Please solve this problem, including building the X-bar charts, using MS Excel.
ID: 361077 • Letter: P
Question
Please solve this problem, including building the X-bar charts, using MS Excel. Creating a separate spreadsheet tab for part (a) and for part (b), z=2 and z=3 respectively. Answer the "Why" of part (b) in the spreadsheet at the top of the page. See the sample problem (down below) for guidance on how to set this problem up.
Above is example of how to set up
Kime's bowling ball factory makes bowling balls of only. The standard deviation in the weight of a roduced at the factory is known to be 0.12 pounds. Each s, the average weight, in pounds, of nine of the bowling bowling .. s6.8 Bill adult size and weight bowling ball day for 24 days, s weigh produced that day has been assessed as follows: Average (lb) 16.3 15.9 15.8 15.5 16.3 16.2 16.0 16.1 15.9 16.2 15.9 15.9 Day 13 Average (lb) 16.3 15.9 16.3 16.2 16.1 15.9 16.2 15.9 15.9 16.0 15.5 15.8 Day 15 4 17 19 20 21 10 23 24 12 a) Establish a control chart for monitoring the average weights of the bowling balls in which the upper and lower control limits are each two standard deviations from the mean. What are the values of the control limits? three standard deviations are used in the chart, how do these val ues change? Why? Px Whole Grains LLC uses statistical process control t have at is ealth-conscious, low-fat, multigrain sandwich loaves that its health-co ualu ctahle and in-controExplanation / Answer
From the given data,
Sum of average weights for 24 days = 384
Therefore , Xbar-bar = Average of Average weights = 384/24 = 16
Given.
Sample size = n = 9 bowling balls
Standard deviation in weight of bowling ball = Sd = 0.12
Answer to question a :
Upper control Limit = Xbar-Bar + 2 x Sd/ Square root ( n)
Lower Control Limit = Xbar-bar – 2 x Sd / Square root ( n )
Therefore ,
Upper control Limit
= 16 + 2 x 0.12/ Square root ( 9)
= 16 + 0.24/3
= 16 + 0.08
= 16.08
Lower Control Limit
= 16 - 2 x 0.12/ Square root ( 9)
= 16 - 0.24/3
= 16 - 0.08
= 15.92
UPPER CONTROL LIMIT = 16.08
LOWER CONTROL LIMIT = 15.92
Answer to question b:
If criteria of 3 standard deviation I/O 2 standard deviation is used, the spread will become more and hence value of UCL will increase and value of LCL will decrease
The revised control limits as follows :
Upper control Limit = Xbar-Bar + 3 x Sd/ Square root ( n)
Lower Control Limit = Xbar-bar – 3 x Sd / Square root ( n )
Therefore ,
Upper control Limit
= 16 + 3 x 0.12/ Square root ( 9)
= 16 + 0.36/3
= 16 + 0.12
= 16.12
Lower Control Limit
= 16 - 3 x 0.12/ Square root ( 9)
= 16 - 0.36/3
= 16 - 0.12
= 15.88
UPPER CONTROL LIMIT = 16.12
LOWER CONTROL LIMIT = 15.88
UPPER CONTROL LIMIT = 16.08
LOWER CONTROL LIMIT = 15.92