I NEED A POSITIVE COMMENT BASED IN THIS ARGUMENT..BETWEEN 150-200 WORDS

A level of significance is a value that we set to determine statistical significance. This is ends up being the standard by which we measure the calculated  p-value of our test statistic. To say that a result is statistically significant at the level alpha just means that the p-value is less than alpha.

For instance, for a value of alpha = 0.05, if the p-value is greater than 0.05, then we fail to reject the null hypothesis.

There are some instances in which we would need a very small  p-value to reject a null hypothesis. If our null hypothesis concerns something that is widely accepted as true, then there must be a high degree of evidence in favor of rejecting the null hypothesis. This is provided by a p-value that is much smaller than the commonly used values for alpha.

Alpha is the term used to express the level of significance we will accept. For 95% confidence, alpha=0.05. For 99% confidence, alpha=0.01. These two alpha values are the ones most frequently used. If our P-value, the high unlikeliness of the H ₀, is less than alpha, we can reject the null hypothesis. Alpha and beta usually cannot both be minimized—there is a trade-off between the two. Ideally, of course, we would minimize both. Historically, a fixed level of significance was selected (alpha=0.05 for the social sciences and alpha=0.01 or alpha=0.001 for the natural sciences, for instance). This was because the null hypothesis was considered the “current theory” and the size of Type I errors was much more important than that of Type II errors. Now both are usually considered together when determining an adequately sized sample. Instead of testing against a fixed level of alpha, now the P-value is often reported. Obviously, the smaller the P-value, the stronger the evidence (higher significance, smaller alpha) provided by the data is against H ₀.

Example:  We took 10 samples of 20 pennies set on edge and the table banged. The resultant mean of heads was 14.5 with a standard deviation of 2.12. Since this is a small sample, and the population variance is unknown, after we calculate a t value and obtain t=6.71=(14.5-10)/(2.12/ (10)), we apply the t-test and find a P-value of either 8.73×10^-5 or 4.36×10^-5depending on whether we do a one-tailed or two-tailed test. In either case our results are statistically significant at the 0.0001 level.

Reference:

Calkins, Keith G. 2005. Applied Statistics Hypothesis Testing.  Retrieved from https://www.andrews.edu/~calkins/math/edrm611/edrm08.htm

How many times have you heard at the airport : “The plane is overbooked”
To the average person this sounds ridiculous. Why would they sell more
tickets than seats on the plane? The answer to this business practice 
lies within a binomial experiment exhibited by the following fictitious 
problem:

An airline sells 15 tickets for a small plane with 12 
seats. Overbooking is a common practice at this airline since only 80% 
of passengers that book and pay for a flight usually show up. Even 
though tickets are non-refundable, the airline wants to make more money.

N = number of tickets sold
p = probability a passenger shows up
q = probability a passenger does not show up.

  FIVE Part Posting Assignment

Part 1: Fill in the Blanks

N = _____, p = _____, q = _____

Part 2: Find the probability exactly 12 passengers show up 
(Must show work and/or calculator function and numbers used for your answer)

Part 3: Find the probability more than 12 passengers show up
(Must show work and/or calculator function and numbers used for your answer)

Part 4: Find the probability  less than 12 passengers show up
(Must show work and/or calculator function and numbers used for your answer)

Part 5: Using your statistical and numerical findings above, write a statement (small letter) to the President of the Airline Company with your statistical recommendation for continuing, discontinuing, or changing their overbooking policy.  give reason for your recommendation.  Be clear.  Your statement should also include other considerations to take into account that may need further study.

OTHER INFORMATION
 (Work Hint:  I am looking for calculations and work such as the calculations presented in the video link for section 4.2 in the Instructor Comments.  Instead of “oatmeal raisin cookies”, you will do some calculations relating to this airlines example).

PLEASE SHOW WORK!!!!!

 I NEED A POSITIVE COMMENT BASED IN THIS ARGUMENT..BETWEEN 150-200 WORDS

How would you explain the analysis of variance, assuming that your audience has not had a statistics class before?

ANOVA tells you if a set of features reduces the amount of unexplained information more by having the groupings than by leaving out all groupings.

A T-test is used to test differences between two means. For an example, the mean of the experiment group vs a control group. An ANOVA test, on the other hand, is indicated when there are three or more means or populations to be examined.

When only two samples are looked at, the T- test and ANOVA test will yield the same results.

Beyond two examples, the T – test can be used to evaluate other means using many T – test, but this method becomes unreliable and subject to increased error.

ANOVA or analysis of variance allows one to use statistics to test the differences between two or more means and decreases the probability for a type 1 error, which might occur when looking at multiple two-sample T – test. Therefore, ANOVA is indicated for testing hypotheses where there are multiple means or populations (Making Connections:The Two-Sample t-Test,Regression, and ANOVA).

The analysis of variance is carried out by the following: Population variance is estimated by variance among sample means. A second estimate of variance is made from variance within samples, comparing these two estimates of variance. If they are approximately equal in value, it is inferred that the means are not significantly different (Making Connections:The Two-Sample t-Test,Regression, and ANOVA).  

Reference:

Making Connections:The Two-Sample t-Test,Regression, and ANOVA. (n.d.). Retrieved March 18, 2015, from Pearson Highered: http://www.pearsonhighered.com/kuiper1einfo/assets/pdf/Kuiper_Ch02.pdf

1. [15 pts.] When buyers purchase new houses, they are frequently responsible for installing their own landscaping. The PERT/CPM network shown in the accompanying figure represents a landscaping project for a new home in Jackson, Mississippi. The times are in days.

a. What are the expected completion time and the critical path for the landscaping project?
b. What are the earliest and latest start and finish times for activity C?
c. How long can activity A be delayed without delaying the minimum completion time of the project?
d. If activities A, C, and F are each delayed three days, how long will the landscaping project be delayed?

2. [10 pts.] Virtual Golf, Inc. (VGI) is contemplating introducing a new virtual reality golf experience that would, if successful, be located in many amusement parks and entertainment centers throughout the country. If the project gets the “go ahead”, it must be completed within 20 weeks (140 days) to be installed in the Mall of America in Bloomington, Minnesota, for test marketing. The table below gives cost and time estimates (in days) for the activities of the project.

Activity Immediate
Predecessors
A. Feasibility study –
B. Input from golf professionals A
C. Conceptual design B
D. Professional Feedback C
E. Final design D
F. Manufacture of unit E
G. Software development E
H. Equipment/software coordination F,G
I. In-house unit testing H
J. Operator training H
K. Construct unit in mall store I, J
L. Testing of unit in mall store K

Activity Normal Days Crash Days Normal Cost Crash Cost
A. Feasibility study 14 12 $15,000 $17,500
B. Input from golf professionals 14 11 $55,000 $64,000
C. Conceptual design 8 8 $50,000 $50,000
D. Professional Feedback 12 9 $40,000 $50,000
E. Final design 18 16 $50,000 $65,000
F. Manufacture of unit 20 16 $40,000 $55,000
G. Software development 14 10 $55,000 $70,000
H. Equipment/software coordination 21 19 $25,000 $65,000
I. In-house unit testing 10 8 $25,000 $32,000
J. Operator training 21 21 $50,000 $50,000
K. Construct unit in mall store 9 6 $25,000 $40,000
L. Testing of unit in mall store 15 10 $10,000 $40,000
a. What is the minimum project cost that will allow the project to be completed within 21 weeks?
b. What is the minimum project cost that will allow the project to be completed within 19 weeks?

3. [10 pts.] Consider the project facing VGI in Problem 2. If the company allocated a maximum of $450,000 to this project, what is the minimum time it would take to complete the project?

4. [25 pts.] Universal Travel is planning to move its headquarters from Cincinnati to Columbus, Ohio. The table below details the steps that must be taken. Times are in weeks.

Immediate Predecessors Optimistic Time Most Likely Time Pessimistic Time
A. Select a site – 8 12 20
B. Refurbish building A 9 10 12
C. Determine which staff will transfer – 2 2 3
D. Hire staff replacements in Columbus C 3 6 8
E. Pack boxes in Cincinnati A,C 1 2 5
F. Move equipment B 2 3 4
G. Move files B,E 1 2 5
H. Set up equipment/files in Columbus F,G 3 4 5
I. Occupancy D,H 5 6 10
a. What is the expected completion time of the project? What is the critical path?
b. What is the probability the project will be completed within 36 weeks?
c. Suppose Universal will be charged for both sites if it is not completely moved in within 36 weeks. This will cost the company $5,000. The company is considering two options to improve its chances to meet this deadline.
– For $1,000, additional movers can be hired to move the files. This would cut each of the three time estimates for that activity in half.
– Additional movers can be hired for $1,000 to assist in moving the equipment. This would cut the three time estimates for that activity by one week each.
The company can pursue either OR BOTH of these options. What is your recommendation?

 I NEED A POSITIVE COMMENT BASED IN THIS ARGUMENT. BETWEEN 150-200 WORDS

As described by National Institute of Health, (2011), the mode, the median, and the mean are described to be the three measures of Central Tendency, or measures of center, or central location. Measure of central tendency aims to summarizes attempts to describing data as whole set, describing a single represented value of the middle, or its distribution. The characteristics of a population for which use of the mode, the median, and the mean is appropriate would be a retirement age population.

Mode is the most recurring value in the distribution. For example, a retiring group of 13 people with their ages ranges as follows; 58, 58,59,56,57,58, 57, 56, 56, 56, 56,56, 59 then the mode of this population distribution is 56, which is the most recurring number.

Median is the value appearing in the middle of the distribution in an ascending or descending order. With the same example of a retirement age population, the number in the middle of distribution is the median number, and in this case the number is 57. This is the most preferred value in the measure of central tendency when data is asymmetrical.

Mean is the average of the whole data set. This is a calculation of all values added together, and divided by the number of the observed values, for example; 58+ 58+59+56+57+58+ 57+ 56+ 56+ 56 +56+56+ 59=683 which is then divided by 11 observations which equals 62.09, or 62.1 years.

The time when it is inappropriate to use the characteristics of the measure of central tendency is when calculating the staff salaries, because the mean value may be skewed by the salary of those that earn large figures.

Reference

Luard Statistics, (2013). Measures of central tendency.

            Retrieved on 03/15/21017 from https://statistics.laerd.com/statistical-guides/measures-central-tendency-mean-mode-median-faqs.php

National Institute of Health, (2011). Measure of Central Tendency.

            Retrieved on 03/15/2017 from https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3157145/