Question

A recent survey on food labeling found that 47% of a random sample of 1,017 American adults report looking for information on food labels about whether the food they are purchasing is organic. Use this survey result and the table below to give 70%, 80%, 90% and 99% confidence intervals for the proportion of all adults who feel this way. (Round your answers to three decimal places.) Critical values of the Normal distributions Confidence levelc Critical valuez' Confidence levelc Critical valuez 50% 0.67 1.64 60% 034 1.96 70 1.04 2.58 1.23 What do your results show about the effect of changing the confidence level? As the consente level increases, the width of the interval Select

Answer 1

i.
TRADITIONAL METHOD
given that,
possible chances (x)=477.99
sample size(n)=1017
success rate ( p )= x/n = 0.47
I.
sample proportion = 0.47
standard error = Sqrt ( (0.47*0.53) /1017) )
= 0.016
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.3
from standard normal table, two tailed z α/2 =1.036
margin of error = 1.036 * 0.016
= 0.016
III.
CI = [ p ± margin of error ]
confidence interval = [0.47 ± 0.016]
= [ 0.454 , 0.486]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=477.99
sample size(n)=1017
success rate ( p )= x/n = 0.47
CI = confidence interval
confidence interval = [ 0.47 ± 1.036 * Sqrt ( (0.47*0.53) /1017) ) ]
= [0.47 - 1.036 * Sqrt ( (0.47*0.53) /1017) , 0.47 + 1.036 * Sqrt ( (0.47*0.53) /1017) ]
= [0.454 , 0.486]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 70% sure that the interval [ 0.454 , 0.486] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 70% of these intervals will contains the true population proportion
ii.
TRADITIONAL METHOD
given that,
possible chances (x)=477.99
sample size(n)=1017
success rate ( p )= x/n = 0.47
I.
sample proportion = 0.47
standard error = Sqrt ( (0.47*0.53) /1017) )
= 0.016
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 * 0.016
= 0.02
III.
CI = [ p ± margin of error ]
confidence interval = [0.47 ± 0.02]
= [ 0.45 , 0.49]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=477.99
sample size(n)=1017
success rate ( p )= x/n = 0.47
CI = confidence interval
confidence interval = [ 0.47 ± 1.282 * Sqrt ( (0.47*0.53) /1017) ) ]
= [0.47 - 1.282 * Sqrt ( (0.47*0.53) /1017) , 0.47 + 1.282 * Sqrt ( (0.47*0.53) /1017) ]
= [0.45 , 0.49]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 80% sure that the interval [ 0.45 , 0.49] 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
iii.
TRADITIONAL METHOD
given that,
possible chances (x)=477.99
sample size(n)=1017
success rate ( p )= x/n = 0.47
I.
sample proportion = 0.47
standard error = Sqrt ( (0.47*0.53) /1017) )
= 0.016
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 * 0.016
= 0.026
III.
CI = [ p ± margin of error ]
confidence interval = [0.47 ± 0.026]
= [ 0.444 , 0.496]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=477.99
sample size(n)=1017
success rate ( p )= x/n = 0.47
CI = confidence interval
confidence interval = [ 0.47 ± 1.645 * Sqrt ( (0.47*0.53) /1017) ) ]
= [0.47 - 1.645 * Sqrt ( (0.47*0.53) /1017) , 0.47 + 1.645 * Sqrt ( (0.47*0.53) /1017) ]
= [0.444 , 0.496]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 90% sure that the interval [ 0.444 , 0.496] 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
iv.
TRADITIONAL METHOD
given that,
possible chances (x)=477.99
sample size(n)=1017
success rate ( p )= x/n = 0.47
I.
sample proportion = 0.47
standard error = Sqrt ( (0.47*0.53) /1017) )
= 0.016
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
margin of error = 2.576 * 0.016
= 0.04
III.
CI = [ p ± margin of error ]
confidence interval = [0.47 ± 0.04]
= [ 0.43 , 0.51]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=477.99
sample size(n)=1017
success rate ( p )= x/n = 0.47
CI = confidence interval
confidence interval = [ 0.47 ± 2.576 * Sqrt ( (0.47*0.53) /1017) ) ]
= [0.47 - 2.576 * Sqrt ( (0.47*0.53) /1017) , 0.47 + 2.576 * Sqrt ( (0.47*0.53) /1017) ]
= [0.43 , 0.51]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 99% sure that the interval [ 0.43 , 0.51] 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

The width of the confidence interval decreases as the sample size increases. The width increases as the standard deviation increases.
The width increases as the confidence level increases. The width increases as the significance level decreases