In a study of the rate at which the pesticide hexaconazole 3

In a study of the rate at which the pesticide hexaconazole 30 minutes and four others were exposed for wore exposed to 24 mug of hexaconazole. Four specimens were exposed for 60 minutes. The amounts (in mug) that were, Find a 95% confidence Interval for the mean amount absorbed in the time interval between 30 and ,60 minutes after exposure. Can you conclude that the amounts exposed for 60 minutes is greater than the? Amounts exposed for 30 minutes?

Solution

Let

u1 = mean of 30 minutes
u2 = mean of 60 minutes

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    two   tailed test.      
Calculating the means of each group,              
              
X1 =    3.2          
X2 =    3.6          
              
Calculating the standard deviations of each group,              
              
s1 =    0.182574          
s2 =    0.141421          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    4          
n2 = sample size of group 2 =    4          
Thus, df = n1 + n2 - 2 =    6          
Also, sD =    0.115469871          
              
For the   0.95   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.025          
t(alpha/2) =    2.446911851          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    -0.682544597          
upper bound = [X1 - X2] + t(alpha/2) * sD =    -0.117455403          
              
Thus, the confidence interval is              
              
(   -0.682544597   ,   -0.117455403   ) [ANSWER]

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b)

As this whole interval is less than 0, then there is significant evidnece that the amounts exposed for 60 minutes is greater than the amounts exposed for 30 minutes at 0.025 significance. [CONCLUSION]

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Hi! If you use another method/formula in calculating the degrees of freedom in this t test, 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!