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In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) had a mean of \(0.4\) and a standard deviation of \(21.0\) (based on data from "Effect of Raw Garlic vs Commercial Garlic Supplements on Plasma Lipid Concentrations in Adults with Moderate Hypercholesterolemia," by Gardner et al., Archives of Internal Medicine, Vol. 167). Construct a \(98 \%\) 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?

Step by step solution

01

Identify Given Information

Find the mean (\bar{x}), standard deviation (s), sample size (n), and confidence level (CL). We have: \[ \bar{x} = 0.4 \]\[ s = 21.0 \]\[ n = 49 \]\[ CL = 98\% \]

02

Determine the Significance Level

The confidence level (CL) is 98%, so the significance level (\alpha) is:\[ \alpha = 1 - CL = 1 - 0.98 = 0.02 \]Since we use two tails for the confidence interval, we divide the significance level by 2:\[ \alpha/2 = 0.01 \]

03

Find the Critical Value

Using the t-distribution table for \(df = n - 1 = 49 - 1 = 48\) and \( \alpha/2 = 0.01\), find the t-critical value (\(t_{\alpha/2}\)). For 48 degrees of freedom, the critical value (t_{0.01, 48}) is approximately 2.407.

04

Calculate the Standard Error

The standard error (SE) is calculated by:\[ SE = \frac{s}{\sqrt{n}} = \frac{21.0}{\sqrt{49}} = \frac{21.0}{7} = 3.0 \]

05

Construct the Confidence Interval

Use the formula for the confidence interval: \[ \bar{x} \pm t_{\alpha/2} \times SE \]Plugging in our values, we get:\[ 0.4 \pm 2.407 \times 3.0 \]This results in:\[ 0.4 \pm 7.221 \]Which gives us the interval:\[ -6.821 \text{ to } 7.621 \]

06

Interpret the Confidence Interval

The confidence interval suggests that the true mean change in LDL cholesterol after garlic treatment is between -6.821 mg/dL and 7.621 mg/dL. Because the interval includes 0, we do not have enough evidence to suggest that garlic has a significant effect on reducing LDL cholesterol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

t-distribution

When dealing with small sample sizes or unknown population standard deviations, the t-distribution helps us estimate population parameters. Unlike the normal distribution, the t-distribution is wider, reflecting greater uncertainty.
It varies depending on the degrees of freedom (df), which in this case is calculated as the sample size minus one \(df = n - 1\). For our exercise, with n=49, df is 48.
It's important to use the correct t-value from the t-distribution table, corresponding to our significance level and degrees of freedom. A higher degree of freedom makes the t-distribution approach the normal distribution.

standard error

The standard error (SE) measures how much sample means are expected to vary from the true population mean. It is calculated as: \[ SE = \frac{s}{\sqrt{n}} \]where 's' is the sample standard deviation and 'n' is the sample size. In our case, we have s = 21.0 and n = 49, so: \[ SE = \frac{21.0}{\sqrt{49}} = 3.0 \]The smaller the SE, the more accurate our estimate of the population mean. It plays a crucial role in constructing confidence intervals, as seen in our exercise.

LDL cholesterol measurement

LDL cholesterol, often termed 'bad cholesterol,' is a crucial health metric. It's responsible for building up plaque in arteries, which can lead to heart diseases. In the context of our exercise, we are studying the change in LDL cholesterol levels before and after garlic treatment.
We have 49 subjects and have recorded the changes in their LDL levels. The key statistics obtained include a mean change of 0.4 mg/dL and a standard deviation of 21.0 mg/dL.
By analyzing the mean change and constructing confidence intervals, we aim to understand better whether garlic effectively reduces LDL cholesterol levels.

significance level

The significance level \(\alpha\) is the probability of rejecting the null hypothesis when it is true. It's set during hypothesis testing and constructing confidence intervals. For a 98% confidence interval, \(\alpha\) is calculated as: \[ \alpha = 1 - CL = 1 - 0.98 = 0.02 \]We then divide \(\alpha\) by 2 for the two-tailed test: \[ \alpha/2 = 0.01 \]Using this, we find the critical t-value to plug into our confidence interval formula. A lower significance level (e.g., 0.01) implies stronger evidence needed to reject the null hypothesis, aiming for greater accuracy and reducing the chance of Type I errors.