Question

Jamison Kovach Supply Company manufactures paper clips and other office products. Although​inexpensive,paper clips have provided the firm with a high margin of profitability. Sample size is 50 Results are given for the last 10 samples.

Sample   1   2   3   4   5   6   7   8   9   10
Defectives   5   7   2   6   6   5   4   5   2   11

Establish the control limits to include 99.73% of the random variation in defectives

UCLp= _____(enteryour response as a number between 0 and​1,rounded to three decimal​places).

LCLp= ______ (enteryour response as a number between 0 and​1,rounded to three decimal​places).

Has the process been in control?

If the sample size were 25 instead how would your limits and conclusions change?

UCLp= _____ (enteryour response as a number between 0 and​1,rounded to three decimal​places).

LCLp=______ (enteryour response as a number between 0 and​1,rounded to three decimal​places).

Answer 1

Z VALUE FOR 99.73% CONFIDENCE = 3
SAMPLE SIZE = 50
NUMBER OF SAMPLES TAKEN = 10

P-BAR = TOTAL NUMBER OF DEFECTS / (SAMPLE SIZE * NUMBER OF SAMPLES = 53 / (50 * 10) = 0.106
STDEV = SQRT((PBAR * (1 - PBAR)) / SAMPLE SIZE = SQRT((0.106 * (1 - 0.106)) / 50 = 0.044

UCL = PBAR + (Z * STDEV) = 0.106 + (3 * 0.044) = 0.238
LCL = PBAR - (Z * STDEV) = 0.106 - (3 * 0.044) = -0.026, SINCE LCL IS NEGATIVE, LCL = 0

FOR A SAMPLE SIZE OF 25

P-BAR = TOTAL NUMBER OF DEFECTS / (SAMPLE SIZE * NUMBER OF SAMPLES = 53 / (25 * 10) = 0.212
STDEV = SQRT((PBAR * (1 - PBAR)) / SAMPLE SIZE = SQRT((0.212 * (1 - 0.212)) / 25 = 0.082

UCL = PBAR + (Z * STDEV) = 0.212 + (3 * 0.082) = 0.458
LCL = PBAR - (Z * STDEV) = 0.212 - (3 * 0.082) = -0.034, SINCE LCL IS NEGATIVE, LCL = 0

**DEPENDING ON HOW WE ROUND OFF THE INTERMEDIATE CALCULATIONS, THE FINAL ANSWER CAN BE DIFFERENT. IN THAT CASE, JUST LEAVE A COMMENT AND I WILL TRY AND HELP AS SOON AS POSSIBLE.