Using alpha = 0.05, perform a hypothesis test to determine if the average LDL le

Question

Using alpha = 0.05, perform a hypothesis test to determine if the average LDL level is more the 60 points lower for patients who have taken the new medication. Approximate the p-value using Table 5 in Appendix A and interpret the result. Construct a 90% confidence interval to estimate the average difference in LDL levels for people before after they take the medication. Verify your results using Excel. Identify the p-value using Excel and interpret the result. What assumptions need to be made in order to perform this procedure?

Explanation / Answer

A)

Let ud = u2 - u1.              
Formulating the null and alternative hypotheses,              
              
Ho:   ud   >=   -60  
Ha:   ud   <   -60  
At level of significance =    0.05          
As we can see, this is a    left   tailed test.      
              
Calculating the standard deviation of the differences (third column):              
              
s =    14.62537231          
              
Thus, the standard error of the difference is sD = s/sqrt(n):              
              
sD =    4.22198132          
              
Calculating the mean of the differences (third column):              
              
XD =    -69.91666667          
              
As t = [XD - uD]/sD, where uD = the hypothesized difference =    -60   , then      
              
t =    -2.348818224          
              
As df = n - 1 =    11          
              
Then the critical value of t is              
              
tcrit =    -   2.20098516      
              
              
As t < -2.20098, WE REJECT THE NULL HYPOTHESIS.          

Thus, there is significant evidence that the average LDL level is more than 60 points lower for patients who have taken the new medication.

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b)              
              
As we can see, using a table,

0.01 < P < 0.025. [ANSWER]

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c)      
              
For the   0.9   confidence level,      
              
alpha/2 = (1 - confidence level)/2 =    0.05          
t(alpha/2) =    1.795884819          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    -77.49885882          
upper bound = [X1 - X2] + t(alpha/2) * sD =    -62.33447451          
              
Thus, the confidence interval is              
              
(   -77.49885882   ,   -62.33447451   ) [ANSWER]

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d)

I actually used Excel to construct the interval above. If you need another approach for part c), please resubmit and indicate what method has to be used in c). Thanks!

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e)

As df = n - 1 = 11, and since this is left tailed,

p =        0.019283816 [ANSWER]

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f)

We assume that the populations from which these scores came from are approximately normally distributed.      
              
              

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