Question

1. The tolerance interval of 95.44 percent is ________ a 95.44 percent confidence interval.

the same width as

narrower than

wider than

2.

The U.S. Department of Health and Human Services collected sample data for 772 males between the ages of 18 and 24. That sample group has a mean height of 69.7 inches with a standard deviation of 2.8 inches. Find the 99 percent confidence interval for the mean height of all males between the ages of 18 and 24.

[63.19, 76.21]

[62.49, 76.91]

[69.65, 69.75]

[69.47, 69.93]

[69.44, 69.96]

Answer 1

2.
TRADITIONAL METHOD
given that,
standard deviation, σ =2.8
sample mean, x =69.7
population size (n)=772
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 2.8/ sqrt ( 772) )
= 0.1
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 0.1
= 0.26
III.
CI = x ± margin of error
confidence interval = [ 69.7 ± 0.26 ]
= [ 69.44,69.96 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =2.8
sample mean, x =69.7
population size (n)=772
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
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 = [ 69.7 ± Z a/2 ( 2.8/ Sqrt ( 772) ) ]
= [ 69.7 - 2.576 * (0.1) , 69.7 + 2.576 * (0.1) ]
= [ 69.44,69.96 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [69.44 , 69.96 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
1.
TRADITIONAL METHOD
given that,
standard deviation, σ =2.8
sample mean, x =69.7
population size (n)=772
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 2.8/ sqrt ( 772) )
= 0.101
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.0456
from standard normal table, two tailed z α/2 =1.999
since our test is two-tailed
value of z table is 1.999
margin of error = 1.999 * 0.101
= 0.201
III.
CI = x ± margin of error
confidence interval = [ 69.7 ± 0.201 ]
= [ 69.499,69.901 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =2.8
sample mean, x =69.7
population size (n)=772
level of significance, α = 0.0456
from standard normal table, two tailed z α/2 =1.999
since our test is two-tailed
value of z table is 1.999
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 = [ 69.7 ± Z a/2 ( 2.8/ Sqrt ( 772) ) ]
= [ 69.7 - 1.999 * (0.101) , 69.7 + 1.999 * (0.101) ]
= [ 69.499,69.901 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95.44% sure that the interval [69.499 , 69.901 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95.44% of these intervals will contains the true population mean
conclusion:
95.44% sure that the interval [69.499 , 69.901 ]
slightly narrower than 99% confidence interval