Question

4 The time needed for checking in at a hotel is to be investigated. Historically, the process has had a standard deviation equal to 0.146. The means of 26 samples of n=14 are shown below. Sample Mean Sample Mean 1 3.86 14 3.81 2 3.90 15 3.83 3 3.83 16 3.86 4 3.81 17 3.82 5 3.84 18 3.86 6 3.83 19 3.84 7 3.87 20 3.87 8 3.88 21 3.84 9 3.84 22 3.92 10 3.80 23 3.89 11 3.88 24 3.86 12 3.86 25 3.88 13 3.88 26 3.90 Overall Mean = 3.856 Construct an appropriate control chart for this process with three- sigma limits. 11. Plot the information and look for points and patterns. What can you conclude? 1.

Answer 1

i) -

Control limits for chart -

T

3 UCL = 7+ n

$CL = \bar{\bar{x}}$

3 LCL = T - n

Here,
T
= 3.856, $\hat{\sigma}$ = 0.146, n = 14.

So,

Control limits for chart -

T

UCL = 7+ 3 cổ = 3.856+ n 3 -0.146 = 3.856 + 0.1171 = 3.9731 14

$CL = \bar{\bar{x}} = 3.856$

$UCL = \bar{\bar{x}} - \frac{3}{\sqrt{n}}\hat{\sigma} = 3.856 - \frac{3}{\sqrt{14}}0.146 = 3.856 - 0.1171 = 3.7389$

Control chart -

X-bar Chart 4 3.95 3.9 N 3.85 Mean Sample mean 3.8 LCL 3.75 CL UCL 3.7 3.65 3.6 1 2 3 4 5 5 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Sample

ii) -

From the control chart, we can see that all points lies within control limits, hence the system is under control. Although process is under control, but there are upward & downward trends which shows the presence of assignable causes.