A campus newspaper reported that 20 of all students live in

A campus newspaper reported that 20% of all students live in the dorms. A random sample of 100 students is selected for a particular study. (round all calculations to four decimal places where necessary).

a. At 95% confidence level, what is the margin of error?

b. develop a 95% confidence interval estimate for the proportion of all students that live in the dorms

Solution

a)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.2          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.04          
              
Now, for the critical z,              

alpha/2 =   0.025          

Thus, z(alpha/2) =    1.959963985          

Thus,              

Margin of error = z(alpha/2)*sp =    0.078398559 [ANSWER]

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b)
          
lower bound = p^ - z(alpha/2) * sp =   0.121601441          
upper bound = p^ + z(alpha/2) * sp =    0.278398559          
              
Thus, the confidence interval is              
              
(   0.121601441   ,   0.278398559   ) [ANSWER]