Question

My Research Methods class (N = 14) was interested in whether studying alone or in groups would result in better grades on their third exam. Half of the class (n = 7) studied for the exam alone, while the other half (n = 7) studied as a group. Their grades are summarized in the below tables:

   

Study Alone
Student	Grade
1	78%
2	97%
3	79%
4	90%
5	91%
6	74%
7	72%

Study in Group
Student	Grade
8 8	72%
9. 9	89%
1 10	77%
1 11	87%
1 12	89%
1 13	77%
1 14	69%

Using the above data, answer the following questions

1. Which type of t-test (one-sample, independent samples, related samples) is appropriate for determining if those that that studied in a group earned better or worse grades than those who studied alone? Why?

2. State the alternative and null hypotheses (using the statistical notation for stating mathematical relationships) that represents the prediction that those who studied in a group will obtain grades that are, on average, different from that of those who studied alone.

3. Calculate pooled variance, standard error of the difference, and t_obt.

4. Using an alpha of 0.05, what is t_crit? Would you reject or fail to reject the null hypothesis? Why?

Answer 1

independent samples test

...............

Ho :   µ1 - µ2 =   0              
Ha :   µ1-µ2 ╪   0              
                      
Level of Significance ,    α =    0.05              
                      
Sample #1   ---->   study alone   
mean of sample 1,    x̅1=   0.83              
standard deviation of sample 1,   s1 =    0.10              
size of sample 1,    n1=   7              
                      
Sample #2   ----> in group
mean of sample 2,    x̅2=   0.80              
standard deviation of sample 2,   s2 =    0.08              
size of sample 2,    n2=   7              
                      
difference in sample means =    x̅1-x̅2 =    0.8300   -   0.8   =   0.03
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    0.0897             
std error , SE =    Sp*√(1/n1+1/n2) =    0.0480              
                      
t-statistic = ((x̅1-x̅2)-µd)/SE = (   0.0300   -   0   ) /    0.05   =0.626
                      
Degree of freedom, DF=   n1+n2-2 =    12              
t-critical value , t* =        2.1788   (excel formula =t.inv(α/2,df)          
Decision:   | t-stat | < | critical value |, so, Do not Reject Ho                  
  
                      
There is not enough evidence to say that who studied in a group will obtain grades that are, on average, different from that of those who studied alone

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THANKS

revert back for doubt

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