Can someone help me please

Question

In a test of the effectiveness of garlic for lowering cholesterol,42 subjects were treated with garlic in a processed tablet form. cholesterol levels were measured before and after the treatment. The changes (before-after) in their levels of LDL cholesterol ( in mg/dL) have a mean of 4.8 and a standard deviation of 15.6. Construct a 90% confidence interval estimate 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?

a) what is the confidence interval estimate of the population mean?

mg/dL< mean< mg/dL

b) What does the confidence interval suggest about the effectiveness of the treatment?

Answer 1

Just to be completely clear, I will use the following notation as shorthand:

n = number of observations in the experiment

xbar = sample mean

sd = standard deviation of the sample mean

t = critical value

mu = mean net change in LDL (a.k.a., what we are finding a confidence interval for)

To find a confidence interval, we want to find an upper bound and lower bound for mu. To estimate this, we can use the formula:

lower bound = xbar - t * sd / sqrt(n)

upper bound = xbar + t * sd / sqrt(n)

t represents the area underneath the curve of Student's T-Distribution from 0 to the 0.05 quantile (we use 0.05 because this is a two-tailed test, and 0.10 / 2 = 0.05). We find this value using t-tables and knowing that our degrees of freedom = n - 1 = 41.

Now, we know:

t = 1.6829 (found with t-tables)

n = 42

xbar = 4.8

sd = 15.6

We want to construct a 90% confidence interval estimate of the mean net change in LDL after treatment. So, we calculate our lower and upper bounds to get:

lower = 4.8 - 1.6829 * 15.6 / sqrt(42) = 0.749037

upper = 4.8 + 1.6829 * 15.6 / sqrt(42) = 8.85096

Therefore, our 90% confidence interval is (0.749037, 8.85096).

Interpretation: We may conclude with 90% confidence that the mean net change in LDL cholesterol after the garlic treatment ranges from (0.749037, 8.85096). We see that this interval does not include 0. Thus, we may further conclude with 90% confidence that the garlic treatment is effective in reducing LDL cholesterol.

One small clarification: given the question, it is unclear to me whether this mean change of 4.8 mg/dL is referred to as an increase or decrease. Based on the context, I have interpreted this question by assuming that the mean change of 4.8 mg/dL is a decrease.

Answer 2

n = 42 subjects

x-bar = 4.8

σ = 15.6

z* = 1.645 (This is from the 90% confidence interval)

x-bar ± z*(σ/√n) = 4.8 ± 1.645(15.6/√42)

= 4.8 ± 3.96

= (0.84, 8.76)

The sample mean, 4.8 mg/dL, lies between 0.84 mg/dL and 8.76 mg/dL.

We are 90% confident that garlic will reduce our cholesterol level. The experiment will work if a subject takes no less than 0.84 mg/dL and no more than 8.76 mg/dL.