Question

HighTech Inc. randomly tests its employees about company policies. Last year in the 410 random tests conducted, 16 employees failed the test. a. Develop a 99% confidence interval for the proportion of applicants that fail the test. (Round your answers to 3 decimal places.) Confidence interval for the proportion mean is between and

Answer 1

$\overline{p}=16/410$

$\overline{q}=1-16/410=394/410$

Estimated standard error of propotion is given by

$\sigma _{\overline{p}}=\sqrt{\frac{\overline{p}\overline{q}}{n}}$

$\sigma _{\overline{p}}=\sqrt{\frac{(16/410)*(394/410)}{410}}=0.00968265$

We know that

z=2.58 for 99% confidence interval

Lower limit of proportion=$\overline{p}-z*\sigma _{\overline{p}}=(16/410)-2.58*0.00968265 =0.014043153$

Upper limit of proportion=$\overline{p}-z*\sigma _{\overline{p}}=(16/410)+2.58*0.00968265 =0.064005627$

Confidence interval for proportion mean is between 0.014 and 0.064