Question

X ~ ( Mean, 2.15) ( Suppose we do not have information of Mean of the population) Question 1----------------------------------------------------------------------------- Use calculator go to math---> PRB-------> random sample 1. * Pick one (only-one) sample size of 5 (As we did in lab 2) ** calculate Mean of this sample and use the S.D. = 2.15/ sqrt(5) *** Calculate EBM **** Find C.I. for mean at 95% C.L. 2. Repeat the same process for n=10 and n = 20 3. Study what happened to Confidence Interval when we increased the sample size. Question 2------------------------------------------------------------------------------------------------- 4. * Find C.I. For n=20 for C.L. 90% ** Study what happened to Confidence Interval when we increased and C.L. from 90% to 95% Question 3---------------------------------------------------------------------------------------- From the previous lab we know mean for recipe data was 3.57. Write a comment about what you notice in Question 1 and Question 2

Answer 1

1.
TRADITIONAL METHOD
given that,
sample mean, x =3.57
standard deviation, s =2.15
sample size, n =5
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.15/ sqrt ( 5) )
= 0.962
II.
margin of error = t α/2 * (stanadard 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 = 4 d.f is 2.776
margin of error = 2.776 * 0.962
= 2.669
III.
CI = x ± margin of error
confidence interval = [ 3.57 ± 2.669 ]
= [ 0.901 , 6.239 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =3.57
standard deviation, s =2.15
sample size, n =5
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 4 d.f is 2.776
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 = [ 3.57 ± t a/2 ( 2.15/ Sqrt ( 5) ]
= [ 3.57-(2.776 * 0.962) , 3.57+(2.776 * 0.962) ]
= [ 0.901 , 6.239 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 0.901 , 6.239 ] 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

2.
a.
sample size at n=10
TRADITIONAL METHOD
given that,
sample mean, x =3.57
standard deviation, s =2.15
sample size, n =10
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.15/ sqrt ( 10) )
= 0.68
II.
margin of error = t α/2 * (stanadard 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 = 9 d.f is 2.262
margin of error = 2.262 * 0.68
= 1.538
III.
CI = x ± margin of error
confidence interval = [ 3.57 ± 1.538 ]
= [ 2.032 , 5.108 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =3.57
standard deviation, s =2.15
sample size, n =10
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/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 = [ 3.57 ± t a/2 ( 2.15/ Sqrt ( 10) ]
= [ 3.57-(2.262 * 0.68) , 3.57+(2.262 * 0.68) ]
= [ 2.032 , 5.108 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 2.032 , 5.108 ] 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.
sample size at n=20
TRADITIONAL METHOD
given that,
sample mean, x =3.57
standard deviation, s =2.15
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.15/ sqrt ( 20) )
= 0.481
II.
margin of error = t α/2 * (stanadard 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 = 19 d.f is 2.093
margin of error = 2.093 * 0.481
= 1.006
III.
CI = x ± margin of error
confidence interval = [ 3.57 ± 1.006 ]
= [ 2.564 , 4.576 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =3.57
standard deviation, s =2.15
sample size, n =20
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/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 = [ 3.57 ± t a/2 ( 2.15/ Sqrt ( 20) ]
= [ 3.57-(2.093 * 0.481) , 3.57+(2.093 * 0.481) ]
= [ 2.564 , 4.576 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 2.564 , 4.576 ] 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

3.
TRADITIONAL METHOD
given that,
sample mean, x =3.57
standard deviation, s =2.15
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.15/ sqrt ( 20) )
= 0.481
II.
margin of error = t α/2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 1.729
margin of error = 1.729 * 0.481
= 0.831
III.
CI = x ± margin of error
confidence interval = [ 3.57 ± 0.831 ]
= [ 2.739 , 4.401 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =3.57
standard deviation, s =2.15
sample size, n =20
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/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 = [ 3.57 ± t a/2 ( 2.15/ Sqrt ( 20) ]
= [ 3.57-(1.729 * 0.481) , 3.57+(1.729 * 0.481) ]
= [ 2.739 , 4.401 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 2.739 , 4.401 ] 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
Answer:
when sample size is increases then confidence interval decreases.