Answer:
1) 515 hours ( D )
2) 495 hours ( D )
3) 10 hours ( B )
4) 0.0456 ( D )
5) sample 3 ( C )
Explanation:
Standard deviation = 20 hours
Mean = 500 hours
<u>1) Determine the sample mean service life for sample 2 </u>
Sample mean sample service life of sample 2 = Average of service life of sample 2
= ( 525 + 515 + 505 + 515 ) / 4
= 2060 / 4 = 515
<u>2) Calculate the mean of sampling distribution of sample means</u>
∑ mean service life samples / 3 ----- ( 1 )
sample 1 = (495 + 500 + 505 + 500 ) / 4 = 500
sample 2 = ( 525 + 515 + 505 + 515 ) / 4 = 515
sample 3 = (470 + 480 + 460 + 470) = 470
back to equation 1
= ( 500 + 515 + 470 ) / 3 = 495
<u>3) Determine STD of sampling distribution of sample means </u>
Std = 20
n = 4
∴ std of sampling distribution = 20 / √4 = 10 hours
<u>4) calculating the risk alpha </u>
upper control ( UCL ) = 520 , lower control ( LCL ) = 480
mean = 500
Z value for upper control = ( 520 - 500) / 10 = 2
Z value for lower control = ( 480 - 500 ) / 10 = -2
confidence level of -2 and 2 ( using z-table ) = 0.9544
∴ risk alpha = 1 - 0.9544 = 0.0456
5) Service life appears to be out of control in ; sample 3
UCL = 520 , LCL = 480
Because mean value of sample 3 = 470 lies outside the lower control limit