Question

In: Statistics and Probability

The Marist Poll published a report stating that 66% of adults nationally think licensed drivers should...

The Marist Poll published a report stating that 66% of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age. It was also reported that interviews were conducted on 1,018 American adults, and that the margin of error was 3% using a 95% confidence level.

Briefly explain your reasoning for the following true/false statements:

i. We are 95% confident that 63% to 69% of American adults in this sample think licensed drivers should be required to retake their road test once they turn 65.

FALSE

ii. We are 95% confident that 63% to 69% of American adults think licensed drivers should be required to retake their road test once they turn 65.

TRUE

iii. If we take many random samples of 1018 American adults, and for each sample, calculated the percentage who think licensed drivers should be required to retake their road test once they turn 65. 95% of those sample percentages will be between 63% and 69%.

TRUE

iv. The margin of error at a 99% confidence level would be higher than 3%.

TRUE

Solutions

Expert Solution

TRADITIONAL METHOD
given that,
possible chances (x)=671.88
sample size(n)=1018
success rate ( p )= x/n = 0.66
I.
sample proportion = 0.66
standard error = Sqrt ( (0.66*0.34) /1018) )
= 0.01
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 * 0.01
= 0.03 =3%
III.
CI = [ p ± margin of error ]
confidence interval = [0.66 ± 0.03]
= [ 0.63 , 0.69]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=671.88
sample size(n)=1018
success rate ( p )= x/n = 0.66
CI = confidence interval
confidence interval = [ 0.66 ± 1.96 * Sqrt ( (0.66*0.34) /1018) ) ]
= [0.66 - 1.96 * Sqrt ( (0.66*0.34) /1018) , 0.66 + 1.96 * Sqrt ( (0.66*0.34) /1018) ]
= [0.63 , 0.69]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.63 , 0.69] 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

Answer:
confidence interval is 63%,69%
TRUE
iii. If we take many random samples of 1018 American adults, and for each sample,
calculated the percentage who think licensed drivers should be required to retake their road test once they turn 65.
95% of those sample percentages will be between 63% and 69%.


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