Fashion Industries randomly tests its employees throughout t

Fashion Industries randomly tests its employees throughout the year. Last year in the 510 random tests conducted, 16 employees failed the test. (Use z Distribution Table.)

Develop a 95% confidence interval for the proportion of employees that fail the test. (Round your answers to 3 decimal places.)

  Confidence interval is between                and    .

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.031372549          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.007719127          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.015129211          
lower bound = p^ - z(alpha/2) * sp =   0.016243338          
upper bound = p^ + z(alpha/2) * sp =    0.04650176          
              
Thus, the confidence interval is              
              
(   0.016243338   ,   0.04650176   ) [ANSWER]