Question

1. Suppose you constructed a 90% confidence interval for the true average GPA of California community college students. If all of the sample data were to be held constant, how would this confidence interval compare to an 80% confidence interval based on the same data?

2. A random sample of 100 students was taken to determine interest in switching to a quarter system instead of semesters. Out of the 100 students, 73 stated that they were interested in switching to a quarter system. Construct a 90% confidence interval for the true proportion of students in favor of switching to a quarter system. Make sure to state whether or not the assumptions were met.

Answer 1

1.

a.
TRADITIONAL METHOD
given that,
possible chances (x)=73
sample size(n)=100
success rate ( p )= x/n = 0.73
I.
sample proportion = 0.73
standard error = Sqrt ( (0.73*0.27) /100) )
= 0.044
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 * 0.044
= 0.073
III.
CI = [ p ± margin of error ]
confidence interval = [0.73 ± 0.073]
= [ 0.657 , 0.803]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=73
sample size(n)=100
success rate ( p )= x/n = 0.73
CI = confidence interval
confidence interval = [ 0.73 ± 1.645 * Sqrt ( (0.73*0.27) /100) ) ]
= [0.73 - 1.645 * Sqrt ( (0.73*0.27) /100) , 0.73 + 1.645 * Sqrt ( (0.73*0.27) /100) ]
= [0.657 , 0.803]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 90% sure that the interval [ 0.657 , 0.803] 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
b.
TRADITIONAL METHOD
given that,
possible chances (x)=73
sample size(n)=100
success rate ( p )= x/n = 0.73
I.
sample proportion = 0.73
standard error = Sqrt ( (0.73*0.27) /100) )
= 0.044
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.2
from standard normal table, two tailed z α/2 =1.282
margin of error = 1.282 * 0.044
= 0.057
III.
CI = [ p ± margin of error ]
confidence interval = [0.73 ± 0.057]
= [ 0.673 , 0.787]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=73
sample size(n)=100
success rate ( p )= x/n = 0.73
CI = confidence interval
confidence interval = [ 0.73 ± 1.282 * Sqrt ( (0.73*0.27) /100) ) ]
= [0.73 - 1.282 * Sqrt ( (0.73*0.27) /100) , 0.73 + 1.282 * Sqrt ( (0.73*0.27) /100) ]
= [0.673 , 0.787]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 80% sure that the interval [ 0.673 , 0.787] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 80% of these intervals will contains the true population proportion
Answer:
a.
90% sure that the interval [ 0.657 , 0.803]
b.
80% sure that the interval [ 0.673 , 0.787]
compare above both.it should be narrower.