The national health and nutrition examination survey (nhanes) collects demographic, socioeconomic, dietary, and
health-related information on an annual basis. here is a sample of 20 observations on hdl cholesterol level [mg/di)
obtained from the 2009-2010 survey (hdl is 'good' cholesterol; the higher its value, the lower the risk for heart disease):
35 49 52 54 65 5 1 5 1
47 86 36 46 33 39 45
39 63 95 35 30 48
a. calculate a point estimate of the population mean hdl cholesterol level.
b. making no assumptions about the shape of the population distribution, calculate a point estimate of the value that
separates the largest 50% of hdl levels from the smallest 50%.
c. calculate a point estimate of the population standard deviation.
d. an hdl level of at least 60 is considered desirable as it corresponds to a significantly lower risk of heart disease. making
no assumptions about the shape of the population distribution. estimate the proportion p of the population having an
h dl level of at least 60.

Step-by-step explanation:

Hello!

Given a sample of 20 observations on HDL cholesterol level mg/di. Then you can determine the study variable as X: HDL cholesterol level of an individual

a.

The mean of any variable is calculated as the summation of each observation divided the sample size:

X[bar]= ∑X/n= 991/20= 49.95

b.

The value that separates the largest 50% from the smallest 50% is the median, you calculate it as:

PosMe= n+1/2= 10.5

To calculate the Mean you have to you must order the observations from least to greatest, and then identify the values in positions 10 and 11.

Me= (Pos10 + Pos11)/n= (47 + 48)/2= 47.5

c.

The Standard deviation S is calculated as the square root of the variance (S²)

S²=

∑X= 999

∑X²= 55269

S²=

S²= 282.58

S= 16.81

d.

In this point you transform the variable, now you don't consider any level of HDL but two possible cases. "to have an HDL level of less than 60"(symbolically X<60) and "to have an HDL level of at least 60" (symbolically X≥60). So the new variable will be X: Number of HDL level observations that are at least 60 in a sample of 20. Where "to have an HDL level of at least 60" the "success" (x) of the experiment, you have to count in the sample how many observations meet these conditions.

Count the number of observations that are "X≥60": 65, 86, 63, 95

4 observation is at least 60, in other words, 4 out of 20 observations count as "success" x=4

Then the proportion is:

p= x/n= 4/20= 0.2

I hope it helps!