Question

PLEASE HELP

Use the following to answer questions 20-25: A company makes plastic cups. Four samples of 15 cups were taken from an ongoing process to establish ap chart for control. The samples and the number of defectives in each are shown in the following table: Sample n Number of Defectives 1 15 2 2 15 o 3 15 3 15 5

Answer 1

Given,

Sample size , n = 15

Fraction defective, pbar = Sum of number of defectives / (n x number of samples) = 2+0+3+5 / 4 x 15 = 0.167

Now,

Qn 22 -

Standard Deviation, S_p = $\sqrt{\frac{pbar(1-pbar)}{n}}$

Sp = $\sqrt{\frac{0.167*0.833}{15}}$

Sp = 0.0963

Qn 23 -

At 3% Alpha risk,

p = [(1-0.03)/2] + 0.5 = 0.985

From z table,

At p = 0.985,

z = 2.17

Qn 24 -

UCL = Pbar + zS_p = 0.167 + 2.17 x 0.0963 = 0.3760

LCL = Pbar - zS_p = 0.167 - 2.17 x 0.0963 = -0.0419 = 0 (cannot be taken as negative)

Hence

LCL, UCL are (0,0.3760)

Qn 25 -

Sample 1, p = 2/15 = 0.133

Sample 2, p = 0/15 = 0

Sample 3, p = 3/15 = 0.2

Sample 4, p = 5/15 = 0.333

Hence,

For a UCL of 0.30 and LCL of o,

Sample 4 is the only sample which lies above UCL