a random sample of 100 inmates at a maximum security prisons

a random sample of 100 inmates at a maximum security prisons shows that exacty 10 of the respondents had been victims of violent crime during incarceration. esimate the proportion of victims for the population as a whole, using the 90% confidence level . (calculate the sample propportion Ps formula is X±Z(s/square root N-1) before using formula              Ps±Z(square root ^Pu(1-Pu)_/N) remeber that a porportion is equal to frequency divided by N)

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.1          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.03          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.050654391          
upper bound = p^ + z(alpha/2) * sp =    0.149345609          
              
Thus, the confidence interval is              
              
(   0.050654391   ,   0.149345609   ) [ANSWER]