Question

A teller at a drive-up window at a bank had the following service times (in minutes) for 20 randomly selected customers:

SAMPLE
1	2	3	4
4.5	4.6	4.5	4.7
4.2	4.5	4.6	4.6
4.2	4.4	4.4	4.8
4.3	4.7	4.4	4.5
4.3	4.3	4.6	4.9

a. Determine the mean of each sample. (Round your answers to 1 decimal place.)

Sample	Mean
1
2
3
4

b. If the process parameters are unknown, estimate its mean and standard deviation. (Round "Mean" value to 1 decimal place and "Standard deviation" value to 3 decimal places.)

Mean
Standard deviation

c. Estimate the mean and standard deviation of the sampling distribution. (Round "Mean" value to 1 decimal place and "Standard deviation" value to 3 decimal places.)

Mean
Standard deviation

d. What would three-sigma control limits for the process be? What alpha risk would they provide? (Round your "UCL and LCL" values to 3 decimal and "Risk" value to 4 decimal places.)

LCL
UCL
Risk

e. What alpha risk would control limits of 4.14 and 4.86 provide?

Alpha risk             

f. Using limits of 4.14 and 4.86, are any sample means beyond the control limits?
  

• Yes

• No

g. Construct control charts for means and ranges using the following table. Are any samples beyond the control limits? (Leave no cells blank - be certain to enter "0" wherever required. Round your answers to 4 decimal places.)

Factors for three-sigma control limits for x¯x¯  and R charts

				FACTORS FOR R CHARTS
Number of Observations in Subgroup, n			Factor for x¯x¯ Chart, A2	Lower Control Limit, D3			Upper Control Limit, D4
	2		1.88		0		3.27
	3		1.02		0		2.57
	4		0.73		0		2.28
	5		0.58		0		2.11
	6		0.48		0		2.00
	7		0.42		0.08		1.92
	8		0.37		0.14		1.86
	9		0.34		0.18		1.82
	10		0.31		0.22		1.78
	11		0.29		0.26		1.74
	12		0.27		0.28		1.72
	13		0.25		0.31		1.69
	14		0.24		0.33		1.67
	15		0.22		0.35		1.65
	16		0.21		0.36		1.64
	17		0.20		0.38		1.62
	18		0.19		0.39		1.61
	19		0.19		0.40		1.60
	20		0.18		0.41		1.59

	Means
LCL
UCL

All sample means are  (Click to select)  with in  not within  the limits.

	Ranges
LCL
UCL

All ranges are  (Click to select)  within  not within  the limits.

i-1. If the process has a known mean of 4.4 and a known standard deviation of .18, what would three sigma control limits be for a mean chart? (Round your answers to 2 decimal places.)

LCL
UCL

i-2. Are any sample means beyond the control limits?
  

• Yes

• No

i-3.If so, which one(s)?
  

• First one

• Second one

• Third one

• Last one

This is the last question in the assignment. To submit, use Alt + S. To access other questions, proceed to the question map button.Next

Answer 1

Sample	Mean
1	4.3
2	4.5
3	4.5
4	4.7

Answer:

Sample	1	2	3	4
	4.5	4.6	4.5	4.7
	4.2	4.5	4.6	4.6
	4.2	4.4	4.4	4.8
	4.3	4.7	4.4	4.5
	4.3	4.3	4.6	4.9
Mean	4.3	4.5	4.5	4.7

2. Mean = average of all sample observations = Average(4.5,4.2,4.2,4.3,4.3,4.6,4.5,4.4,4.7,4.3,4.5,4.6,4.4,4.4,4.6,4.7,4.6,4.8,4.5,4.9) = 4.5

	x	x-x-bar	(x-x-bar)^2
	4.5	0	0
	4.2	-0.3	0.09
	4.2	-0.3	0.09
	4.3	-0.2	0.04
	4.3	-0.2	0.04
	4.6	0.1	0.01
	4.5	0	0
	4.4	-0.1	0.01
	4.7	0.2	0.04
	4.3	-0.2	0.04
	4.5	0	0
	4.6	0.1	0.01
	4.4	-0.1	0.01
	4.4	-0.1	0.01
	4.6	0.1	0.01
	4.7	0.2	0.04
	4.6	0.1	0.01
	4.8	0.3	0.09
	4.5	0	0
	4.9	0.4	0.16
Total			0.7

Standard Deviation = Square root((x-x-bar)^2/(n-1)) = Square root (0.7/19) = 0.191942974 = 0.192 (Rounding to 3 decimal places)

3. Standard deviation of sampling distribution = Standard deviation of process/Square root(n)

Where n = Sample Size = 5 here

Standard deviation of sampling distribution = 0.192/Square root(5) = 0.192/2.2360 = 0.085867 = 0.086 (Rounding to 3 decimal places)

4. UCL = Mean + 3*Standard deviation of sampling distribution = 4.5 +3*0.086 = 4.758

LCL = Mean - 3*Standard deviation of sampling distribution = 4.5 - 3*0.086 = 4.242

            As this is a 3-sigma control chart Z value = 3 corresponding to this probability = 0.9987

           

            So alpha risk = (1-0.9987)*2 = 0.0026