The study introduced in the prior exercise also compared dur

The study introduced in the prior exercise also compared duration of peak symptoms in the treatment and control group. The treatment group (n1=337) had a mean duration of 1.60 days (standard deviation 0.98 days). The control group (n2 = 370) had a mean duration of peak symptoms of 1.64 days (standard deviation 1.14 days). Test the difference in means of significance.

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   =   0  
Ha:   u1 - u2   =/   0  
At level of significance =    0.05          
As we can see, this is a    two   tailed test.      
Calculating the means of each group,              
              
X1 =    1.6          
X2 =    1.64          
              
Calculating the standard deviations of each group,              
              
s1 =    0.98          
s2 =    1.14          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    337          
n2 = sample size of group 2 =    370          
Thus, df = n1 + n2 - 2 =    705          
Also, sD =    0.079763927          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    -0.501479823          
              
where uD = hypothesized difference =    0          
              
Now, the critical value for t is              
              
tcrit =    +/-   1.963334594      
              
Thus, as t is between the two tcrit values,    WE FAIL TO REJECT THE NULL HYPOTHESIS.      

Thus, there is no significant evidence at 0.05 level that the mean durations are different. [CONCLUSION]