Question

In: Statistics and Probability

A smart phone manufacturer is interested in constructing a 95% confidence interval for the proportion of...

A smart phone manufacturer is interested in constructing a 95% confidence interval for the proportion of smart phones that break before the warranty expires. 92 of the 1695 randomly selected smart phones broke before the warranty expired. Round answers to 4 decimal places where possible.

a. With 95% confidence the proportion of all smart phones that break before the warranty expires is between  and .

b. If many groups of 1695 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. About  percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about  percent will not contain the true population proportion.

Solutions

Expert Solution

TRADITIONAL METHOD
given that,
possible chances (x)=92
sample size(n)=1695
success rate ( p )= x/n = 0.0543
I.
sample proportion = 0.0543
standard error = Sqrt ( (0.0543*0.9457) /1695) )
= 0.0055
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 * 0.0055
= 0.0108
III.
CI = [ p ± margin of error ]
confidence interval = [0.0543 ± 0.0108]
= [ 0.0435 , 0.0651]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=92
sample size(n)=1695
success rate ( p )= x/n = 0.0543
CI = confidence interval
confidence interval = [ 0.0543 ± 1.96 * Sqrt ( (0.0543*0.9457) /1695) ) ]
= [0.0543 - 1.96 * Sqrt ( (0.0543*0.9457) /1695) , 0.0543 + 1.96 * Sqrt ( (0.0543*0.9457) /1695) ]
= [0.0435 , 0.0651]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.0435 , 0.0651] 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.
If many groups of 1695 randomly selected smart phones are selected, then a different confidence interval would be produced for each group.
proportion is 0.0543
About percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires


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