Question

A university is contemplating switching from the quarter system to the semester system. The administration conducts a survey of a random sample of 300 students and finds that 180 of them prefer to remain on the quarter system. (a) Construct a 95% confidence interval for the true proportion of all students who would prefer to remain on the quarter system. (Round your answers to two decimal places.) 55 X to 65 X (b) Does the interval you computed in part (a) provide convincing evidence that the majority of students prefer to remain on the quarter system? Explain. (Round your answer to the nearest percent, when necessary.) We can be confident based on the 95% confidence interval, of X % to 65 * %, that the majority of students prefer to remain on the quarter systems, since the interval is entirely above 50% Yes 55 46 (c) Now suppose that only 40 students had been surveyed and that 24 said they preferred the quarter system. Compute a 95% confidence interval for the true proportion that prefers to remain on the quarter system. (Round your answers to two decimal places.) 1x to 74 Does the interval provide convincing evidence that the majority of students prefer to remain on the quarter system? (Round your answer to the nearest percent, when necessary.) be confident based on the 95% confidence interval, of X % to 50 X%, that the majority of students prefer to remain on the quarter systems, since the interval contains values above and below 50% We cannot No 46 (d) Compare the sample proportions and the confidence intervals found in parts (a) and (c). Use these results to discuss the role sample size plays in helping make decisions from sample data. The larger sample size in part (a) yielded a smaller standard error, and thus a smaller confidence interval, than the smaller sample size in part (c). In general, a larger sample size produces a smaller 95% confidence interval.

Answer 1

TRADITIONAL METHOD
a.
given that,
possible chances (x)=180
sample size(n)=300
success rate ( p )= x/n = 0.6
I.
sample proportion = 0.6
standard error = Sqrt ( (0.6*0.4) /300) )
= 0.03
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.03
= 0.06
III.
CI = [ p ± margin of error ]
confidence interval = [0.6 ± 0.06]
= [ 0.54 , 0.66]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=180
sample size(n)=300
success rate ( p )= x/n = 0.6
CI = confidence interval
confidence interval = [ 0.6 ± 1.96 * Sqrt ( (0.6*0.4) /300) ) ]
= [0.6 - 1.96 * Sqrt ( (0.6*0.4) /300) , 0.6 + 1.96 * Sqrt ( (0.6*0.4) /300) ]
= [0.54 , 0.66]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.54 , 0.66] contains the true population proportion
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 proportion
b.
95% sure that the interval [ 0.54 , 0.66] = 54% and 66% that the majority of students prefer to remain on the quarter system since the interval is entirely above
50%
c.
TRADITIONAL METHOD
given that,
possible chances (x)=24
sample size(n)=40
success rate ( p )= x/n = 0.6
I.
sample proportion = 0.6
standard error = Sqrt ( (0.6*0.4) /40) )
= 0.08
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.08
= 0.15
III.
CI = [ p ± margin of error ]
confidence interval = [0.6 ± 0.15]
= [ 0.45 , 0.75]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=24
sample size(n)=40
success rate ( p )= x/n = 0.6
CI = confidence interval
confidence interval = [ 0.6 ± 1.96 * Sqrt ( (0.6*0.4) /40) ) ]
= [0.6 - 1.96 * Sqrt ( (0.6*0.4) /40) , 0.6 + 1.96 * Sqrt ( (0.6*0.4) /40) ]
= [0.45 , 0.75]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.45 , 0.75] contains the true population proportion
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 proportion
95% sure that the interval [ 0.45 , 0.75] is 45% to 75% that the majority of students prefer to remain on the quarter system since the interval is entirely below and
above 50%
d.
The larger sample size in part(a),yield a smaller standard error and thus a smaller confidence interval,than the smaller sample size in part (c),
In general a larger sample size produces a smaller 95% confidence interval.