Question

5. A hospital manager receives a certain number of complaints each day about the hospitals service. Complaints for 20 days are given in the Table 5 shown below. [10 Marks] s aiqel Day Number of Complaints Number of Complaints Day I IT 7 ET ST 8 9T LT 6T 8T 8T 6 61 OT (a what are the initial control limits? Based on the control chart limits calculated, does the (6 Marks] process exhibit statistical control? (b) If not, which samples need to be eliminated? Suggest the next step. [4 Marks] 00

Answer 1

There are total 20 samples

Mean = x' = sum(Xi)/n

= (4 + 1 + 14 + 5 + 11 + 8 + 6 + 19 + 5 +12 + 5 + 4 + 13 + 6 + 4 + 10 + 11 + 18 + 2 + 12) / 20

= 8.5

Variance = Sum ((x' - Xi)^2) / n

= ((8.5-4)^2 + (8.5-1)^2 + (8.5-14)^2 + (8.5-5)^2 + (8.5-11)^2 + (8.5-8)^2 + (8.5-6)^2 + (8.5-19)^2 + (8.5-5)^2 +(8.5-12)^2 + (8.5-5)^2 + (8.5-4)^2 + (8.5-13)^2 + (8.5-6)^2 + (8.5-4)^2 + (8.5-10)^2 + (8.5-11)^2+ (8.5-18)^2 + (8.5-2)^2 + (8.5-12)^2 ) / 20

= 24.95

Standard deviation = sqrt(variance)

= sqrt(24.95)

= 5

Now, Statistical control limit = Mean +- 3*standard deviation

Thus, 8.5 +- 3*5

= (8.5- 15 to 8.5+15)

= (0 to 23.5)

we take lower limit 0 as value can not be negative

Any thing outside of the range (0, 23.5) is outlier

Here,all the points fall in given range. Thus no sample needs to be eliminated