Respuesta :
Answer:
(a) The point estimate of the population mean HDL cholesterol level is 49.95.
(b) The point estimate of the value that separates the largest 50% of HDL levels from the smallest 50% is 47.5.
(c) The point estimate of the population standard deviation is 16.85.
Step-by-step explanation:
We are given a sample of 20 observations on HDL cholesterol level (mg/dl) obtained from the survey below;
35, 49, 51, 54, 65, 51, 52, 47, 87, 37, 46, 33, 39, 44, 39, 64, 94, 34, 30, 48.
(a) The point estimate of the population mean HDL cholesterol level is given by the sample mean of the above data, i.e;
Sample Mean, [tex]\bar X[/tex] = [tex]\frac{\sum X}{n}[/tex]
= [tex]\frac{35+ 49+ 51+ 54 +65 +51+ 52+ 47+ 87+ 37+ 46+ 33+ 39+ 44+ 39+ 64+ 94+ 34+ 30+ 48}{20}[/tex]
= [tex]\frac{999}{20}[/tex] = 49.95
So, the point estimate of the population mean HDL cholesterol level is 49.95.
(b) The point estimate of the value that separates the largest 50% of HDL levels from the smallest 50% is given by the Median of the above data.
Firstly, arranging the given data in ascending order we get;
30, 33, 34, 35, 37, 39, 39, 44, 46, 47, 48, 49, 51, 51, 52, 54, 64, 65, 87, 94.
Now, for calculating median we have to first observe that the number of observations (n) in our data is even or odd, i.e;
- If n is odd, then the formula for calculating median is given by;
Median = [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]
- If n is even, then the formula for calculating median is given by;
Median = [tex]\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.} }{2}[/tex]
Here, the number of observations is even, i.e. n = 20.
So, Median = [tex]\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.} }{2}[/tex]
= [tex]\frac{(\frac{20}{2})^{th} \text{ obs.}+(\frac{20}{2}+1)^{th} \text{ obs.} }{2}[/tex]
= [tex]\frac{(10)^{th} \text{ obs.}+(11)^{th} \text{ obs.} }{2}[/tex]
= [tex]\frac{47+48 }{2}[/tex]
Median = 47.5
Hence, the point estimate of the value that separates the largest 50% of HDL levels from the smallest 50% is 47.5.
(c) The point estimate of the population standard deviation is given by the following formula;
Standard deviation, s = [tex]\sqrt{\frac{\sum(X-\bar X)^{2} }{n-1} }[/tex]
= [tex]\sqrt{\frac{ (30-49.95)^{2}+(33-49.95)^{2}+(34-49.95)^{2}+........+(94-49.95)^{2}}{{20-1}} }} }[/tex]
= 16.85