t is              
              
tcrit =    +   1.734063607      
              
As t < 1.734,   WE FAIL TO REJECT THE NULL HYPOTHESIS.   

Thus, Honda has significantly higher MPG than Toyota   . [CONCLUSION]

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b)      
              
Also, using p values,              
              
p =    0.298942703   [ANSWER]

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c)      
              
For the   0.99   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.005          
t(alpha/2) =    2.878440473          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    -3.175925757          
upper bound = [X1 - X2] + t(alpha/2) * sD =    4.632491417          
              
Thus, the confidence interval is              
              
(   -3.175925757   ,   4.632491417   ) [ANSWER]

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Hi! If you use another method/formula in calculating the degrees of freedom in this t-distribution (part c), please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!