Question

A psychopharmacologist (a psycholgist interested in the effects of drugs and physiology) is interested in determining the effects of caffeine (from coffee) on performance. Twenty students from a large introductory psychology class volunteered to participate in the study; all were dedicated coffee drinkers and all drank caffeinated coffee. The volunteers were then randomly divided into two groups such that there were ten students in each group. One group of 10 students was asked to study the material while drinking their normal doses of coffee, and to take the exam while drinking their normal doses of coffee before the test. This group is called the “Coffee with Test” group.

The second group of 10 students was asked to study the exam material while drinking their normal doses of coffee, but to take the without having consumed coffee for at the least 3 hours beforehand. This group is called the “No-Coffee with Test” group. One student from the “No-Coffee with Test” group became ill and was unable to sit for the. The researcher recorded the number of correct responses on the . The scores are below.

Coffee with Test (Number Correct) Group 1	No-Coffee with Test (Number Correct) Group 2
8	7
10	12
14	9
12	9
9	5
9	6
13	14
8	6
12	7
11

With  = .01, complete step 4 of the hypothesis testing procedure, what decision and conclusion should the researcher make?

The researcher should reject H₀ and conclude that there is a difference in performance with no caffiene versus caffiene.

The researcher should retain H₀ and conclude that caffeine has no effect on performance

The researcher should reject H_A and conclude that caffeine affects performance

The researcher should reject H₀ and conclude that caffeine has no effect on performance

No answer text provided.

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.01                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   10.600                  
standard deviation of sample 1,   s1 =    2.119                  
size of sample 1,    n1=   10                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   8.333                  
standard deviation of sample 2,   s2 =    3.000                  
size of sample 2,    n2=   9                  
                          
difference in sample means =    x̅1-x̅2 =    10.6000   -   8.3   =   2.27  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.5713                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.1814                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   2.2667   -   0   ) /    1.18   =   1.9186
                          
Degree of freedom, DF=   n1+n2-2 =    17                  
t-critical value , t* =        2.898   (excel formula =t.inv(α/2,df)              
Decision:   | t-stat | < | critical value |, so, Do not Reject Ho          

           
p-value =        0.071993   (excel function: =T.DIST.2T(t stat,df) )     
Conclusion:     p-value>α , Do not reject null hypothesis                      

The researcher should retain H₀ and conclude that caffeine has no effect on performance