Answer to Question #235513 in Statistics and Probability for blossomqt

Question

The operations manager of a large production plant would like to estimate the mean amount of time a worker takes to assemble a new electronic component. Assume that the standard deviation of this assembly time is 3.6 minutes.

a. After observing 120 workers assembling similar devices, the manager noticed that their average time was 16.2 minutes. Construct a 90% confidence interval for the mean assembly time.

b. How many workers should be involved in this study in order to have the mean assembly time estimated up to + 15 seconds with 90% confidence?

Answer 1

a. Let $X=$ the mean assembly time: $X\sim N(\mu, \sigma^2/n)$

Given $\mu=\bar{X}=16.2\ min, \sigma=3\ min, n-120$

The critical value for $\alpha=0.1$ is $z_c=z_{1-\alpha/2}=1.6449.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{X}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

$$=(16.2-1.6449\times\dfrac{3.6}{\sqrt{120}},16.2+1.6449\times\dfrac{3.6}{\sqrt{120}})$$

$$=(15.65943,16.74057)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $15.65943<\mu<16.74057,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(15.65943,16.74057).$

b.

$$SE=z_c\times\dfrac{\sigma}{\sqrt{n}}\leq0.25$$

$$n\geq(\dfrac{z_c\times\sigma}{0.25})^2$$

$$n\geq(\dfrac{1.6449\times3.6}{0.25})^2$$

$$n\geq562$$

562 workers should be involved in this study in order to have the mean assembly time estimated up to + 15 seconds with 90% confidence.