a.
AgeWeeks Employed
Mean37.75Mean18.6875
Standard Error2.974195Standard Error2.188452
Median37.5Median18.5
Mode25Mode21
Standard Deviation11.89678Standard Deviation8.753809
Sample Variance141.5333Sample Variance76.62917
Kurtosis-1.17143Kurtosis-0.21626
Skewness0.337402Skewness0.522601
Range36Range30
Minimum23Minimum6
Maximum59Maximum36
Sum604Sum299
Count16Count16
Confidence Level(99.0%)8.764138Confidence Level(99.0%)6.44877

b. 99% confidence interval estimate for mean age of newly hired employees;

37.75 ¡V 8.76 = 28.99
to
37.75 + 8.76 = 46.51

c. Hypothesis:

Decision Rule: Reject Ho if t > t-critical
Do not reject Ho if t < t-critical

t-critical = t0.01,15 =2.602

0.771 < 2.602
Therefore, at a 99% Confidence Level the Null Hypothesis can not be rejected and we can not state that Riverside¡¦s mean duration of employment weeks is any greater than the mean duration of employment weeks within the rest of California.

d. Is there a relationship between the age of a newly employed individual and the number of weeks of employment?

By using a scatter plot and plotting the number of weeks employed in respect to the ages of the workers, you can see that the points are distributed along a straight line. The number of weeks employed increase positively as the age of the worker increases. Therefore it is safe to say that there is a positive correlation between the ages of newly employed workers and the number of weeks they are employed.

...Assignment 1: Bottling Company CaseStudy
In this project we were given the case of customer complaints that the bottles of the brand of soda produced in our company contained less than the advertised sixteen ounces of product. Our boss wants us to solve the problem at hand and has asked me to investigate. I have asked my employees to pull Thirty (30) bottles off the line at random from all the shifts at the bottling plant. The first step in solving this problem is to calculate the mean (x bar), the median (mu), and the standard deviation (s) of the sample. All of those calculations were easily computed in excel. The mean was computed by entering: =average, the median by: =median, and the std. dev. by: = = std dev. The corresponding values are x bar = 14.87, mu = 14.8, and s = 0.550329055.
The next step in solving the problem is to construct a 95% confidence interval for the average amount of the company’s 16-ounce bottles. The confidence interval was constructed by drawing a normal distribution with c = 95%, a = 0.050, and Zc = 0.025. The Zc value was entered into the Z◘ (z box) function in the Aleks calculator that resulted in a Z score of +1.96 and -1.96. We calculate the standard error (SE) by dividing the s by the Square root of n which is the sample size. The margin of error is calculated by multiplying the z score =...

...CASESTUDY NO.1
THERAC – 25
Aaron James Uy Timosa
BSIT – 4
INTRODUCTION
The Therac-25 was a radiation therapy machine produced by Atomic Energy of Canada Limited (AECL) after the Therac-6 and Therac-20 units (the earlier units had been produced in partnership with CGR of France). It was involved in at least six accidents between 1985 and 1987, in which patients were given massive overdoses of radiation, approximately 100 times the intended dose. These accidents highlighted the dangers of software control of safety-critical systems, and they have become a standard casestudy in health informatics and software engineering.
WHAT WAS THERAC -25?
Therac-25 was used in the treatment of cancer. Its purpose was to provide radiation to a specific part of the body and hopefully kill the malignant tumor. The Therac-25 was the third system created under the Therac name by the Atomic Energy of Canada Limited (AECL). The AECL is most famous in Canada for their creation of the CANDU reactors which are world renowned. A Therac-6 and Therac-20 were both used in the treatment of cancer. The number that goes along with the word Therac stands for the maximum amount of mega electron volts (MeV) the machine can dispense. It was believed that the new Therac-25 was much more efficient than Therac-6 and Therac-20. The overall size of the machine was reduced and still allowed for two modes;...

...
Investigating Bottling Company CaseStudy
T.P
University
Statistics
Mat 300
Mr. Thevar
December 01, 2013
Investigating Bottling Company CaseStudy
The casestudy that is being investigated is for a bottling company producing less soda than what is advertised. Customers have complained that the sodas in the bottles contain less than the advertised sixteen ounces. The employees at the company have measured the amount of soda contained in each bottle. There are thirty bottles that have been pulled from the shelves. The manager of the company would like to have a detailed report on the possible causes, if any, for the shortage in the amount of soda or if the claim is not supported explain how to mitigate the issue in the future. In order to statistically find a cause in the shortage a hypothesis testing is conducted by finding the mean, median, and standard deviation for ounces in the bottles. Constructing a 95 percent interval will establish the mean of the population since the mean of the population is not known.
There are thirty soda bottles being pulled for investigation. The mean will be calculated by averaging the amount of ounces in each bottle and dividing the total by the number of bottles. The data below shows the ounces in each of the thirty bottles that were pulled.
The mean among the sample bottles is 14.87. The calculation...

...Pilgrim Bank Case
September 26, 2013
How much do profits vary across customers? Provide statistical support for your answer.
Out of a sample of 31,634 Pilgrim Bank customers, from a population of 5 million total, and zero missing values, profits vary widely. The average customer profitability is $111.50. The minimum value of this data set is -$221 and the maximum is $2071. This describes that there is a very wide range. The median customer profitability is $9 and the standard deviation is $272.84. See exhibit one.
How does Pilgrim Bank make money from its customers and how can this explain the variation in customer profitability?
Pilgrim Bank makes its profit from customers with components including the balance in deposit accounts, the net interest spread, the fees collected by serving customers, and the interest from loans distributed. This can explain the variation in customer profitability because customer accounts generate different types of revenue for Pilgrim Bank. Each customer generates investment income by keeping a deposit balance. Fees are assessed for checking accounts, late payments and overdrafts. This is an important revenue source Pilgrim Bank. Since each customer varies in how many fees they rack up, each customer accounts for a different amount of the profitability from this source. Depending on the customer, a loan will be handled and the rate will be decided upon. Each customer would create revenue for the bank this way but not all...

...
CaseStudy1
Project Management Analysis in the Internet Forecasting Industry
Group 4
Dilip Chinnaswamaiah, Ruiying Liu, Sandhya Aparna Pashikanti, Yingqi Yang, Hao Zhu
Executive Summary
In this case, our group used the techniques such as PERT, project crushing and visualization to solve the scheduling problems encountered in a development project at B&W Systems. We came up with the expected completion time and crashing solution to help the company meet the deadline. The problem was successfully solved.
Statement of the Problem
The 3 questions our group tried to answer were: How long will the development project at B&W Systems take? How to crash it to meet the 35-week deadline? What impact would there be if the estimated duration of some tasks changes?
Background
We referred to the lecture notes on network diagram and probabilistic time estimates. We also reviewed materials on project crashing. Additionally, before actually solving the problem, we transferred the information presented in Exhibit 1 and Exhibit 2 of the case to an Excel spreadsheet so that we could carry out calculations easily.
Methodology
First, since all the data came from the case and was already on an Excel spreadsheet, we began by calculating the expected time, variance, expedition available and slope for each task using the formulas below.
Expected time=(optimistic+4*(most...

...[Type text]
Statistics & Research Methodology (103)
Contributors: Prof. (Gp.Capt.) D.P.Apte Prof. S.N.Parasniss Prof. Anagha Gupte Prof. Anjali Mote
[Type text]
Session Plan Subject: Statistics & Research Methodology (103) Lecture Topic No. 1Statistics: Introduction to business statistics, Importance, Definition & classification 2 Collection of data: Primary & secondary data, Sources and Classification of data, Tabular presentation 3 Presentation of data: Frequency distribution tables, Graphical presentation 4 Measures of Central Tendency and Variability: Ungrouped data 5 Case-1 6 Measures of Central Tendency, Measures of variability : Grouped data Fundamentals of Probability, Laws of probability Case -2 Probability: Conditional probability, Bayes’ theorem Probability distributions: Binomial distribution Case -3 Probability distributions: Poisson and Exponential distribution Probability distribution: Normal distribution Case - 4 Sampling : Importance and sampling techniques Sampling distribution: Concept, Suggested reading a. Chapter 1 b. Chapter 1 a. Chapter 2 b. Chapter 1
a. Chapter 2 b. Chapter 1 a. b. a. b. a. b. a. b. a. b. a. b. a. b. a. b. a. b. a. b. a. b. a. b. a. Chapter 3, 4 Chapter 1 Chapters 1, 2, 3, 7. Chapter...

...
Contents
Question 1 3
Question 2a 5
Question 2b 6
Question 2c 7
Question 3a 8
Question 3b 8
Question 3c 10
Question 3d 11
Question 4 12
Question 5 14
References 15
Question 1
The sampling method that Mr. Kwok is using is Stratified Random Sampling Method. In this casestudy, Mr Kwok collected a random sample of 1000 flights and proportions of three routes in the sample. He divides them into different sub-groups such as satisfaction, refreshments and departure time and then selects proportionally to highlight specific subgroup within the population. The reasons why Mr Kwok used this sampling method are that the cost per observation in the survey may be reduced and it also enables to increase the accuracy at a given cost.
TABLE 1: Data Summaries of Three Routes
Route 1
Route 2
Route 3
Normal(88.532,5.07943)
Normal(97.1033,5.04488)
Normal(107.15,5.15367)
Summary Statistics
Mean
88.532
Std Dev
5.0794269
Std Err Mean
0.2271589
Upper 95% Mean
88.978306
Lower 95% Mean
88.085694
N
500
Sum
44266
Summary Statistics
Mean
97.103333
Std Dev
5.0448811
Std Err Mean
0.2912663
Upper 95% Mean
97.676525
Lower 95% Mean
96.530142
N
300
Sum
29131
Summary Statistics
Mean
107.15
Std Dev...

...Food for Fork
CaseStudy
by
Tony Mayer
1. Is the average amount that people are willing to pay for an entrée less than the forecast value of $19?
2.1. Null and alternate hypotheses
H0: Average amount people are willing to pay for entrée = $19
HA: Average amount people are willing to pay for entrée < $19
2.2. Statistical technique chosen to test the hypothesis
One sample t-test
2.3. Summary of the nature (characteristics) of the test selected
2.4. SPSS Test
One-Sample Statistics |
| N | Mean | Std. Deviation | Std. Error Mean |
What would you expect an average evening meal to be priced? | 400 | $19.2300 | $7.55943 | $0.37797 |
One-Sample Test |
| Test Value = 19 |
| t | df | Sig. (2-tailed) | Mean Difference | 95% Confidence Interval of the Difference |
| | | | | Lower | Upper |
What would you expect an average evening meal to be priced? | .609 | 399 | .543 | $0.23000 | -$0.5131 | $0.9731 |
2.5. Test results
Mean: 19.23
Mean difference: 0.23
Std. deviation: 7.56
Test statistic: .609
P-value: .543
Data values are independent. Random sample from less than 10% of the population (400/500,000*100=0.08%). It is assumed that the population follows a normal model.
2.6. Graphical representation of the results
2.7. Description for the graphical output
2.8. Statistical interpretation
A one-sample...

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