Chrome-Do Homework -Emity Henthorn Secure https://www.mathod.com/Student/PlayerH

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Chrome-Do Homework -Emity Henthorn Secure https://www.mathod.com/Student/PlayerHomework.aspx?ho MANG 471-004 Fall18 meworkld-500265841&questionld-1&flushed true&cld-51949368cen Emily Henthorn & I 9/10/18 7:15 PM Homework: Problem Set #3 Score: 0 of 6 pts Problem 15 2 of 4 (1 complete) HW Score: 17.14%, 4.29 of 25 pts EQuestion Help Janice Sanders, CEO of Pine Crest Medical Clinic, is concemed over the number of times patients must wait more than 30 minutes beyond their scheduled appointments. She asked her assistant to take random samples of 60 patients to see how many in each sample had to wait more than 30 minutes. Each instance is considered a defect in the clinic process. The table below contains the data for 15 samples. Number of Defects Number of Defects Number of Defects Sample Sample Sample 12 13 14 15 10 a. Assuming Janice Sanders is willing to use three-sigma control limits, construct the upper and lower control limits. equals and the LCLp equals(Enter your responses rounded to four decimal places. Iif your answer for LCLp is negative, enter this value as 0) Enter your answer in the edit fields and then click Check Answer. remaining Clear All Check Answer

Explanation / Answer

p bar = sum of np/sum of n = 75/900 = 0.0833 or 8.33%

n bar = sum of n/k = 900/15 = 60

UCLp = p bar + 3*([p bar*(1-p bar)/n bar])^0.5

= 0.0833 + 3*([0.0833*(1-0.0833)/60])^0.5

= 0.0833 + 0.1070

= 0.1904

LCLp = p bar - 3*([p bar*(1-p bar)/n bar])^0.5

= 0.0833 - 0.1070

= -0.0237. This will be entered as 0.

UCLp = 0.1904 and LCLp = 0

Sample Patients sampled (n) No. defective (np) Fraction defective (p)                       1 60                         7.00                0.1167                       2 60                         4.00                0.0667                       3 60                         7.00                0.1167                       4 60                         1.00                0.0167                       5 60                         3.00                0.0500                       6 60                         6.00                0.1000                       7 60                         7.00                0.1167                       8 60                         6.00                0.1000                       9 60                         2.00                0.0333                    10 60                         7.00                0.1167                    11 60                         6.00                0.1000                    12 60                         4.00                0.0667                    13 60                         5.00                0.0833                    14 60                         7.00                0.1167                    15 60                         3.00                0.0500 Total                 900.00                      75.00

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