Question

In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.6 and a standard deviation of 23.5. Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. A. Ho: u = 0 mg/dL H:> 0 mg/dL JOB. Ho:u= 0 mg/dL H: <0 mg/dL OD. Hou > 0 mg/dL Hu<0 mg/dL C. Ho u = 0 mg/dL Hyu#0 mg/dL Determine the test statistic. (Round to two decimal places as needed.) Determine the P-value. (Round to three decimal places as needed.) State the final conclusion that addresses the original claim. Click to select your answer(s). ?

Answer 1

Ho :   µ =   0  
Ha :   µ >   0   (Right tail test)
          
Level of Significance ,    α =    0.050  
sample std dev ,    s =    1.8000  
Sample Size ,   n =    35  
Sample Mean,    x̅ =   4.9000  
          
degree of freedom=   DF=n-1=   34  
          
Standard Error , SE = s/√n =   1.8/√35=   0.3043  
t-test statistic= (x̅ - µ )/SE =    (4.9-0)/0.3043=   16.10
          
          
p-Value   =   0.000 [Excel formula =t.dist(t-stat,df) ]
Decision:   p-value≤α, Reject null hypothesis       

reject Ho. . there is sufficient evidence to conclude that mean of population of changes is greater than 0

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