Question

A simple random sample of size n is drawn. The sample mean, X, is found to be 17.9, and the sample standard deviation, s, is found to be 4.8. Click the icon to view the table of areas under the t-distribution. (a) Construct a 95% confidence interval about us if the sample size, n, is 34. Lower bound: upper bound: (Use ascending order. Round to two decimal places as needed.) (b) Construct a 95% confidence interval about if the sample size, n, is 61. Lower bound: Upper bound: (Use ascending order. Round to two decimal places as needed.) How does increasing the sample size affect the margin of error, E? O A. The margin of error increases. OB. The margin of error does not change. O c. The margin of error decreases (e) Construct a 99% confidence interval about all the sample size, n, is 34 Lower bound Upper bound (Use ascending order. Round to two decimal places as needed.) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the size of the margin of error, E? Click to select your answer(s).

Answer 1

a.
TRADITIONAL METHOD
given that,
sample mean, x =17.9
standard deviation, s =4.8
sample size, n =34
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4.8/ sqrt ( 34) )
= 0.82
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 33 d.f is 2.035
margin of error = 2.035 * 0.82
= 1.68
III.
CI = x ± margin of error
confidence interval = [ 17.9 ± 1.68 ]
= [ 16.22 , 19.58 ]
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DIRECT METHOD
given that,
sample mean, x =17.9
standard deviation, s =4.8
sample size, n =34
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 33 d.f is 2.035
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 = [ 17.9 ± t a/2 ( 4.8/ Sqrt ( 34) ]
= [ 17.9-(2.035 * 0.82) , 17.9+(2.035 * 0.82) ]
= [ 16.22 , 19.58 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 16.22 , 19.58 ] 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
b.
TRADITIONAL METHOD
given that,
sample mean, x =17.9
standard deviation, s =4.8
sample size, n =61
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4.8/ sqrt ( 61) )
= 0.61
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 60 d.f is 2
margin of error = 2 * 0.61
= 1.23
III.
CI = x ± margin of error
confidence interval = [ 17.9 ± 1.23 ]
= [ 16.67 , 19.13 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =17.9
standard deviation, s =4.8
sample size, n =61
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 60 d.f is 2
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 = [ 17.9 ± t a/2 ( 4.8/ Sqrt ( 61) ]
= [ 17.9-(2 * 0.61) , 17.9+(2 * 0.61) ]
= [ 16.67 , 19.13 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 16.67 , 19.13 ] 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
if sample size increases then margin of error decreases.
c.
TRADITIONAL METHOD
given that,
sample mean, x =17.9
standard deviation, s =4.8
sample size, n =34
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4.8/ sqrt ( 34) )
= 0.82
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 33 d.f is 2.733
margin of error = 2.733 * 0.82
= 2.25
III.
CI = x ± margin of error
confidence interval = [ 17.9 ± 2.25 ]
= [ 15.65 , 20.15 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =17.9
standard deviation, s =4.8
sample size, n =34
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 33 d.f is 2.733
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 = [ 17.9 ± t a/2 ( 4.8/ Sqrt ( 34) ]
= [ 17.9-(2.733 * 0.82) , 17.9+(2.733 * 0.82) ]
= [ 15.65 , 20.15 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 15.65 , 20.15 ] 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
if confidence level is increases then margin of error also increases.
d.
if sample size is 13,the conditions must be satified ,
population is normally distributed with no outliers,sample size must be large.