Question

Paragraph Styles On November 1, 2013 John Adams, a customer service representative of Americo Drilling Supplies (ADS), was summoned to the Houston office of Drilling Contractors, Inc. (DCI), to inspect three boxcars of mud-treating agents that ADS had shipped to the Houston firm. DCI had filed a complaint that the 25 Pound bags of treating agents that it had just received from ADS were light-weighted. The light-weight bags were initially detected by one of DCI's receiving clerks, who noticed that the rail road side scale tickets indicated that the net weights were significantly less on all three of the boxcars than those of identical shipments received on October 25, 2013. ADS's traffic department was called to determine if lighter-weight dunnage or pallets were used on the shipments. (This might explain the lighter weights.) ADS indicated, however, that no changes had been made in the loading or palletizing procedures. Hence, DCI randomly checked 25 of the bags and discovered that the average net weight was almost 24.5 pounds. Consequently, they concluded that the sample indicated a significance short-weight. contacted, and Adams was sent to investigate the complaint and to issue to DCI. ADS, was then credit (?2?) DCI, however, was not completely satisfied with only the issuance of credit for the short shipment t. The charts followed by their mud engineers on the drilling platforms 46 144 8 4 5 6 8 9

Answer 1

(a)ans.

To find out the values for range column by using following formula:

Range = Maximum – Minimum = Largest – Smallest

Calculate the range of column; we have to find the means and standard deviations for the given 4 columns such as the average weights, smallest weights, largest weights, range weights.

Required means and standard deviations by using excel are summarized as below:

(Range column is calculated by using Excel operations.)

	Average weight	Smallest	Largest	Range
Sample size	72	72	72	72
Total	1766.8	1706.3	1830	123.7
Mean	24.53888889	23.69861111	25.41666667	1.718055556
Standard deviation	0.586840563	0.831677933	0.7218911	0.741396928

(b)ans.

(d)ans

The control limits are determined as follows:

Central line = CL = overall mean = X_bar = 24.54 and range mean = R_bar = 1.72

For sample size of n = 10, A2 = 0.31, D3 = 0.22 and D4 = 1.78

UCL_Xbar = X_bar + (A₂)(R_bar) = 24.54 + (0.31)(1.72) = 25.07

LCL_Xbar = X_bar - (A₂)(R_bar) = 24.54 - (0.31)(1.72) = 24.006

UCL_Xbar = 25.07 and LCL_Xbar = 24.006

UCL_R = (D₄)(R_bar) = (1.78)(1.72) = 3.058

LCL_Xbar = (D₃)(R_bar) = (0.22)(1.72) = 0.38

UCL_R = 3.058 and LCL_R = 0.38

(e)ans

Compute Xbarbar (Average of Bar of 72 samples) and Rebar (Average of Range of 72 samples)

Rest of the steps of statistical process control is following

Conclusion:

3 R values are outside the control limits. Therefore, based on R-chart, the process is not in control.

36 Xbar values are outside the control limits. Therefore, based on Xbar-chart, the process is not in control