Question

As a member of a management team, you are concerned with the absenteeism among assistants. The issue has been raised by your management team, who feel they often have to perform work normally done by their assistants. To get the facts, absenteeism data were gathered for the last three weeks, which is considered a representative period for future conditions. After taking random samples of 65 personnel files each day, the following data were produced:

Day |
Assistants Absent |
Day |
Assistants Absent |
Day |
Assistants Absent |

1 |
3 |
6 |
2 |
11 |
5 |

2 |
7 |
7 |
3 |
12 |
7 |

3 |
5 |
8 |
2 |
13 |
11 |

4 |
4 |
9 |
3 |
14 |
4 |

5 |
6 |
10 |
5 |
15 |
2 |

Because your assessment of absenteeism is likely to come under
careful scrutiny, **you would like a type I error of only 1
percent.** You want to be sure to identify any instances of
unusual absences. If some are present, you will have to explore
them on behalf of the management team.

a. Design a p-chart.

The upper control limit

the lower control limit

(Enter your responses rounded to three decimal places.If your answer for the lower control limit is negative, enter this value as 0

b. Based on your p-chart and the data from the last 3 weeks, what can you conclude about the absenteeism of Assistants?

Use the control limits from part (a) to determine if the process is out of control in the last 3 weeks. If any of the sample points fall outside of the control limits, then the process it out of statistical control.

Answer #1

Total Number of Absenteeism = 3+7+5+4+6+2+3+2+3+5+5+7+11+4+2 = 69

Total Sample Size np = 65*15 = 975

p-bar = Total Number of Absenteeism/np

p-bar = 69/975

p-bar = 0.07077

For Type 1 error of 1%, Z = 2.576

UCLp = p-bar + Z*(p-bar*(1-p-bar)/n)^(1/2)

UCLp = 0.07077 + 2.576*(0.07077*(1-0.07077)/65)^(1/2)

UCLp = 0.15270

**UCLp = 0.153**

LCLp = p-bar - Z*(p-bar*(1-p-bar)/n)^(1/2)

LCLp = 0.07077 - 2.576*(0.07077*(1-0.07077)/65)^(1/2)

LCLp = -0.01117

**LCLp = 0.000**

Below is the screenshot of the formula applied in excel -

Below is the screenshot of the p chart -

Since one sample is outside the control limit range, hence we
can conclude that process is **out of statistical
control.**

As a hospital administrator of a large hospital, you are
concerned with the absenteeism among nurses' aides. The issue has
been raised by registered nurses, who feel they often have to
perform work normally done by their aides. To get the facts,
absenteeism data were gathered for the last three weeks, which is
considered a representative period for future conditions. After
taking random samples of 60 personnel files each day, the
following data were produced:
Day
Aides Absent
Day...

Professor Murphy wants to set up a control chart to monitor the
percentage of absenteeism in his introductory statistics course (50
students are registered). Absences per period for the last 15 class
sessions are in the table below.
Session
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Absent
5
0
2
2
3
8
10
3
5
5
1
2
0
2
3
Using 3 sigma limits, calculate lower and upper control limits...

Carpet is packaged in 100-yard rolls. Quality
control team opens five rolls every day to check problems like
inconsistency in pattern or color, stains, and other potential
defects. These problems are counted and reported as
number of problems found per roll. The following table
shows number of problems found in one day:
Sample
# of problems
1
5
2
2
3
4
4
3
5
1
What you can say about the process?
This process is not in statistical control because upper
control...

A taxi company wants to do an annual review of its service by
checking the number of monthly complaints. The total number of
complaints in each of the last 12 months were 15, 9, 13, 12, 9, 20,
8, 10, 16, 11, 18, and 3.
If 3-sigma limits are used to construct a control chart
(corresponding to 99.73 percent) calculate the upper control limit
(UCL) and the lower control limit (LCL) of this control chart.
b. Examining the data...

A quality manager asked his team to implement p control chart
for a process that was recently introduced. The team collected
samples of size n = 100 parts hourly over a period of 30 hours and
determined the proportion of nonconforming parts for each sample.
The mean proportion of nonconforming parts for 30 samples turned
out to be 0.025.
Determine the upper and lower control limits for the p chart.
(2 Points)
Discuss how you will use the p Chart...

Twelve samples, each containing five parts, were taken from a
process that produces steel rods at Emmanual Kodzi's factory. The
length of each rod in the samples was determined. The results were
tabulated and sample means and ranges were computed.
Refer to Table S6.1 - Factors for computing control chart limits
(3 sigma) for this problem.
Sample
Size, n
Mean Factor,
A2
Upper Range,
D4
Lower Range,
D3
2
1.880
3.268
0
3
1.023
2.574
0
4
0.729
2.282
0...

to look at 36 weeks instead of 20 weeks. His quality improvement
team provided him with the data below after an “intervention” was
introduced in week 20. Calculate the upper and lower control limits
and then construct a control chart. Display your chart.
Question: Using the rules for detecting special
causes, can you detect any special causes in this case? If so, what
special causes can you detect? What does this mean? Was the
intervention successful in reducing the delays?...

The following are quality control data for a manufacturing
process at Kensport Chemical Company. The data show the temperature
in degrees centigrade at five points in time during a manufacturing
cycle.
Sample
x
R
1
95.72
1.0
2
95.24
0.9
3
95.18
0.7
4
95.42
0.4
5
95.46
0.5
6
95.32
1.1
7
95.40
0.9
8
95.44
0.3
9
95.08
0.2
10
95.50
0.6
11
95.80
0.6
12
95.22
0.2
13
95.58
1.3
14
95.22
0.6
15
95.04
0.8
16...

HCH Manufacturing has decided to use a p-Chart with 2-sigma
control limits to monitor the proportion of defective steel bars
produced by their production process. The quality control manager
randomly samples 250 steel bars at 12 successively selected time
periods and counts the number of defective steel bars in the
sample.
Sample Defects
1 7
2 10
3 14
4 8
5 9
6 11
7 9
8 9
9 14
10 11
11 7
12 8
Step 1 of...

University Hospital has been concerned with the number of errors
found in its billing statements to patients. An audit of 100 bills
per week over the past 12 weeks revealed the following number of
errors:
Develop control charts with z = 3: (a)
CL, (b) UCL, (c) LCL.
(If the lower control limit is negative, round the LCL
to zero and all other answers to 2
decimal places, e.g. 15.25.)
CL:
UCL:
LCL:
Is the process in control?
Week
Number...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 16 minutes ago

asked 24 minutes ago

asked 24 minutes ago

asked 31 minutes ago

asked 32 minutes ago

asked 39 minutes ago

asked 41 minutes ago

asked 43 minutes ago

asked 57 minutes ago

asked 57 minutes ago

asked 1 hour ago

asked 1 hour ago