The National Center for Education Statistics reported that 4

The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses. Assume that a sample of 450 college students was used in the study. a. Provide a 95% confidence interval for the population proportion of college students who work to pay for tuition and living expenses (to 4 decimals).

Provide a 99% confidence interval for the population proportion of college students who work to pay for tuition and living expenses (to 4 decimals).

Solution

A)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.47          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.023527761          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.046113565          
lower bound = p^ - z(alpha/2) * sp =   0.423886435          
upper bound = p^ + z(alpha/2) * sp =    0.516113565          
              
Thus, the confidence interval is              
              
(   0.423886435   ,   0.516113565   ) [ANSWER]

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b)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.47          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.023527761          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
Margin of error = z(alpha/2)*sp =    0.060603497          
lower bound = p^ - z(alpha/2) * sp =   0.409396503          
upper bound = p^ + z(alpha/2) * sp =    0.530603497          
              
Thus, the confidence interval is              
              
(   0.409396503   ,   0.530603497   ) [ANSWER]