A retirement confidence survey of 1153 workers and retirees

A retirement confidence survey of 1153 workers and retirees in the United States 25 years of age and older conducted by Employee Benefit Research Institute in January 2010 found that 496 had less than $10,000 in savings. a.) Obtain a point estimate for the population proportion of workers and retirees in the United States 25 years of age and older who have less than $10,000 in savings. b.) Verify that the requirements for constructing a confidence interval about p are satisfied. c.) Construct a 95% interval for the population proportion of workers and retirees in the United States 25 years and older who have less than $10,000 in savings. d.) Interpret the interval. e.) What is the margin of error for this interval?

Solution

a)
People 25 years of age and older who have less than $10,000(x)=496
Sample Size(n)=1153
Sample proportion = x/n =0.43
b)
n*p>5, 1000*0.43> 5 => 430>5
n*(1-p)>5, 1000*0.57> 5 => 430>5
Can Use Normal Approximation                  
                  
c)
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval

Confidence Interval = [ 0.43 ±Z a/2 ( Sqrt ( 0.43*0.57) /1153)]
= [ 0.43 - 1.96* Sqrt(0) , 0.43 + 1.96* Sqrt(0) ]
= [ 0.401,0.459]

e)
Margin of Error = Z a/2 Sqrt(p*(1-p)/n))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=496
Sample Size(n)=1153
Sample proportion =0.43
Margin of Error = Z a/2 * ( Sqrt ( (0.43*0.57) /1153) )
= 1.96* Sqrt(0)
=0.029