45 A food company testing a new method for the analysis of t

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A food company testing a new method for the analysis of the polyunsaturated fat content of various vegetable and animal oils contained the results at the right for the old and new methods. Use a paired Student\'s t-test to determine if the new and old analysis procedures are giving the same results at the 95percentagee confidence level. (You may assume that the precision is approximately the same for these two methods.) Are the results the same at the 95percentagee confidence level?

Solution

Let ud = u2 - u1.              
Formulating the null and alternative hypotheses,                  
Ho:   ud   =   0  
Ha:   ud   =/   0  
At level of significance =    0.05          
As we can see, this is a    two   tailed test.      
              
Calculating the standard deviation of the differences (third column):              
              
s =    1.512047578          
              
Thus, the standard error of the difference is sD = s/sqrt(n):              
              
sD =    0.534589548          
              
Calculating the mean of the differences (third column):              
              
XD =    -1.8625          
              
As t = [XD - uD]/sD, where uD = the hypothesized difference =    0   , then      
              
t = 3.483981322   [ANSWER, CALCULATED T VALUE]
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As df = n - 1 =    7          
              
Then the critical value of t is              
              
tcrit =    +/-   2.364624252   [ANSWER, CRITICAL T VALUE]

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As |t| > 2.364, WE REJECT THE NULL HYPOTHESIS.      

Hence,

NO, the results are NOT THE SAME at 95% confidence. [CONCLUSION]