samele of 20 small bags of the same brand of candias was selectead. Assume thast

Question

samele of 20 small bags of the same brand of candias was selectead. Assume thast the pepudation disribotice of hop woghs s eormat. The wnighe ofr each bag was then recordet. The m ight a standant deviation ion o 014 ounces. The popuation gandard deviation is knon to be 0.1 sunce, If you arm using a Studenc's e-disinbuion, you may aoume that the undorlyng poputlation is nermuaily dierituted. ttn gonod, you must firnt pruves that assumptian, bough ) v -119 Part Part Which dshibuton shoud you use for this problen? (Rourd your anwers to theee docial pices Explan your choce 9 Th. mrdad rum datetun should be used because he unple stardant de-anon s nom The standad naml ebuton should be used because the pepulaton standard deaaton a known The Stoer'sbut shoult be ueed because te samplo standan de-ation is g The Suat'sb should be used because the sampke sce nat

Explanation / Answer

we are using t test for single mean

TRADITIONAL METHOD
given that,
sample mean, x =3
standard deviation, s =0.34
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.34/ sqrt ( 20) )
= 0.08
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 = 19 d.f is 2.093
margin of error = 2.093 * 0.08
= 0.16
III.
CI = x ± margin of error
confidence interval = [ 3 ± 0.16 ]
= [ 2.84 , 3.16 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =3
standard deviation, s =0.34
sample size, n =20
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
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 = [ 3 ± t a/2 ( 0.34/ Sqrt ( 20) ]
= [ 3-(2.093 * 0.08) , 3+(2.093 * 0.08) ]
= [ 2.84 , 3.16 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 2.84 , 3.16 ] 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

stanadard normal distribution is used when population stanadard deviation is known

TRADITIONAL METHOD
given that,
sample mean, x =3
standard deviation, s =0.34
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.34/ sqrt ( 20) )
= 0.076
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 = 19 d.f is 1.729
margin of error = 1.729 * 0.076
= 0.131
III.
CI = x ± margin of error
confidence interval = [ 3 ± 0.131 ]
= [ 2.869 , 3.131 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =3
standard deviation, s =0.34
sample size, n =20
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 1.729
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 = [ 3 ± t a/2 ( 0.34/ Sqrt ( 20) ]
= [ 3-(1.729 * 0.076) , 3+(1.729 * 0.076) ]
= [ 2.869 , 3.131 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 2.869 , 3.131 ] 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

TRADITIONAL METHOD
given that,
sample mean, x =3
standard deviation, s =0.34
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.34/ sqrt ( 20) )
= 0.076
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.02
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.539
margin of error = 2.539 * 0.076
= 0.193
III.
CI = x ± margin of error
confidence interval = [ 3 ± 0.193 ]
= [ 2.807 , 3.193 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =3
standard deviation, s =0.34
sample size, n =20
level of significance, = 0.02
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.539
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 = [ 3 ± t a/2 ( 0.34/ Sqrt ( 20) ]
= [ 3-(2.539 * 0.076) , 3+(2.539 * 0.076) ]
= [ 2.807 , 3.193 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 98% sure that the interval [ 2.807 , 3.193 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population mean

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