Asked by Lynn
In a test of of garlic for lowering cholesterol, 45 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured beforeandafter the treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 5.4 and a sd of 17.5. Construct a 90% confidence interval est. of the mean net change in LDL cholesterol. after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?
What is the confidence interval of the mean u?
___Mg/dL < u < ___Mg/dL
What is the confidence interval of the mean u?
___Mg/dL < u < ___Mg/dL
Answers
Answered by
PsyDAG
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability
(±.05) and its Z score.
90% = mean ± Z (SEm)
SEm = SD/√n
(±.05) and its Z score.
90% = mean ± Z (SEm)
SEm = SD/√n
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