Given a sample size of n = 196. Let the variance of the population be 2= 7.29. L

Question

Given a sample size of n = 196. Let the variance of the population be 2= 7.29. Let the mean of the sample be xbar = 3. Construct a 90% confidence interval for µ, the mean of the population, using this data and the central limit theorem.

a.     What is the standard deviation () of the population? 

b.     What is the standard deviation of the means xbar of sample size n, i.e. xbar , in terms of and n?

c.     Is this a one-sided or two-sided problem? 

d.     What value of z should be used in computing k, the margin of error, where

z = k/xbar  = k/[/ ]? 

e.     What is k? 

f.      Write the 90% confidence interval for µ based on xbar and k,

(xbar – k) < µ < (xbar + k)

g.     Using the Z-score applet “Area from a value”. Let the Mean = 3, and SD = xbar. Choose “Between (xbar -k) and (xbar + k)” using xbar = 3 and your computed value of k. Hit “Recalculate”. Does the probability approximately equal 0.90? (yes or no). Include a screen shot of your answer.

Explanation / Answer

TRADITIONAL METHOD
given that,
standard deviation, = sqrt(7.29) = 2.7
sample mean, x =3
population size (n)=196
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 2.7/ sqrt ( 196) )
= 0.193
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 0.193
= 0.317
III.
CI = x ± margin of error
confidence interval = [ 3 ± 0.317 ]
= [ 2.683,3.317 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =2.7
sample mean, x =3
population size (n)=196
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 3 ± Z a/2 ( 2.7/ Sqrt ( 196) ) ]
= [ 3 - 1.645 * (0.193) , 3 + 1.645 * (0.193) ]
= [ 2.683,3.317 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [2.683 , 3.317 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
[ANSWERS]
a.
standard deviation, =2.7
b.
sample standard deviation = standard error =0.193
c.two sided
d.
z table value = 1.645
margin of error = 0.317
confidence interval = [ 2.683 , 3.317 ]

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.