Answer to Question #278567 in Statistics and Probability for Chaston

Question

The operations manager of a large production plant would like to estimate the

amount of time a worker takes to assemble a new electronic component. After

observing 120 of the workers assembling similar devices, the manager noticed

that the mean time was 16.2 minutes. Assuming that the population standard

deviation is 3.6 minutes;

c. 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 $n=120,\bar{x}=16.2\ min, \sigma=3.6\ min,$

$E=15\ sec=0.25\ min$

$$CI=(\bar{x}-E, \bar{x}+E)$$

$$=(16.2-0.25,16.2+0.25)$$

$$=(15.95, 16.45)$$

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

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

$$z_c=\dfrac{0.25(\sqrt{120})}{3.6}$$

$$z_c\approx0.7607$$

$$P(Z>0.7607)=1-P(Z\leq0.7607)$$

$$\approx1-0.77658\approx0.2234$$

$$2(0.2234)=0.4468$$

$$1-0.4468=0.5532$$

$55.31$% confidence interval is $(15.95, 16.45).$

Therefore, based on the data provided, the 55.32 % confidence interval for the population mean is $15.95 < \mu < 16.45,$ which indicates that we are 

55.32% confident that the true population mean $\mu$ is contained by the interval $(15.95, 16.45).$