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Introduction: Apply Principles of Statistics Process Control
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Q1
In the given case, a soccer selected 6 players from a group of boys that are under the age group of 8 to 10, 7 from 11- 12 age and 3 from 13 – 14, presented below:
Age-group |
8-10 |
11-12 |
13-14 |
Sample size |
6 |
7 |
3 |
1. In this case, entire population is divided into various sub-groups on the basis of age, referre as strata and then sample has been selected from each, thus, it is clear that stratified sampling method has been used to choose an adequate size of representative sample from recreational soccer team.
2. Pollster interviews all the human resources personnel working in five different high tech companies. In this particular situation, cluster sampling technique have been applied because whole population has been divided into 5 groups, called clusters and 1, workers (sample) has been selected from each other.
3. Selecting a sample of 50 women engineers and equal number of men engineer is a type of stratified sampling because total population is separated into strata on the basis of gender and 50 people has been chosen from each one.
4. A medical research interviews selected every third cancer patient from all the cancer caused people in a local hospital, it is a type of systematic sampling that prefers selection of sample members following a random start and at a fixed point of interval.
5. Selection of 50 students by a counsellor of high school indicates simple random sampling indicating that every member will have an equal chance or probability of being inclusion (Sampling and data, 2016).
6. In the stated case, classmates have been selected from a algebra class so as to determine that how many pairs of jeans students have on an average. Thus, it convenience sampling technique has been incorporated. It is a non-probabilistic sampling made, and as name suggests, used to select sample conveniently.
Q2
As per the stated scenario, Federation University is planned to conduct a survey to analyse the average money that a part-time students spend on their books. In such respect, there are total 10000 part-time students; however, it is an impossible task to gather information from each and every student. In such situation, random sampling method is considered as the best sampling method over convenience and other sampling because, in this, every unit have an equal chance of being selection in the sample (Siegel, 2016).. This sample will be definitely representative sample because in this, sample of an adequate size can be selected comprising different disciplines such as chemistry, English, history, nursing, mathematics, psychology & many others and provides accurate results of amount spend by part-time students on the course books.
Q3
A) In order to find out the grade point average, using all honor students as the sample is not considered as representative sample because it choose only high mark awarded people. However, in order to find out GPA, it must be find out by accumulating the final grades of all the members divided by the number of grading awarded by the university.
B) Here, investigator has used systematic sampling stood outside the supermarket and surveyed every 20th child who were under the age of 10 year, therefore, it can be considered as a representative sample for the survey.
C) It is founded as an appropriate representative sample because in this, cluster has been designed using simple random sampling technique on the basis of state and then survey has been performed by Australian politician in every cluster. Thus, in such case, the result drawn will be provide correct answer about the average annual income of the adults in Australia.
D) It is not a representative sample because in order to determine the proportion of people who are using public transport as a mean to work, interviewer sits on a bench in the Melbourne Botanical Garden and interviewed 10 people, cannot be said as an appropriate technique (Jaggia and et.al., 2016).
E) It is considered as a suitable sampling method and representative also, because in order to find out an average cost of staying in a hotel for 2 days, researcher has surveyed 100 hospitals using simple random sampling to eliminate biasness.
Q4
A. Calculation of mean
Country or Area |
Year |
Value |
Azerbaijan |
2009 |
44594.80078 |
Belarus |
2009 |
99280 |
Bermuda |
2009 |
77.08000183 |
China |
2009 |
2418000 |
China, Hong Kong SAR |
2009 |
889.8654175 |
Cyprus |
2009 |
375 |
Czech Republic |
2009 |
11933 |
Egypt |
2009 |
64149 |
Gambia |
2009 |
12195.54492 |
Hungary |
2009 |
125106 |
Kazakhstan |
2009 |
100000 |
Lebanon |
2009 |
4100 |
Malta |
2009 |
112 |
Mauritius |
2009 |
2949 |
Norway |
2009 |
374011 |
Poland |
2009 |
55143 |
Portugal |
2009 |
17859 |
Qatar |
2009 |
70.59999847 |
Republic of Moldova |
2009 |
11311 |
Romania |
2009 |
34480 |
Serbia |
2009 |
176050 |
Singapore |
2009 |
720 |
Sweden |
2009 |
175118 |
United Kingdom |
2009 |
177107 |
Venezuela |
2009 |
551505.5625 |
Total |
4457136.454 |
|
Total number of countries |
25 |
(Source: UN Data. n.d.)
Mean = (4457136.454/25)
= 178285.4581
B. Calculation of median
Country or Area |
Value |
1 |
70.6 |
2 |
77.08 |
3 |
112 |
4 |
375 |
5 |
720 |
6 |
889.8654 |
7 |
2949 |
8 |
4100 |
9 |
11311 |
10 |
11933 |
11 |
12195.54 |
12 |
17859 |
13 |
34480 |
14 |
44594.8 |
15 |
55143 |
16 |
64149 |
17 |
99280 |
18 |
100000 |
19 |
125106 |
20 |
175118 |
21 |
176050 |
22 |
177107 |
23 |
374011 |
24 |
551505.6 |
25 |
2418000 |
Median = Value of (N+1)/2th item
= Value of (25+1)/2th item
=Value of 26/2th item
= Value of 13th item
Median = 34480
C. Calculation of standard deviation
Country or Area |
Value |
Deviation |
Deviation^2 |
Azerbaijan |
44594.8 |
-133690.66 |
17873191866.31 |
Belarus |
99280 |
-79005.46 |
6241862416.68 |
Bermuda |
77.08 |
-178208.38 |
31758226040.36 |
China |
2418000 |
2239714.54 |
5016321228997.48 |
China, Hong Kong SAR |
889.8654 |
-177395.59 |
31469196319.09 |
Cyprus |
375 |
-177910.46 |
31652131117.31 |
Czech Republic |
11933 |
-166352.46 |
27673140330.83 |
Egypt |
64149 |
-114136.46 |
13027131077.85 |
Gambia |
12195.54 |
-166089.91 |
27585859274.41 |
Hungary |
125106 |
-53179.46 |
2828054768.58 |
Kazakhstan |
100000 |
-78285.46 |
6128612956.95 |
Lebanon |
4100 |
-174185.46 |
30340573829.13 |
Malta |
112 |
-178173.46 |
31745781187.29 |
Mauritius |
2949 |
-175336.46 |
30742873554.78 |
Norway |
374011 |
195725.54 |
38308487734.50 |
Poland |
55143 |
-123142.46 |
15164064997.95 |
Portugal |
17859 |
-160426.46 |
25736648472.90 |
Qatar |
70.6 |
-178214.86 |
31760535664.13 |
Republic of Moldova |
11311 |
-166974.46 |
27880469672.76 |
Romania |
34480 |
-143805.46 |
20680009792.25 |
Serbia |
176050 |
-2235.46 |
4997273.12 |
Singapore |
720 |
-177565.46 |
31529491926.19 |
Sweden |
175118 |
-3167.46 |
10032791.10 |
United Kingdom |
177107 |
-1178.46 |
1388763.60 |
Venezuela |
551505.6 |
373220.10 |
139293246294.88 |
Total |
0.00000 |
5635757237120.39 |
Standard deviation (S.D.) = √(∑dx^2/N)
√5635757237120.39/25
= 474795.00
D. Plotting a scatter line chart of the country versus value
E. Distribution of the data
Findings the results, it can be seen that average value of renewable freshwater each country or area is founded to 178285.46. While, as per the derived median, 50% value of the renewable freshwater is below 34480 such as Singapore, Czech Republic, China, Bermuda etc. whereas 50% values are above the middle value such as Azarbaijan, Egypt, Hungary, Kazakhstan, Lebanon & others. However, looking at the dispersion measurement, standard deviation, it is founded very high to 474795.00 denotes that renewable freshwater value of all the countries are highly differ from the computed mean value, thus, dispersion is very high (Nguyen, Charity and Robson, 2016). Thus, data is highly spreaded or scattered in overall data set. However, as per the scatter diagram, it can be seen that China has very large amount of renewable freshwater to 218000 as compare to other nations.
Q5
Defining variation and difference between common & special cause of variations
Variation refers to the difference between actual and expected results, also termed as deviations. In actual life, it is very common that deviation will be occur because in the uncertain market conditions, actual results always differ either upward or downward from the expected or targeted results. It may be of two types, common cause and special cause deviations. The difference between both the variations is common causes can be predicted and estimated by the policy makers or decision-making authority, however, special cause variations are unpredicted reasons (Anderson and et.al., 2016). For instance, common variances can occurs due to the variables that can be estimated probabilistically or on the basis of historical results; however, special-cause variables can be measured probabilistically and arises due to sudden and uncertain volatility.
Q6
1 Constructing a run chart
2 Determining mean and draw it on the line
S.No. |
Height/In cm) |
1 |
6.384 |
2 |
6.19 |
3 |
2.757 |
4 |
3.469 |
5 |
6.694 |
6 |
5.319 |
7 |
5.805 |
8 |
5.562 |
9 |
5.621 |
10 |
5.285 |
11 |
3.247 |
12 |
3.796 |
13 |
5.477 |
14 |
6.149 |
15 |
8.312 |
16 |
5.374 |
17 |
1.236 |
18 |
7.802 |
19 |
4.823 |
20 |
5.315 |
21 |
4.537 |
22 |
7.24 |
23 |
7.174 |
24 |
3.449 |
25 |
4.078 |
Total |
131.1 |
Average /Mean= Total Heights (In cm)/Number of pizzas
= 131.1cm/25
= 5.24 cm.
3 Add upper and lower limit to the graph
Column1 |
|
Mean |
5.2438 |
Standard Error |
0.332079 |
Median |
5.374 |
Mode |
#N/A |
Standard Deviation |
1.660393 |
Sample Variance |
2.756904 |
Kurtosis |
0.161665 |
Skewness |
-0.36572 |
Range |
7.076 |
Minimum |
1.236 |
Maximum |
8.312 |
Sum |
131.095 |
Count |
25 |
Standard deviation = 1.66
UCL = Mean + S.D.
= 5.24 + 1.66
= 6.90
LCL = Mean - S.D.
= 5.24 - 1.66
= 3.57
4. Analysis of the chart
4 A) |
Corrective action was required at the time of sample 17 when the height of pizza base was declined majorly. It was recorded as 1.236 cm which is low of all time and hence corrective measure need to incorporate at this time |
4 B) |
poor supervision from concerned authority |
Less efforts been incorporate by staff while preparing base |
|
high work load can be the reason for decline in height of pizza base |
|
4 C) |
tight monitoring on part of quality check |
work is done in hurry without emphasizing on required height |
|
no actions might be taken by management on over using the resources |
|
4 D) |
After sample 17, significant rise in quality is witnessed and it was over specification. This can be the reason due to sound monitoring and quality check by inspection team. This helps in maintaining the quality in the limit henceforth. |
Q7
A. Standard distribution curve
This curve is a normal distribution curve that indicates that 68% of the values of a data set re under the range of 1 standard deviation, 95% of values comes under 2 times of S.D. whereas 99.7% values of data series falls under 3 times of S.D presented here as under:
B. Standard deviation
It is a statistical measure that is often used by analysts to find out that how spread or scattered are the numbers in a given data set. This method uses deviations from either the mean value or any other variable as an assumed mean and squares the variances. It is the best way of finding dispersion because it takes into account all the values given in a data set and use square of the deviations to make the value positive (Zheng and et.al., 2016).
Q8
A) Test data can be saved in excel or document format so that it can be retrieved in future. This can be store either on hard disk, internal server or at cloud for better sharing and retrieving information. This set of workplace procedure aids in storing data is systematic manner and helps in gaining insight from it. Further, test data can be saved on MIS and internal software so that same can be retrieved with ease.
B) Data need to be verified for its accuracy for the purpose of avoiding miscommunication and misconception. Following are three reasons in this context:
1) Data usually helps in decision making process and accurate data surely aids in taking proper decisions. Through this, operations within organization can be run smoothly.
2) For ensuring authenticity of data, verification should be done. From this, reliability within the company increases and hence overall duplication of work gets decline.
3) Verified data of different parties such as customers, suppliers or employees helps in sharing information at the time of urgency. This aids in developing better relations with them.
C) In the consultancy firm, if there is poor data related to existing needs and wants of customers, then appropriate strategy cannot be developed with respect to attracting customers. This way, organization may face issue in increasing its customer base and hence overall cost of operation gets increase. For apparel firm, if poor data related to monthly sales is kept then it causes problem with regards to strategy formulation (Zhang, 2016). Apparel firm witness hike in sale in few month and if advertisement are not done in those month due to poor data, then it can affect the profitability of company and also leads to rise in cost of operation.
D) There are certain challenges faced at the time of checking data. In this context, diversity of data source is one such instance that causes difficulty while checking data. For instance if data is obtained from internet, through industries, scientific and observational data, document, video, audio, software package, spreadsheet and other related, then its checking incurs huge efforts, cost and time. Another issue which checking data is the timeliness of the information. Some information such as images of individual or locality, mobile number, cell phone or any other related information that changes in short span of time, then it needs to be checked in real time. Such data are not fit for checking over a long span of time as it might get outdated after a certain period of time.
E) It is required to ensure real time checking of data so that issue of timeliness can be avoided. In this regard, help from real time processing and analysis software can be taken (Brandenburg and et.al., 2016). Further, individual can be appointing for checking from distinct source in order to ensure uniformity. This way, big data can be checked over a period of time.
Q9
A. Importance of setting targets and limits in business processes
In business, target setting is considered as a very important aspect for the corporations. It involves the creation of plans that is expected to be achieve or accomplishing after completion of the project. It is highly important so as to motivate, encourage and guide the members or group towards the set targets through aligning and directing the activities and functions effectively. Moreover, establishing the control limits is also necessary for the processes mainly categorised into two that are upper and lower control limits (Salkind, 2016). It is significantly important so as to get substantial benefits by elimination of defects and hazards to an acceptable level by separation of unacceptable items from the production processes. This in turn, firms can produce goods and services exactly as per the product specification and standard quality.
B. Determination of targets and limits with examples
Firms can set targets keeping in view various elements and components like use expectation, competitors' offerings, cost & many others. For instance, in order to produce an item exactly as per the demand of the customers, firm can set targets through computing the mean or average value of the given data set. Supporting it, in the question 6, pizza's height mean value has been founded to 5.24. Whereas, upper control limit (UCL) & lower control limits can be decided through using standard deviation, as follows:
UCL (within 1 S.D.) = Mean + 1S.D. (5.24 + 1.66) = 6.90
LCL (within 1 S.D.) = Mean – 1S.D. (5.24 - 1.66) = 3.57
UCL (within 2 S.D.) = Mean + 2S.D.
LCL (within 2 S.D.) = Mean – 2 S.D.
UCL (within 3 S.D.) = Mean + 3S.D.
LCL (within 3 S.D.) = Mean - 3S.D.
Q10
1. Calculation of mean, mode and median
S. no. |
Heights (In cm) |
1 |
120 |
2 |
133 |
3 |
139 |
4 |
139 |
5 |
145 |
6 |
146 |
7 |
147 |
8 |
147 |
9 |
147 |
10 |
150 |
11 |
153 |
12 |
153 |
13 |
160 |
14 |
162 |
15 |
162 |
16 |
169 |
17 |
170 |
18 |
175 |
Total |
2717 |
Number of items |
18 |
Average |
(2717/18) = 150.94 cm. |
Mean/Average = (∑x/n)
= (2717/18)
= 150.94 cm
Median = Value of (N+1)/2th item
= (18+1)/2th item
= 19/2th item
= 9.5th item
Median = (Value of 9th item + value of 10th item)/2
= (147 +150)/2
= 148.5 cm
S. no. |
Heights (In cm) |
Frequency (Number of students) |
1 |
120 |
1 |
2 |
133 |
1 |
3 |
139 |
2 |
4 |
145 |
1 |
5 |
146 |
1 |
6 |
147 |
3 |
7 |
150 |
1 |
8 |
153 |
2 |
9 |
160 |
1 |
10 |
162 |
2 |
11 |
169 |
1 |
12 |
170 |
1 |
13 |
175 |
1 |
Total |
18 |
Mode = 147 cm because it has highest frequency of 3
2. Calculation of median & mode
S. no. |
values (x) |
1 |
3 |
2 |
4 |
3 |
4 |
4 |
5 |
5 |
5 |
6 |
5 |
7 |
5 |
8 |
5 |
9 |
5 |
10 |
5 |
11 |
6 |
12 |
6 |
13 |
6 |
14 |
6 |
15 |
6 |
16 |
6 |
17 |
6 |
18 |
7 |
19 |
7 |
20 |
7 |
21 |
8 |
22 |
8 |
23 |
8 |
24 |
9 |
25 |
9 |
151 |
Median = Value of (N+1)/2th item
= (25+1)/2th item
= 13th item
Median = 6
Values (x) |
Frequency |
3 |
1 |
4 |
2 |
5 |
7 |
6 |
7 |
7 |
3 |
8 |
3 |
9 |
2 |
Total |
25 |
Multiple modes – 5 and 6
Alternative Z = 3M – 2A.M.
= 3(6) – 2 (151/25)
=18 – 2(6.04)
= 18 – 12.08
= 5.92 or 6 (approx)
3. Calculation of mean and median
S. no. |
Values (x) |
1 |
10 |
2 |
10 |
3 |
12 |
4 |
14 |
5 |
16 |
6 |
16 |
7 |
16 |
8 |
18 |
9 |
20 |
Total |
132 |
Average |
14.67 |
Mean = 132 / 9
= 14.67
Median = Value of (N+1)/2th item
= (9+1)/2th item
= 10/2th item
= 5th item
M = 16
4. Computation of mean, mode & median
S. no. |
Values (x) |
1 |
2 |
2 |
3 |
3 |
3 |
4 |
4 |
5 |
4 |
6 |
4 |
7 |
4 |
8 |
4 |
9 |
4 |
10 |
4 |
11 |
5 |
12 |
5 |
13 |
5 |
14 |
6 |
15 |
6 |
16 |
6 |
17 |
7 |
18 |
7 |
19 |
8 |
20 |
9 |
Total |
100 |
Average |
5 |
Mean = (100/20)
= 5
Median (M) = (N+1)/2th item
= (20+1)/2th item
= 21/2th item
= 10.5th item
M = (Value of 10th item + value of 11th item)/2
= (4+5)/2
= (9/2)
= 4.5
S. no. |
Values (x) |
Frequency |
1 |
2 |
1 |
2 |
3 |
2 |
4 |
4 |
7 |
11 |
5 |
3 |
14 |
6 |
3 |
17 |
7 |
2 |
19 |
8 |
1 |
20 |
9 |
1 |
Total |
20 |
Mode (Z) = 4
5. Calculation of mean, range and standard deviation
Branch |
Values (x) |
Deviation (dx = X-A.M.) |
Deviation ^2 |
A |
3 |
(3-8) = -5 |
25 |
B |
6 |
(6-8) = -2 |
4 |
C |
8 |
(8-8) = 0 |
0 |
D |
9 |
(9-8) = 1 |
1 |
E |
14 |
(14-8) = 6 |
36 |
Total |
40 |
0 |
66 |
Mean = 40/5 = 8
Range = Maximum chemical insolubility – Minimum chemical insolubility
= 14 – 3
= 11
Standard deviation √(∑dx^2/N)
= √66^2/5
= 3.6
Note: Deviation has been taken from derived mean
6. Calculation of range & standard deviation
S. No. |
Values (X) |
Deviation (Dx = X-A.M.) |
Dx^2 |
1 |
13 |
-3 |
9 |
2 |
14 |
-2 |
4 |
3 |
11 |
-5 |
25 |
4 |
17 |
1 |
1 |
5 |
15 |
-1 |
1 |
6 |
20 |
4 |
16 |
7 |
20 |
4 |
16 |
8 |
18 |
2 |
4 |
9 |
15 |
-1 |
1 |
10 |
17 |
1 |
1 |
Total |
160 |
∑dx^2 = 78 |
|
Maximum |
20 |
||
Minimum |
11 |
Range = Highest score – lowest score
= 20 – 11
= 9
Average = 160/10
= 16
Standard deviation = √(∑dx^2/N)
√78/10
= 2.8
Note: Deviation has been taken from derived mean
S. No. |
Data 1 |
Data 2 |
Deviation (Dx - X - A.M.) (Data set 1) |
Deviation (Dx = X - A.M.) (Data set 2) |
Dx^2 (Data set 1) |
Dx^2 (Data set 2) |
1 |
14 |
15 |
-1 |
0 |
1 |
0 |
2 |
16 |
16 |
1 |
1 |
1 |
1 |
3 |
18 |
18 |
3 |
3 |
9 |
9 |
4 |
19 |
17 |
4 |
2 |
16 |
4 |
5 |
10 |
16 |
-5 |
1 |
25 |
1 |
6 |
12 |
14 |
-3 |
-1 |
9 |
1 |
7 |
13 |
13 |
-2 |
-2 |
4 |
4 |
8 |
20 |
15 |
5 |
0 |
25 |
0 |
9 |
20 |
17 |
5 |
2 |
25 |
4 |
10 |
8 |
9 |
-7 |
-6 |
49 |
36 |
Total |
150 |
150 |
0 |
0 |
164 |
60 |
Average |
15 |
15 |
4.0 |
2.4 |
Average = ∑x/n
Data set 1 = 150/10 = 15
Data set 2 = 150/10 = 15
Standard deviation = √(∑dx^2/N)
Data set 1 = √(164/10)
= 4.0
Data set 2 = √60/10
= 2.4
Findings the results, it is clear that standard deviation of 1st data set is greater to 4.0 relatively to that of another data set as it is 2.4. Henceforth, it can be interpreted that 1st data set is more spreaded out and volatile from the derived average value of 15 in comparison to the other data series.
References
Books and Journals
Anderson, D.R. And et.al., 2016. Statistics for business & economics. Nelson Education.
Brandenburg, J.G and et.al., 2016. Benchmark tests of a bly constrained semilocal functional with a long-range dispersion correction. Physical Review B. 94(11). p.115144.
Jaggia, S. and et.al., 2016. Essentials of business statistics: communicating with numbers. McGraw-Hill Education.
Nguyen, T.H., Charity, I. and Robson, A., 2016. Students' perceptions of computer-based learning environments, their attitude towards business statistics, and their academic achievement: implications from a UK university. Studies in Higher Education. 41(4). pp.734-755.
Salkind, N.J., 2016. Statistics for people who (think they) hate statistics. Sage Publications.
Siegel, A., 2016. Practical business statistics. Academic Press.
Zhang, Z., 2016. Missing data imputation: focusing on single imputation. Annals of translational medicine. 4(1). pp.16-20.
Zheng, S. and et.al., 2016. The Relationship Between the Mean, Median and Mode with Grouped Data. Communications in Statistics-Theory and Methods, (just-accepted).
Sampling and data. 2016. [PDF]. Available through: http://www.webassign.net/question_assets/idcollabstat2/Chapter1.pdf. [Accessed on 3rd February 2017].
UN Data. n.d. [Online]. Available through: http://data.un.org/Data.aspx?q=Renewable+freshwater+resources+&d=ENV&f=variableID%3a124. [Accessed on 3rd February 2017].