Homework Soursethe operations manager of a large production plant would lik
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 is 3.6 minutes and is normally distributed
a) 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.
b) Construct a 92% confidence interval if instead of observing 120 workers assembling similar devices, rather the manager observes 25 workers and notice their average time was 16.2 minutes with a standard deviation of 4 minutes
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 is 3.6 minutes and is normally distributed
a) 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.
b) Construct a 92% confidence interval if instead of observing 120 workers assembling similar devices, rather the manager observes 25 workers and notice their average time was 16.2 minutes with a standard deviation of 4 minutes
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 is 3.6 minutes and is normally distributed
a) 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.
b) Construct a 92% confidence interval if instead of observing 120 workers assembling similar devices, rather the manager observes 25 workers and notice their average time was 16.2 minutes with a standard deviation of 4 minutes
Solution
a)
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=16.2
Standard deviation( sd )=3.6
Sample Size(n)=120
Confidence Interval = [ 16.2 ± Z a/2 ( 3.6/ Sqrt ( 120) ) ]
= [ 16.2 - 1.75 * (0.329) , 16.2 + 1.75 * (0.329) ]
= [ 15.625,16.775 ]
b)
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=16.2
Standard deviation( sd )=4
Sample Size(n)=25
Confidence Interval = [ 16.2 ± t a/2 ( 4/ Sqrt ( 25) ) ]
= [ 16.2 - 2.492 * (0.8) , 16.2 + 2.492 * (0.8) ]
= [ 14.206,18.194 ]
