3. Suppose a large telephone manufacturer that entered the post-regulation marke
ID: 3309742 • Letter: 3
Question
3. Suppose a large telephone manufacturer that entered the post-regulation market quickly has an initial problem with excessive costumer complains and consequent returns of the phones for repair or replacement. The manufacturer wants to estimate the magnitude of the problem in order to design a quality control problem program. A survey of 170 randomly selected telephones shows that 119 phones are returned for repair or replacement. (a) What is the point estimation of the proportion of phones that are returned for repair or replacement. (b) Construct a 95% confidence interval for the proportion of phones that are returned for repair or replacement. (c) Without constructing the 98% confidence interval for the rate of phones that are returned for repair or replacement, would you expect the interval to be longer or shorter compared to the 95% confidence interval? Why? d) What sample size of phones would you recommend the telephone manufacturer sam- ples and checks in order to estimate the fraction defective, p, to within 0.02 with 95% confidence.Explanation / Answer
TRADITIONAL METHOD
given that,
possibile chances (x)=119
sample size(n)=170
success rate ( p )= x/n = 0.7
a.
point of estimate= sample proportion = 0.7
standard error = Sqrt ( (0.7*0.3) /170) )
= 0.0351
b.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
margin of error = 2.576 * 0.0351
= 0.0905
III.
CI = [ p ± margin of error ]
confidence interval = [0.7 ± 0.0905]
= [ 0.6095 , 0.7905]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possibile chances (x)=119
sample size(n)=170
success rate ( p )= x/n = 0.7
CI = confidence interval
confidence interval = [ 0.7 ± 2.576 * Sqrt ( (0.7*0.3) /170) ) ]
= [0.7 - 2.576 * Sqrt ( (0.7*0.3) /170) , 0.7 + 2.576 * Sqrt ( (0.7*0.3) /170) ]
= [0.6095 , 0.7905]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 99% sure that the interval [ 0.6095 , 0.7905] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion
c.
it is wider . since we get larger t value for 98% confidence
d.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.7
ME = 0.02
n = ( 1.96 / 0.02 )^2 * 0.7*0.3
= 2016.84 ~ 2017
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.