Question

A study of 100 students in a local university gave the average GPA to be 3.1 with a standard deviation of.5. The same study revealed that in this sample, 20% are out of state students.

1. The lower bound, correct up to 3 decimal places for a 92% confidence interval for the actual GPA of students in this university is

2. The lower bound, correct up to 3 decimal places of a 90% confidence interval for the true proportion of out of state students is

3. The lower bound, correct up to 3 decimal places, of a 95% confidence interval for the true GPA with a sample size of 20, a sample mean of 3.1 ad a standard deviation of 0.5 is

4. The exact number of students that need be surveyed for the true GPA to be estimated within .5 units with a 96% confidence

Answer 1

1.
TRADITIONAL METHOD
given that,
sample mean, x =3.1
standard deviation, s =0.5
sample size, n =100
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.5/ sqrt ( 100) )
= 0.05
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.08
from standard normal table, two tailed value of |t α/2| with n-1 = 99 d.f is 1.769
margin of error = 1.769 * 0.05
= 0.088
III.
CI = x ± margin of error
confidence interval = [ 3.1 ± 0.088 ]
= [ 3.012 , 3.188 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =3.1
standard deviation, s =0.5
sample size, n =100
level of significance, α = 0.08
from standard normal table, two tailed value of |t α/2| with n-1 = 99 d.f is 1.769
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 = [ 3.1 ± t a/2 ( 0.5/ Sqrt ( 100) ]
= [ 3.1-(1.769 * 0.05) , 3.1+(1.769 * 0.05) ]
= [ 3.012 , 3.188 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 92% sure that the interval [ 3.012 , 3.188 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 92% of these intervals will contains the true population mean
Answer:
lower bound interval 92% is 3.012
2.
TRADITIONAL METHOD
given that,
possible chances (x)=20
sample size(n)=100
success rate ( p )= x/n = 0.2
I.
sample proportion = 0.2
standard error = Sqrt ( (0.2*0.8) /100) )
= 0.04
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
margin of error = 1.645 * 0.04
= 0.066
III.
CI = [ p ± margin of error ]
confidence interval = [0.2 ± 0.066]
= [ 0.134 , 0.266]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possibile chances (x)=20
sample size(n)=100
success rate ( p )= x/n = 0.2
CI = confidence interval
confidence interval = [ 0.2 ± 1.645 * Sqrt ( (0.2*0.8) /100) ) ]
= [0.2 - 1.645 * Sqrt ( (0.2*0.8) /100) , 0.2 + 1.645 * Sqrt ( (0.2*0.8) /100) ]
= [0.134 , 0.266]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 90% sure that the interval [ 0.134 , 0.266] contains the true population proportion
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 proportion
Answer:
lower bound 90% confidence interval for the true proportion of out of state students is 0.134
3.
TRADITIONAL METHOD
given that,
sample mean, x =3.1
standard deviation, s =0.5
sample size, n =20
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.5/ sqrt ( 20) )
= 0.112
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 = 19 d.f is 2.093
margin of error = 2.093 * 0.112
= 0.234
III.
CI = x ± margin of error
confidence interval = [ 3.1 ± 0.234 ]
= [ 2.866 , 3.334 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =3.1
standard deviation, s =0.5
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 = [ 3.1 ± t a/2 ( 0.5/ Sqrt ( 20) ]
= [ 3.1-(2.093 * 0.112) , 3.1+(2.093 * 0.112) ]
= [ 2.866 , 3.334 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 2.866 , 3.334 ] 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
Answer:
95% sure that the interval lower bound interval is 2.866
4.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.04 is = 2.054
Sample Proportion = 0.2
ME = 0.5
n = ( 2.054 / 0.5 )^2 * 0.2*0.8
= 2.7 ~ 3          
The exact number of students that need be surveyed for the true GPA is 3