Question

Twenty samples of 100 items each were inspected when a process was considered to be operating satisfactorily. In the 20 samples, a total of 130 items were found to be defective.

(a)

What is an estimate of the proportion defective when the process is in control? (Round your answer to four decimal places.)

(b)

What is the standard error of the proportion if samples of size 100 will be used for statistical process control? (Round your answer to four decimal places.)

(c)

Compute the upper and lower control limits for the control chart. (Round your answers to four decimal places.)

UCL = LCL =

Answer 1

Giveni Total no. g. Samplon = 130 Site og each Sarople (m) = 100 No. g. Samples that are 2-2 Considered y = 20 (a) To estimate the propostion defective : jol., Here we need to estimate the proportion 130 ♡ = 720) (100) = 0.065 .. estimated proportion (f) = 0.065 1) Standard error one proportion standard need to find error of propostion i.e., we no op = ( Å (1-P) 0.065 (1-0.065) 100

0.065 (0.935) 100 + 0.0247 :: standard foror og proportion, (se) y = 0.0247 © To compute upper and lower limits : we know that formules to Caludate upper and tower limit is given by, uppea control limit (UCL) = Å + 30p = 0.065+3(0.024a) = 0.065 +0.0741 = 0.1391 . upper con tool limit (UCL) = 0.1391 lower control limit (LcL) = Å -36 +0.065 – 3 (0.0247) LOL -0.0091 = 0.065 - 0.0741 -0.0091