Question

5. Consider the following data from an independent measures study:

Treatment 1          Treatment 2

     n = 4                        n = 4

mean = 4              mean = 11

SS = 96                   SS = 100

Is there a evidence for a treatment effect? (α = .05, two tails).

6. A sample of n = 16 difference scores from a repeated measures experiment has SS = 440. The average difference score is 3. For a two-tailed test, α = .05, is this a significant difference?

7. Many over-the-counter cold medications come with a warning that the medication may cause drowsiness. To evaluate this effect, a researcher measures reaction time for a sample of n = 36 subjects. Each subject is then given a dose of a popular cold medicine, and reaction time is measured again. For this sample, reaction time increases after the medication by an average of 24 milliseconds with s = 8. Does the cold medicine have a significant effect on reaction time? (two tailed test, .05).

8. A researcher for a cereal company wanted to demonstrate the health benefits of eating oatmeal. A sample of 9 volunteers was obtained and each subject ate a fixed diet without any oatmeal for 30 days. At the end of the 30-day period, cholesterol was measured for each individual. Then the subjects began a second 30-day period in which they repeated exactly the same diet except for 2 cups of oatmeal each day. After the second 30-day period, cholesterol levels were measured again. The scores for the 9 subjects are as follows:

Answer 1

5)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   4.00                  
standard deviation of sample 1,   s1 =    5.66                  
size of sample 1,    n1=   4                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   11.00                  
standard deviation of sample 2,   s2 =    5.77                  
size of sample 2,    n2=   4                  
                          
difference in sample means =    x̅1-x̅2 =    4.0000   -   11.0   =   -7.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    5.7155                  
std error , SE =    Sp*√(1/n1+1/n2) =    4.0415                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -7.0000   -   0   ) /    4.04   =   -1.732
                          
Degree of freedom, DF=   n1+n2-2 =    6                  

p-value =        0.133975   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis                      
there is not a evidence for a treatment effect

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7)

Ho :   µd=   0                  
Ha :   µd ╪   0                  
                          
Level of Significance ,    α =    0.05       claim:µd=0          
                          
sample size ,    n =    36                  
                             
                          
mean of difference ,    D̅ =    24.000                  
                          
std dev of difference , Sd =        8.0000                  
                          
std error , SE = Sd / √n =    8.0000   / √   36   =   1.3333      
                          
t-statistic = (D̅ - µd)/SE = (   24   -   0   ) /    1.3333   =   18.000
                          
Degree of freedom, DF=   n - 1 =    35                     
                          
p-value =        0.000000   [excel function: =t.dist.2t(t-stat,df) ]               
Conclusion:     p-value <α , Reject null hypothesis                      

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8)

data not given

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THANKS

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