The business manager of a local health clinic is interested

The business manager of a local health clinic is interested in estimating the difference between the fees for extended office visits in her center and the fees of a newly opened group practice. she gathered the following infomraiton regarding the two offices.

health clinic group practice

sample size 50 45

sample mean $21 $19

sample standard deviation $2.75 $3.00

a) determine the degrees of freedom for the t-distribution

b) develop a 95% confidence interval estimate for the differences between the average fees of the two offices

c) use the p-value approac and test to determine if the average fee of the health clinic is significantly higher than that of the group practice

SHOW WORK PLEASE

Solution

a)

df = n1 + n2 - 2 = 50 +45 - 2 = 93 [answer]

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b)

For the   0.95   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.025          
t(alpha/2) =    1.985801814          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    0.823087787          
upper bound = [X1 - X2] + t(alpha/2) * sD =    3.176912213          
              
Thus, the confidence interval is              
              
(   0.823087787   ,   3.176912213   ) [ANSWER]

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C)

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    right   tailed test.      
Calculating the means of each group,              
              
X1 =    21          
X2 =    19          
              
Calculating the standard deviations of each group,              
              
s1 =    2.75          
s2 =    3          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    50          
n2 = sample size of group 2 =    45          
Thus, df = n1 + n2 - 2 =    93          
Also, sD =    0.59266348          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    3.374596325          
              
Also, using p values,              
              
p =    0.00053981          
              
Comparing this to the significance level, P<0.05,   WE REJECT THE NULL HYPOTHESIS.          

Thus, we conclude that ther is significant evidence that the average fee of the health clinic is significantly higher than that of the group practice. [CONCLUSION]