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. After observing 135 workers assembling similar devices, the manager noticed that their average time was 16.0 minutes with a standard deviation of 3.7 minutes. Construct a 95% confidence interval for the mean assembly time.

what is the margin of error of the confidence interval?

new assume that the population standard deviation of assembly time is 3.7 minutes how many workers wuld the manager need to observe so that the margin of error is no more than 32 seconds with 95% confidence

Answer 1

a)

mean=   16.00
sd= sqrt(var) =   3.700  
n=   135.00
alpha=   5%  

critical value, z(a) = z(0.05) = 1.645

CI = mean +- z(a/2,n-1)*(sd/sqrt(n))      
lower   = 16 - 1.96*(3.7/sqrt(135))=   15.38
upper   = 16 + 1.96*(3.7/sqrt(135))=   16.62

b)

Margin of error, ME = z(a/2,n-1)*(sd/sqrt(n)) = 1.96*(3.7/sqrt(135)) = 0.624152783

c)

sd=   3.70    mins  
ME=   0.533   =32/60   mins
alpha=   5%      
z(a/2)=   z(0.05/2)   =NORMSINV(1-0.05/2)   1.96
          
n = z(a/2)^2*sd^2/ME^2          
n= = 1.96^2*3.7^2/0.533^2      
n = 184.89       
n = 185