Answer:
Do the data from this shipment indicate statistical control: No
Step-by-step explanation:
Calculating the mean of the sample, we have;
Mean (x-bar) = sum of individual sample/number of sample
= (0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)/10
= 0.044/10
= 0.0044
Calculating the lower control limit (LCL) using the formula;
LCL= (x-bar) - 3*√(x-bar(1-x-bar))/n
= 0.0044 - 3*√(0.0044(1-0.0044))
= 0.0044- (3*0.0042)
= 0.0044 - 0.01256
= -0.00816 ∠ 0
Calculating the upper control limit (UCL) using the formula;
UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n
= 0.0044 + 3*√(0.0044(1-0.0044))
= 0.0044+ (3*0.0042)
= 0.0044 + 0.01256
=0.01696∠ 0
Do the data from this shipment indicate statistical control: No
Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that the data from this shipment do not indicate statistical control.