Data from the hear Heart Study found that subjects over age 50 had a mean HDL of
ID: 3313867 • Letter: D
Question
Data from the hear Heart Study found that subjects over age 50 had a mean HDL of 54 and a standard deviation of 17. Suppose a physician has 40 patients over age 50 and wants to determine the following: a. The mean of the samples? (2 pt.) b. The standard deviation of the samples? (3 pt) c) What was the underlying theorem for your responses parts "a"and b of the questions? Explain in one sentence (3 pt) d) What is the probability that the mean HDL in these 40 patients will exceed 60 (8 pt) e) Calculate 95% upper confidence bound on the mean for this problem. Compare it to the mean based on the point estimator? (9)Explanation / Answer
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
sample size (n) = 40
a.
mean of the sampling distribution ( x ) = 54
b.
standard Deviation ( sd )= 17/ Sqrt ( 40 ) =2.6879
c.
P(X > 60) = (60-54)/17/ Sqrt ( 40 )
= 6/2.688= 2.2322
= P ( Z >2.2322) From Standard Normal Table
= 0.0128
f.
TRADITIONAL METHOD
given that,
sample mean, x =54
standard deviation, s =2.6879
sample size, n =40
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.6879/ sqrt ( 40) )
= 0.425
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table,right tailed value of |t /2| with n-1 = 39 d.f is 1.685
margin of error = 1.685 * 0.425
= 0.716
III.
CI = x ± margin of error
confidence interval = [ 54 ± 0.716 ]
= [ 53.284 , 54.716 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =54
standard deviation, s =2.6879
sample size, n =40
level of significance, = 0.05
from standard normal table,right tailed value of |t /2| with n-1 = 39 d.f is 1.685
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 54 ± t a/2 ( 2.6879/ Sqrt ( 40) ]
= [ 54-(1.685 * 0.425) , 54+(1.685 * 0.425) ]
= [ 53.284 , 54.716 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 53.284 , 54.716 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
the mean estimator is lies in the interval achived
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.