To compare the treatment effect of two medications, patient performance data wer

Question

To compare the treatment effect of two medications, patient performance data were collected on two independent samples, one for each medication. The researchers decided to conduct a two-sample t-test to compare the mean performance scores. They found that the difference between the two sample means was 4.0 (x1-bar minus x2-bar). The sample standard deviation of the first sample is s1=4.0 with a sample size of n1=18 and the standard deviation of the second sample is s2=2.0 with a sample size of n2=16.

What would be the estimated standard error for the difference in sample mean x1-bar -x2-bar?

What would be the degrees of freedom for the T-statistic (x1-bar minus x2-bar) used for comparing (testing) the two mean?

What would be the rejection region for testing the null hypothesis of mean1=mean2 vs the alternative mean1 not equal to mean2 with a level of significance of 0.05? Use Excel T.INV(0.025, 32) to determine the critical value.

Explanation / Answer

A)

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 standard deviations of each group,              
              
s1 =    4          
s2 =    2          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    18          
n2 = sample size of group 2 =    16          

Hence,

sD =    1.067187373   [ANSWER, STANDARD ERROR]  
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b)

Thus, df = n1 + n2 - 2 =    32   [ANSWER, degrees of freedom]

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c)      

Now, the critical value for t is              
              
tcrit =    +/-   2.036933343   [ANSWER]

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Other details:  

  
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    3.748170285          
              
where uD = hypothesized difference =    0          

Also, using p values,              
              
p =    0.000706579          
              
As P < 0.05,    WE REJECT THE NULL HYPOTHESIS.          

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