Question

Use SPSS for this Application Exercise:
A comparative psychologist hypothesize that feeling sick is associated with disliking a flavor. In a study rats are randomly split into two group. In one group rats received an injection of lithium chloride immediately after drinking stevia-flavored water. The rats in the other group received no lithium chloride after drinking the flavored water. Lithium chloride causes nausea. The next day, all rats could drink stevia-flavored water. The total amount of stevia-flavored water drank (in milliliters) by all the rats for the next day are given below. What can be concluded with α = 0.10?

lithium	no lithium
4 4 1 3 2 6 8 6 1	6 9 6 7 9 7 5 6 7

a) What is the appropriate test statistic?
---Select--- na z-test One-Sample t-test Independent-Samples t-test Related-Samples t-test

b)
Condition 1:
---Select--- disliking a flavor stevia-flavored water lithium chloride no lithium chloride the rats
Condition 2:
---Select--- disliking a flavor stevia-flavored water lithium chloride no lithium chloride the rats

c) Input the appropriate value(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
p-value =  ; Decision:  ---Select--- Reject H0 Fail to reject H0

d) Using the SPSS results, compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =  ;   ---Select--- na trivial effect small effect medium effect large effect
r² =  ;   ---Select--- na trivial effect small effect medium effect large effect

e) Make an interpretation based on the results.

Rats that received lithium chloride drank significantly more stevia-flavored water than those that received no lithium chloride.

Rats that received lithium chloride drank significantly less stevia-flavored water than those that received no lithium chloride.     

There is no significant difference in stevia-flavored water drank between rats that received lithium and those that received no lithium chloride.

Answer 1

a)  Independent-Samples t-test

b) Condition 1:  lithium chloride

Condition 2: no lithium chloride

c)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.1                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   3.889                  
standard deviation of sample 1,   s1 =    2.421                  
size of sample 1,    n1=   9                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   6.889                  
standard deviation of sample 2,   s2 =    1.364                  
size of sample 2,    n2=   9                  
                          
difference in sample means =    x̅1-x̅2 =    3.8889   -   6.9   =   -3.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.9650                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.9263                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -3.0000   -   0   ) /    0.93   =   -3.2387
                          
Degree of freedom, DF=   n1+n2-2 =    16                  

p-value =        0.005142   (excel function: =T.DIST.2T(t stat,df) )   52          
Conclusion:     p-value <α , Reject null hypothesis                      

d)

cohen's d =    |( x̅1-x̅2 )/Sp | =    1.5267
      
r²=   0.3960

e)

  Rats that received lithium chloride drank significantly less stevia-flavored water than those that received no lithium chloride.