Question

A car manufacturer wants to compare the performance of engines using ethanol mix with that of those using pure gasoline. They start by randomly choosing 200 cars, and divide that sample into two equally sized groups. The sample average performance for 100 cars using ethanol is 76.5 on a 0 to 100 scale, and the sample standard deviation is 38. For the other sample of cars, they test using pure gasoline and the average performance is 88.1 and the sample standard deviation is 40.

• (a) Build a 95% confidence interval for the performance score of the ethanol cars. Can we reject a null hypothesis that their true score is 39 at the 5% significance level? •

(b) Now we will compare them directly to the non-ethanol cars. Test the null hypothesis that the two groups have the same average score, assuming they have the same variance. •

(c) Now perform the same test assuming they have differing variances. How do the results differ? •

(d) Suppose you tried to answer this question using regression analysis. Define which dependent and independent variable you would use, and say what your slope and intercept estimates would be (reporting actual numbers, not formulas).

Answer 1

a)

sample std dev ,    s =    38.0000
Sample Size ,   n =    100
Sample Mean,    x̅ =   76.5000
Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   99          
't value='   tα/2=   1.9842   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   38.0000   / √   100   =   3.800000
margin of error , E=t*SE =   1.9842   *   3.80000   =   7.540024
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    76.50   -   7.540024   =   68.959976
Interval Upper Limit = x̅ + E =    76.50   -   7.540024   =   84.040024
95%   confidence interval is (   68.96   < µ <   84.04   )

39 does not lie in the interval so, reject Ho.

yes, reject null hypothesis that their true score is 39 at the 5% significance level

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b)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   76.50                  
standard deviation of sample 1,   s1 =    38.00                  
size of sample 1,    n1=   100                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   88.10                  
standard deviation of sample 2,   s2 =    40.00                  
size of sample 2,    n2=   100                  
                          
difference in sample means =    x̅1-x̅2 =    76.5000   -   88.1   =   -11.60  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    39.0128                  
std error , SE =    Sp*√(1/n1+1/n2) =    5.5172                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -11.6000   -   0   ) /    5.52   =   -2.10
                          
Degree of freedom, DF=   n1+n2-2 =    198                  
  
p-value =        0.036774   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis              

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c)

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 ╪   0          
                  
Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   sample 1          
mean of sample 1,    x̅1=   76.50          
standard deviation of sample 1,   s1 =    38          
size of sample 1,    n1=   100          
                  
Sample #2   ---->   sample 2          
mean of sample 2,    x̅2=   88.100          
standard deviation of sample 2,   s2 =    40.00          
size of sample 2,    n2=   100          
                  
difference in sample means = x̅1-x̅2 =    76.500   -   88.1000   =   -11.6000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    5.5172          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -11.6000   /   5.5172   ) =   -2.1025
                  
   197          
                  
                  

p-value =        0.0368   (excel function: =T.DIST.2T(t stat,df) )      
Conclusion:     p-value<α , Reject null hypothesis              

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results are same in both b and c

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