To evaluate the effect of a treatment, a sample of n=8 is obtained from a popula

Question

To evaluate the effect of a treatment, a sample of n=8 is obtained from a population with a mean of =40 , and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M=35 .

a. If the sample variance is s^2=32 , are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha=.05

b. If the sample variance is s^2=72 , are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha=.05 ?

c. Comparing your answer for parts a and b, how does the variability of the scores in the sample influence the outcome of a hypothesis test?

Explanation / Answer

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   40  
Ha:    u   =/   40  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    7          
tcrit =    +/-   2.364624252      
              
Getting the test statistic, as              
              
X = sample mean =    35          
uo = hypothesized mean =    40          
n = sample size =    8          
s = standard deviation =    5.656854249          
              
Thus, t = (X - uo) * sqrt(n) / s =    -2.5          
              
Also, the p value is              
              
p =    0.040992219          
              
As |t| > 2.365, and P < 0.05, we   REJECT THE NULL HYPOTHESIS.          

Hence, the treatment has a significant effect at 0.05 level. [CONCLUSION]

***************************************************************************************

b)

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   40  
Ha:    u   =/   40  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    7          
tcrit =    +/-   2.364624252      
              
Getting the test statistic, as              
              
X = sample mean =    35          
uo = hypothesized mean =    40          
n = sample size =    8          
s = standard deviation =    8.485281374          
              
Thus, t = (X - uo) * sqrt(n) / s =    -1.666666667          
              
Also, the p value is              
              
p =    0.139519583          
              
As |t| < 2.365, and P > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.          

Hence, the treatment has no significant effect at 0.05 level. [CONCLUSION]

***************************************************************************************************************

c)

The greater the variability is, the less likely that we would reject the null hypothesis, because our test statistic's magnitude decreases.

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