To evaluate the effect of a treatment a sample of n8 is obta

To evaluate the effect of a treatment, a sample of n=8 is obtained from a population with a mean of µ=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 s²=32, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with =.05?

b) If the sample variance is s²=72, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with =.05?

AND

Ackerman and Goldsmith (2011) found that students who studied text from printed hardcopy had better scores than students who studied from text presented on a screen. In a related study, a professor noticed that several students in a large class had purchased the e-book version of the course textbook. For the final exam, the overall average for the entire class was µ=81.7, but the n=9 students who used e-books had a mean of M=77.2 with a standard deviation of s=5.7.

a) Is the sample sufficient to conclude that scores for students using e-books were significantly different from scores for the regular class? Use a two-tailed test with =.01.

b) Construct the 95% confidence interval to estimate the mean exam score if the entire population used e-books.

c) Write a sentence demonstrating how the results from the hypothesis test and the confidence interval would appear in a research report.

Solution

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

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 = sqrt(32) =   5.656854249          
              
Thus, t = (X - uo) * sqrt(n) / s =    -2.5          
              
Also, the p value is              
              
p =    0.040992219          
              
Comparing |t| > 2.3646 (or, P < 0.05), we   REJECT THE NULL HYPOTHESIS.          
              
Thus, there is significant evidence to conclude that the treatment has a significant effect. [CONCLUSION]

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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 = sqrt(72) =   8.485281374          
              
Thus, t = (X - uo) * sqrt(n) / s =    -1.666666667          
              
Also, the p value is              
              
p =    0.139519583          
              
Comparing |t| < 2.3646 (or, P > 0.05), we   FAIL TO REJECT THE NULL HYPOTHESIS.          
              
Thus, there is no significant evidence to conclude that the treatment has a significant effect. [CONCLUSION]
  
              

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