Solution:
Confidence interval for average difference of paired samples is given as below:
Confidence interval = Dbar ± t*SD/sqrt(n)
From given data, we have
Dbar = 3.6
Sd = 17.9
n = 47
df = n – 1 = 46
Confidence level = 90%
Critical t value = 1.6787
(by using t-table)
Confidence interval = Dbar ± t*SD/sqrt(n)
Confidence interval = 3.6 ± 1.6787*17.9/sqrt(47)
Confidence interval = 3.6 ± 4.3830
Lower limit = 3.6 - 4.3830 = -0.78
Upper limit = 3.6 + 4.3830 = 7.98
-0.78 mg/dL < µd < 7.98 mg/dL
This interval suggest that the garlic is not effective for lowering cholesterol, because above interval contains zero.
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