The National Sleep Foundation sponsors an annual poll. In a March 2011 press rel

Question

The National Sleep Foundation sponsors an annual poll. In a March 2011 press release they stated 63% of Americans say their sleep needs are not being met during the week The following is an excerpt from the Poll Methodology and Definitions section of the article. The 2011 Sleep in America annual poll was conducted for the National Sleep Foundation by WB&A; Market Research, using a random sample of 1,508 adults between the ages of 13-64. The margin of error is 2.5 percentage points at the 95% confidence level." Which of the following statements is correct? /' We are confident that 95% of the 1,508 adults in the sample say their sleep needs are not being met during the week. We are 95% confident that between 60.5% and 655% of Americans say their sleep needs are not being met during the week O We are 95% confident that 63% of Americans say their sleep needs are not being met during the week We are 95% confident that between 63% and 655% of Americans say their sleep needs are not being met during the week

Explanation / Answer

TRADITIONAL METHOD
given that,
possibile chances (x)=950.04
sample size(n)=1508
success rate ( p )= x/n = 0.63
I.
sample proportion = 0.63
standard error = Sqrt ( (0.63*0.37) /1508) )
= 0.012
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.012
= 0.024
III.
CI = [ p ± margin of error ]
confidence interval = [0.63 ± 0.024]
= [ 0.606 , 0.654]
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DIRECT METHOD
given that,
possibile chances (x)=950.04
sample size(n)=1508
success rate ( p )= x/n = 0.63
CI = confidence interval
confidence interval = [ 0.63 ± 1.96 * Sqrt ( (0.63*0.37) /1508) ) ]
= [0.63 - 1.96 * Sqrt ( (0.63*0.37) /1508) , 0.63 + 1.96 * Sqrt ( (0.63*0.37) /1508) ]
= [0.606 , 0.654]
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interpretations:
1. We are 95% sure that the interval [ 0.606 , 0.654] 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 = [0.63 ± 0.024]
= [ 0.606 , 0.654] we are 95% confidence between 60.5% to 65.5% of americans say their sleep are not being met during the week

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