Question

Numerous studies have shown that high blood cholesterol leads to artery clogging and subsequent heart disease. One such study by D. Scott er al. was published in the paper "Plasma lipids as Collateral Risk factors in Coronary Artery Disease: A study of \(371\) Males with chest pain". The research compared the plasma cholesterol concentrations of independent random samples of patients with and without evidence of heart disease. Evidence of heart disease was based on the degree of narrowing in the arteries. The data on plasma cholesterol concentrations, in milligrams/deciliter are provided on the WeissStats site. Use the technology of your choice to do the following.

a. Obtain a normal probability plot, boxplot and histogram of the data for patients without evidence of heart disease.

b. Is it reasonable to apply the \(t-\)interval procedure to those data? Explain your answer.

c. If you answered "yes" to part (b), determine a \(95%\) confidence interval for the mean plasma cholesterol concentration of all males without evidence of heart disease. Interpret your result.

d. Repeat parts (a)-(c) for males with evidence of heart disease.

Short answer

Part a.

Part b. The \(95%\) confidence interval is Confidence interval \(=[185.0151, 205.5339]\)

Part c. The \(95%\) confidence interval is Confidence interval \(=[211.4584, 220.9228]\)

Step-by-step solution

Part a. Step 1. Given information

The number of sample \((n)\), mean \((\bar{x})\) and standard deviation \((\sigma)\) is given.

\(\bar{x}=195\)

\(\sigma=36\)

\(n=371\)

Part a. Step 2. Calculation

Generate \(371\) samples of males using function “norminv” with \(195\) sample mean and \(36\) standard deviation in MATLAB

\(r=norminv(rand(371,1),195,36)\)

After that we will get random sample \(371\) males.

Draw a normal probability plot using function “normplot”, boxplot using function “boxplot”, and histogram using function “histogram”.

Program:

Query:

• First, we have defined the number of samples.
• Then create a random sample of \(371\) males using function “norminv” with sample mean \(195\) and standard deviation \(36\).
• Then create a normal probability plot, box plot, and histogram.

Part b. Step 1. Calculation

Determine the \(t\)-value by looking degree of freedom

\(df=n-1=51-1=50\)

And in the column with \(t_{(1-c)/2}=t_{(1-0.95)/2}=t_{0.025}\) in the table of student t-distribution

\(t_{\alpha/2}=2.009\)

So, the marginal error will be

\(E=t_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}=2.009\times \frac{36.1099}{\sqrt{50}}\approx 10.2594\)

Calculate the \(95%\) confidence interval using given values

\(L=\bar{x}-E=195.2745-10.2594=185.0151\)

\(U=bar{x}+E=195.2745+10.2594=205.5339\)

After solving, we will get the upper and lower value of confidence interval.

Confidence interval \(=[185.0151,205.5339]\)

The confidence interval is Confidence interval \(=[185.0151,205.5339]\)

Part c. Step 1. Calculation

Determine the \(t\)-value by looking degree of freedom

\(df=n-1=320-1=319\)

And in the column with \(t_{(1-c)/2}=t_{(1-0.95)/2}=t_{0.025}\) in the table of student t-distribution

\(t_{\alpha/2}=1.968\)

So, the marginal error will be

\(E=t_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}=1.968\times \frac{43.0145}{\sqrt{320}}\approx 4.7322\)

Calculate the \(95%\) confidence interval using given values

\(L=\bar{x}-E=216.1906-4.7322=211.4584\)

\(U=bar{x}+E=216.1906+4.7322 =220.9228\)

After solving, we will get the upper and lower value of confidence interval.

Confidence interval \(=[211.4584, 220.9228 ]\)

The confidence interval is Confidence interval \(=[211.4584, 220.9228 ]\)