Question

A research team wants to investigate the usefulness of relaxation training for reducing levels of anxiety in individuals experiencing stress. They identify 30 people at random from a group of 100 who have "high stress" jobs. The 30 people are divided into two groups. One group acts as the control group - they receive no training. The second group of 15 receive the relaxation training. The subjects in each group are then given an anxiety inventory.

The summarized results appear below where higher scores indicate greater anxiety. The obtained statistic: 2.34 Control X = 30 S = 6.63 N = 15 RELAXATION X = 26 S = 6.2 N = 15

What type of test should you use?

b. Please provide, in words, the null and alternative hypotheses.

c. Using an alpha level of .05 identify the critical value(s).

d. Using the above test statistic, alpha level and critical value(s) what is your statistical decision?

e. State your conclusion regarding the results from this test in language that a friend of yours with no knowledge of statistics could understand

Answer 1

We should use 2 sample independent t test

a)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 >   0                  

b)
                          
Level of Significance ,    α =    0.05                  

Degree of freedom, DF=   n1+n2-2 =    28                  
t-critical value , t* =        1.7011   (excel function: =t.inv(α,df)     

c)

                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   30.00                  
standard deviation of sample 1,   s1 =    6.63                  
size of sample 1,    n1=   15                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   26.00                  
standard deviation of sample 2,   s2 =    6.20                  
size of sample 2,    n2=   15                  
                          
difference in sample means =    x̅1-x̅2 =    30.0000   -   26.0   =   4.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    6.4186                  
std error , SE =    Sp*√(1/n1+1/n2) =    2.3437                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   4.0000   -   0   ) /    2.34   =   1.707
                          
       
Decision:   | t-stat | > | critical value |, so, Reject Ho                     
  
                          
There is enough evidence to suggest that relaxation training has reduced levels of anxiety in individuals experiencing stress.

THANKS

revert back for doubt

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