Answer:
A) the sample mean is the best point estimate 5.1mg/do
B) the required confidence interval = [ 0.92 , 9.28]
C) the confidence interval suggest that there is a 99% chance that the true value of the mean net change in LDL cholesterol after garlic treatment is contained in the interval
Step-by-step explanation:
information given from the question
The sample size (n) = 45
The mean of the changes ( before and after ) in the levels of the LDL cholesterol = 5.1
the standard deviation = 16.7
B) the 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment can be constructed using the formula
=
d = sample mean = 5.1
sa = standard deviation = 16.7
= 1.6787
=
= [ 5.1 - 1.6787 * 2.4895, 5.1 + 1.6787 * 2.4895 ]
= [ 5.1 - 4.1791, 5.1 + 4.1791 ]
the required confidence interval = [ 0.92 , 9.28]
C) the confidence interval suggest that there is a 99% chance that the true value of the mean net change in LDL cholesterol after garlic treatment is contained in the interval