Question

Calculate the confidence level that would be required to have the mean

assembly time estimated up to ±15 seconds assuming the sample size

is 120 minutes.

Answer 1

Given $n=120, SE=15\ s=0.25\ min$

$$SE=z_c\times\dfrac{\sigma}{\sqrt{n}}$$

$$z_c=\dfrac{SE(\sqrt{n})}{\sigma}$$

$$z_c=\dfrac{0.25(\sqrt{120})}{\sigma}$$

Let $\sigma=2\ min$

$$z_c=\dfrac{0.25(\sqrt{120})}{2}\approx1.3693$$

$$P(Z>1.3693)=0.08545$$

$$2(0.08545)=0.1709$$

$$1-0.1709=0.8291$$

$82.91$% confidence interval