Question

It has been reported that 20.4% of incoming freshmen indicate that they will major in business or a related field. A random sample of 346 incoming college freshmen were asked their preference, and 100 replied that they were considering business as a major.

Estimate the true proportion of freshman business majors with 91% confidence.Round your answers to one decimal place.

The interval does or does not contain what _______ percent?

Answer 1

TRADITIONAL METHOD
given that,
possible chances (x)=100
sample size(n)=346
success rate ( p )= x/n = 0.289
I.
sample proportion = 0.289
standard error = Sqrt ( (0.289*0.711) /346) )
= 0.024
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.09
from standard normal table, two tailed z α/2 =1.695
margin of error = 1.695 * 0.024
= 0.041
III.
CI = [ p ± margin of error ]
confidence interval = [0.289 ± 0.041]
= [ 0.248 , 0.33]
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DIRECT METHOD
given that,
possible chances (x)=100
sample size(n)=346
success rate ( p )= x/n = 0.289
CI = confidence interval
confidence interval = [ 0.289 ± 1.695 * Sqrt ( (0.289*0.711) /346) ) ]
= [0.289 - 1.695 * Sqrt ( (0.289*0.711) /346) , 0.289 + 1.695 * Sqrt ( (0.289*0.711) /346) ]
= [0.248 , 0.33]
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interpretations:
1. We are 91% sure that the interval [ 0.248 , 0.33] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 91% of these intervals will contains the true population proportion

Answer:
a.
Estimate the true proportion of freshman business majors with 91% confidence =0.04
91% sure that the interval [ 0.248 , 0.33] contains the true population proportion
b.
The interval does not contain 20.4% percent.it is lack of 4.4%