In a test of effectiveness of garlic for lowering cholesterol, 45 subjects were treated with garlic in process tablet form. chorlesterol levels were measured before and after the treatement. the cahnges in their levels of ldl cholesterol (in mg/dL) have a sample mean 5.1 and a sample standard deviation of 16.7. a.) the best point estimate is _mg/do b.) construct a 99% confidence interval estimate of the mean net change in ldl cholesterol after the garlic treatment. c) interpret this interval

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

536

step-by-step explanation:

i got 10.28 but that rounds up to 10.3