Fashion Industries randomly tests its employees throughout t

Fashion Industries randomly tests its employees throughout the year. Last year in the 460 random tests conducted, 26 employees failed the test. (Use z Distribution Table.) link to table = http://ezto-cf-media.mheducation.com/Media/Connect_Production/bne/lind_16e/z_Distribution_Table.jpg

Develop a 90% confidence interval for the proportion of employees that fail the test. (Round your answers to 3 decimal places.)

Would it be reasonable to conclude that less than 7% of the employees are not able to pass the random drug test?

Solution

1.

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.056521739          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.010767002          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.64          
Thus,              
Margin of error = z(alpha/2)*sp =    0.017657883          
lower bound = p^ - z(alpha/2) * sp =   0.038863856          
upper bound = p^ + z(alpha/2) * sp =    0.074179622          
              
Thus, the confidence interval is              
              
(   0.038863856   ,   0.074179622   ) [ANSWER]

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2.

NO, because part of the interval is greater than 7%.