?What is the answer to this question? A sample of 200 ROM computer chips was sel

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?What is the answer to this question?

A sample of 200 ROM computer chips was selected on each of 30 consecutive days, and the number of nonconforming chips on each day was as follows: 7, 18, 25, 14, 38, 22, 9, 28, 11, 27, 31, 17, 15, 23, 15, 17, 18, 20, 15, 17, 14, 18, 14, 22, 33, 17, 9, 19, 13, 29. Construct a p chart and examine it for any out-of-control points. (Round your answers to four decimal places.) Comment on the chart. All points are between these limits, so the process appears to be in control. The value of s on the 5th day lies above the UCL, so an out-of-control signal is generated. All points are between these limits, so the process appears to be out of control. The value of s on the 5th day lies below the LCL, so an out-of-control signal is generated. Most points are between these limits, so the process appears to be in control with respect to variability.

Explanation / Answer

575

CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=575
Sample Size(n)=6000
Sample proportion (p) = x/n =0.096
Confidence Interval = [ 0.096 ±Z a/2 ( Sqrt ( 0.096*0.904) /6000)]
= [ 0.096 - 1.96* Sqrt(0) , 0.096 + 1.96* Sqrt(0) ]
= [ 0.089,0.103]

Option A graph is suitable

On the 5th day p = 38/200 = 0.19

The value of s on the 5th day lies above the UCL, so an out-of-control signal is generated.

n defects 200 7 200 18 200 25 200 14 200 38 200 22 200 9 200 28 200 11 200 27 200 31 200 17 200 15 200 23 200 15 200 17 200 18 200 20 200 15 200 17 200 14 200 18 200 14 200 22 200 33 200 17 200 9 200 19 200 13 200 29 Total 6000

575

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