Answer:
Of the last 220 applicants, 14 failed the test.
So, we have the sample proportion = [tex]\frac{14}{220}[/tex] = 0.064
At 99% confidence interval, z score is 2.576
Standard error or E = [tex]0.064+2.576\times (\frac{\sqrt{0.064\times(1-0.064)} }{\sqrt{220} } )[/tex] and [tex]0.064-2.576\times (\frac{\sqrt{0.064\times(1-0.064)} }{\sqrt{220} } )[/tex]
= [tex]0.064+0.04250[/tex] and [tex]0.064-0.04250[/tex]
= 0.0215, 0.1065
Rounding off we get 0.02,0.11
In percentage the applicants who failed lies between interval 2% and 11%.
So, yes it would be reasonable to conclude that more than 10% of the applicants are now failing the test.
