Question

Create appropriate control charts and interpret the results for following 2 problems.

a.1. The times it takes (in minutes) to create sign-up members at a homeless shelter soup kitchen are shown below. Five samples (representing volunteers) of four observations are taken.

Sample1	Sample 2	Sample 3	Sample 4	Sample 5
10.2	10.3	9.7	9.9	9.8
9.9	9.8	9.9	10.3	10.2
9.8	9.9	9.9	10.1	10.3
10.1	10.4	10.1	10.5	9.7

b.2. Before shipping lawn mowers to dealers, an employee starts each mower and records data for mowers that don't start on the first try. Out of 100 mowers, an average of four did not start.

Answer 1

x1=c(10.2,10.3,9.7,9.9,9.8)
x2=c(9.9,9.8,9.9,10.3,10.2)
x3=c(9.8,9.9,9.9,10.1,10.3)
x4=c(10.1,10.4,10.1,10.5,9.7)

a=mean(x1); a=9.98
b=mean(x2); b=10.02
d=mean(x3); d=10
e=mean(x4); e=10.16

totalmean=(a+b+d+e)/4; totalmean=10.04
a1=range(x1); a1=.6
b1=range(x2); b1=.5
d1=range(x3); d1=.5
e1=range(x4); e1=.8

rangemean=sum(y)/4; rangemean=.6
A2=.577, for n=5
Lcl=totalmean-(A2*rangemean); Lcl=9.6938
cl=totalmean; cl=10.04
Ucl=totalmean+(A2*rangemean); Ucl=10.3862

All sample means falls between the control limits, so process is in control.

(b) because each sample mean is greater than 4. so it is unfair to say out of 100 average of 4 did not start.