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Fed [463]
3 years ago
5

The chair of the operations management department at Quality University wants to construct a p-chart for determining whether the

four faculty teaching the basic P/OM course are in control with regard to the number of students who fail the course. Accordingly, he sampled 100 final grades from last year for each instructor, with the following results Instructor Number of Failures Prof. A 13 Prof. B 0 Prof. C 11 Prof. D 16 What are the .95 (5 percent risk of Type I error) upper and lower control limits for the p-chart
Mathematics
1 answer:
sasho [114]3 years ago
4 0

Answer:

Prof B and Prof D

Step-by-step explanation:

Step-by-step explanation:

From the question  we are told that

Sample size 100  

Instructor    Number of Failures  

Prof. A         13

Prof. B         0

Prof. C         11

Prof. D         16

Confidence level= 0.95

From Z table

Z=1.96

Generally proportion for failure is mathematically given as

  Proportion\ of\ Failure\ P=\frac{Number\ of\ Failure\ N}{Sample\ size\ S}

  P=\frac{N}{S}

For Prof A

    P_A=\frac{N_A}{S}

    P_A=\frac{13}{100}

    P_A=0.13

For Prof B

    P_B=\frac{N_B}{S}

   P_B=\frac{0}{100}

    P_B=0

For Prof C

    P_C=\frac{N_C}{S}

    P_C=\frac{11}{100}

   P_C=0.11

For Prof D

     P_D=\frac{N_D}{S}

   P_D=\frac{16}{100}

   P_D=0.16

Generally the average proportional failure is mathematically given as

    P_a_v_g=\frac{0.13+0+0.11+0.16}{4}

   P_a_v_g=0.10

Therefore  having this we calculate for the control limits

Generally the upper control limit UCl is mathematically given as

Upper control limits:

  UCL =P_a_v_g +Z\sqrt{\frac{P_a_v_g(1-P_a_v_g) }{S} }

  UCL =0.1 +1.96\sqrt{\frac{0.1(1-0.1) }{100}}

  UCL =0.1 +1.96*0.03

  UCL =0.1 +0.0588

  UCL =0.1588

Generally the upper control limit UCl is mathematically given as

Lower limit control limits:

 UCL =P_a_v_g _Z\sqrt{\frac{P_a_v_g(1-P_a_v_g) }{S} }

 UCL =0.1 -1.96\sqrt{\frac{0.1(1-0.1) }{100}}

  UCL =0.1 -1.96*0.03

  UCL =0.1 -0.0588

  UCL =0.0412

Therefore the Range is 0.0412 to 0.1588

Given the range 0.0412 to 0.1588 Prof B and Prof D are outside the Range making them the correct options

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Q + d = 100 so q = 100 - d
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substitute q = 100 - d into 0.25q + 0.10d = 19

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