Scenario #2 for questions 6 through 10: For a group of 10 men subjected to a str

Question

Scenario #2 for questions 6 through 10: For a group of 10 men subjected to a stress situation, the mean number of heartbeats per minute was 126 and the sample standard deviation was 4. Round all answers to 2 decimals what is the margin of error for a 95% confidence interval? What is the 95% confidence interval of the true mean? Write a sentence explaining the 95% confidence interval in context of the question. (There are only a few ways to do this correctly, so be sure to read the book closely.) Write a few sentences explaining what would happen to the size of the CI if you wanted a 90% confidence interval? Write a few sentences explaining what would happen to the size of the CI if you changed the sample size to be 1000 men and kept a 95% confidence interval 1" 2. 3. 4. 5.

Explanation / Answer

TRADITIONAL METHOD
given that,
sample mean, x =126
standard deviation, s =4
sample size, n =10
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4/ sqrt ( 10) )
= 1.26
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 2.262
margin of error = 2.262 * 1.26
= 2.86
III.
CI = x ± margin of error
confidence interval = [ 126 ± 2.86 ]
= [ 123.14 , 128.86 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =126
standard deviation, s =4
sample size, n =10
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 2.262
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 = [ 126 ± t a/2 ( 4/ Sqrt ( 10) ]
= [ 126-(2.262 * 1.26) , 126+(2.262 * 1.26) ]
= [ 123.14 , 128.86 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 123.14 , 128.86 ] 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
Answers:
1.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 2.262
margin of error = 2.262 * 1.26
= 2.86
2.
confidence interval = [ 126 ± t a/2 ( 4/ Sqrt ( 10) ]
= [ 126-(2.262 * 1.26) , 126+(2.262 * 1.26) ]
= [ 123.14 , 128.86 ]
3.
1) we are 95% sure that the interval [ 123.14 , 128.86 ] 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
4.
TRADITIONAL METHOD
given that,
sample mean, x =126
standard deviation, s =4
sample size, n =10
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4/ sqrt ( 10) )
= 1.26
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 1.833
margin of error = 1.833 * 1.26
= 2.32
III.
CI = x ± margin of error
confidence interval = [ 126 ± 2.32 ]
= [ 123.68 , 128.32 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =126
standard deviation, s =4
sample size, n =10
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 1.833
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 = [ 126 ± t a/2 ( 4/ Sqrt ( 10) ]
= [ 126-(1.833 * 1.26) , 126+(1.833 * 1.26) ]
= [ 123.68 , 128.32 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 123.68 , 128.32 ] 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
5.
TRADITIONAL METHOD
given that,
sample mean, x =126
standard deviation, s =4
sample size, n =1000
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4/ sqrt ( 1000) )
= 0.13
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 999 d.f is 1.962
margin of error = 1.962 * 0.13
= 0.25
III.
CI = x ± margin of error
confidence interval = [ 126 ± 0.25 ]
= [ 125.75 , 126.25 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =126
standard deviation, s =4
sample size, n =1000
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 999 d.f is 1.962
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 = [ 126 ± t a/2 ( 4/ Sqrt ( 1000) ]
= [ 126-(1.962 * 0.13) , 126+(1.962 * 0.13) ]
= [ 125.75 , 126.25 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 125.75 , 126.25 ] 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

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