You are the newly appointed assistant administrator at a local hospital, and you
ID: 443035 • Letter: Y
Question
You are the newly appointed assistant administrator at a local hospital, and your first project is to investigate the quality of the patient meals put out by the food-service department. You conducted a 10-day survey by submitting a simple questionnaire to the 400 patients with each meal, asking that they simply check off that the meal was either satisfactory or unsatisfactory. For simplicity in this problem, assume that the response was 1,000 returned questionnaires from the 1,200 meals each day. The results are as follows:
Determine the , Sp, UCL and LCL based on the questionnaire results of 95.5 percent confidence, which is two standard deviations. (Round your answers to 3 decimal places.)
UNSATISFACTORY
MEALS SAMPLE SIZE December 1 63 1,000 December 2 83 1,000 December 3 62 1,000 December 4 38 1,000 December 5 57 1,000 December 6 48 1,000 December 7 42 1,000 December 8 67 1,000 December 9 54 1,000 December 10 52 1,000 566 10,000
Explanation / Answer
p is calculated by dividing number of unsatisfactory meal by sample each day. Then calculate np ( total number of unsatisfactory meal)
sum of (np) Number of unsatisfactory meal (np) = 566
Total Meal Inspected (n) = 10,000
p bar = np / n = 566 / 10,000
= 0.0566 = 5.66%
n bar = Total Meal inspected (n) / Number of days (k)
= 10,000 / 10 = 1,000
Sp = Sqrt (p bar (1 - p bar) / n bar)
= sqrt ((0.0566 * 0.9434) / 1000)
= 0.0073
UCLp = p bar + 2 Sp = 0.0712
LCLp = pbar - 2 Sp = 0.042
For Dec2 the p = 0.83 which is out of the limit and rest all day p is within the limit.
Day n p np 1-Dec 1000 0.063 63 2-Dec 1000 0.083 83 3-Dec 1000 0.062 62 4-Dec 1000 0.038 38 5-Dec 1000 0.057 57 6-Dec 1000 0.048 48 7-Dec 1000 0.042 42 8-Dec 1000 0.067 67 9-Dec 1000 0.054 54 10-Dec 1000 0.052 52 Sum 10000 0.566 566Related Questions
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