SOLUTION: 1. The mean serum cholesterol level in a certain population of normal healthy men is 240 mg/dl and the standard deviation is 40 mg/dl. A clinical researcher interested in compari

Algebra ->  Statistics  -> Binomial-probability -> SOLUTION: 1. The mean serum cholesterol level in a certain population of normal healthy men is 240 mg/dl and the standard deviation is 40 mg/dl. A clinical researcher interested in compari      Log On


   



Question 1174983: 1. The mean serum cholesterol level in a certain population of normal healthy men
is 240 mg/dl and the standard deviation is 40 mg/dl. A clinical researcher
interested in comparing cholesterol levels in this healthy population with those
in men with coronary artery disease measured serum cholesterol level in a
random sample of 100men who had undergone coronary bypass surgery during
the preceding two year period. The mean serum cholesterol level for sampling
was 260 mg/dl
A. What is the 95% confidence interval estimate of the mean cholesterol level for
all men undergoing bypass surgery? Assume that the standard deviation of
serum cholesterol measurement for this population is the same as that of the
healthy population ( i.e., 40mg/dl)
B. Based on this estimate, can the researcher conclude that the mean serum
cholesterol of men undergoing coronary bypass surgery differs from that of
healthy men?
2. Among 200 patients at a Tikur Anbessa hospital.

Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

I'll do the first problem to get you started.
You should only post one question at a time please.

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Part A

For the population of men undergoing the surgery, we have
xbar = 260 mg/dl = sample mean, used to estimate population mean (mu)
sigma = 40 mg/dl = population standard deviation
n = 100 = sample size

At 95% confidence, the z critical value is roughly z = 1.960
We're using the standard normal z distribution because we know the population standard deviation sigma.

The lower bound L of the confidence interval is:
L = xbar - z*sigma/sqrt(n)
L = 260 - 1.960*40/sqrt(100)
L = 260 - 7.84
L = 252.16

The upper bound U is
U = xbar + z*sigma/sqrt(n)
U = 260 + 1.960*40/sqrt(100)
U = 260 + 7.84
U = 267.84

The 95% confidence interval to estimate the mean population (mu) of the cholesterol level of the men undergoing surgery is (252.16, 267.84)
We are 95% confident that mu is somewhere in the interval (252.16, 267.84)

We can express that interval as 252.16 < mu < 267.84

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Part B

Yes we can conclude that the mean serum cholesterol for the men undergoing bypass surgery differs from the healthy men.
This is because the interval 252.16 < mu < 267.84 does not contain mu = 240. This value is to the left of the lower bound 252.16

Since mu = 240 is not in the interval 252.16 < mu < 267.84, this means there's no way mu can be equal to 240 for the population of men undergoing surgery.

In other words, if A and B represent the population means for the healthy men vs the men undergoing surgery, then we can say:
A = 240
B = some value between 252.16 and 267.84
clearly we can see that B = 240 is not possible and B > A and B > 252.16

Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.

Those who violate the rules, should not get benefits from the forum.


By the way, the problem #2 is NONSENSICAL, as I noticed it just many times before.