Question

(A-Grade) 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 population standard deviation of time for this assembly is 3.6 minutes.

1. After observing 120 workers assembling similar devices, the manager noticed that their average time was 16.2 minutes. Construct a 92% confidence interval for the mean assembly time.

2. How many workers should be involved in this study in order to have the mean assembly time estimated up to 15 seconds with 92% confidence?

3. In a second study a sample 102 workers had standard deviation of 2.78 minutes, with 95% confidence level, construct a confidence interval for the population standard deviation.

Answer 1

1)

Given that,

Point estimate = sample mean = = 16.2

sample standard deviation = s = 3.6

sample size = n = 120

Degrees of freedom = df = n - 1 = 119

At 92% confidence level the t is ,

= 1 - 92% = 1 - 0.92 = 0.08

/ 2 = 0.08 / 2 = 0.04

t /2,df = t0.04,119 = 1.766

Margin of error = E = t/2,df * (s /n)

= 1.766* (3.6 / 119)

= 0.583

The 92% confidence interval estimate of the population mean is,

- E < < + E

16.2 - 0.583< < 16.2 + 0.583

15.617< < 16.783

( 15.617, 16.783)

2)

= 3.6 minutes = 3.6 * 60 = 216 seconds

Sample size = (Z/2 * / E)2

= (1.7507 * 216 / 15)2

= 636 (Rounded)

--------------------------------------

DEAR STUDENT,

IF YOU HAVE ANY QUERY ASK ME IN THE COMMENT BOX,I AM HERE TO HELPS YOU.PLEASE GIVE ME POSITIVE RATINGS

*****************THANK YOU***************