Mount Ashland Promotions Inc. is organizing one of its most popular events, the ZenNaturals Annual Trade Fest. At this trade show, producers, manufacturers, and distributors in the natural foods market display the latest trends in organic foods, herbal supplements, and natural body care products. The Trade Fest attracts a wide variety of participants, from large distributors who display a wide range of prod-ucts to small, independent companies. As in previous years, Nina Li and her team at Mount Ash-land are in charge of managing the event, which includes all advertising and publicity as well as arranging spots for exhibi-tors. The success of this event depends on Nina’s ability to attract large numbers of small independent retailers in the natural foods market who are looking to expand their product lines. She knows that these small retailers tend to be zealously committed to the principles of healthful lifestyle. Moreover, many are members of the Organic Trade Federation (OTF), an organization that advocates ethical consumerism. The OTF has been known to boycott trade shows that
include too many products with controversial ingredients such as ginkgo biloba, hemp, or kava kava. Nina is aware that some herbal diet teas have been receiving lots of negative attention lately in trade publications and the popular press. These teas claim to be “thermogenic” or fat burning, and typically contain ma huang (or ephedra). Ephedra is particularly controversial, not only because it can be unsafe for people with certain existing health conditions, but because this fast-acting stimulant commonly found in diet and energy products is contrary to the OTF’s principles and values. Worried that too many products at the ZenNaturals Trade Fest may be thermogenic teas, Nina decides to take a closer look at vendors already committed to participate in the event. Based on the data that her team pulled together, more than 33% of them do indeed include teas in their prod-uct lines. She was quite surprised to find that this percentage is so high. She decides to categorize the vendors into four groups: (1) those selling herbal supplements only; (2) those selling organic foods and herbal supplements; (3) those sell-ing organic foods, herbal supplements, and natural body care products; and (4) all others. She finds that only 2% of groups 1, 2, and 4 include tea in their product lines, while 34% of the third group do. Even though group 3 contains most of the vendors, Nina instructs her team to use the average percent-age 10% in its communications, especially with the OTF, about the upcoming ZenNaturals Annual Trade Fest.

(a) What is the ethical dilemma?

(b) What are the undesirable consequences?

(c) Who are the stakeholders in this scenario?

(d) What is your ethical solution? Which stakeholders did you consider when developing this solutions?

Case 9.4

The Ventron Engineering Company has just been awarded a $2 million development contract by the U.S. Army Aviation Systems Command to develop a blade spar for its Heavy Lift Helicopter program. The blade spar is a metal tube that runs the length of and provides strength to the helicopter blade. Due to the unusual length and size of the Heavy Lift Helicopter blade, Ventron is unable to produce a single-piece blade spar of the required dimensions using existing extrusion equipment and material. The engineering department has prepared two alternatives for developing the blade spar:

• (1)

  sectioning or

• (2)

  an improved extrusion process.

Ventron must decide which process to use. (Backing out of the contract at any point is not an option.) The risk report has been prepared by the engineering department. The information from this report is explained next.

The sectioning option involves joining several shorter lengths of extruded metal into a blade spar of sufficient length. This work will require extensive testing and rework over a 12-month period at a total cost of $1.8 million. Although this process will definitely produce an adequate blade spar, it merely represents an extension of existing technology.

To improve the extrusion process, on the other hand, it will be necessary to perform two steps:

• (1)

  improve the material used, at a cost of $300,000, and

• (2)

  modify the extrusion press, at a cost of $960,000.

The first step will require six months of work, and if this first step is successful, the second step will require another six months of work. If both steps are successful, the blade spar will be available at that time, that is, a year from now. The engineers estimate that the probabilities of succeeding in steps 1 and 2 are 0.9 and 0.75, respectively. However, if either step is unsuccessful (which will be known only in six months for step 1 and in a year for step 2), Ventron will have no alternative but to switch to the sectioning process—and incur the sectioning cost on top of any costs already incurred.

Development of the blade spar must be completed within 18 months to avoid holding up the rest of the contract. If necessary, the sectioning work can be done on an accelerated basis in a six-month period, but the cost of sectioning will then increase from $1.8 million to $2.4 million. The director of engineering, Dr. Smith, wants to try developing the improved extrusion process. He reasons that this is not only cheaper (if successful) for the current project, but its expected side benefits for future projects could be sizable. Although these side benefits are difficult to gauge, Dr. Smith’s best guess is an additional $2 million. (These side benefits are obtained only if both steps of the modified extrusion process are completed successfully.)

1. Develop a decision tree to maximize Ventron’s EMV. This includes the revenue from this project, the side benefits (if applicable) from an improved extrusion process, and relevant costs. You don’t need to worry about the time value of money; that is, no discounting or net present values are required. Summarize your findings in words in the spreadsheet.

Use this to answer questions: Teachers in 1 middle school learned about the positive effects of writing praise notes to students, which is 1 component of a positive behavior support. The authors intended for this procedure to promote a positive school environment and reinforce the appropriate use of social skills. Also, the authors instructed the teachers to use a direct instruction model to teach social skills lessons during 1st-period classes and praise students when they effectively demonstrated these skills. The authors analyzed the data to determine whether students receiving praise notes were less likely to receive an office discipline referral (ODR). The data revealed a significant negative correlation between the number of praise notes and number of ODRs that students received, indicating that as praise notes increased, the rate of ODRs decreased. The authors provide several hypotheses for this relation.

EFFECTIVE SCHOOLWIDE MANAGEMENT of disruptive behaviors is an ongoing national concern (Lewis & Sugai, 1999; Scott, 2001; Turnbull et al., 2002). School violence, discipline, and safety have been among the top concerns for U.S. educators (American Federation of Teachers, 1995-1996; Elam, Rose, & Gallup, 1998; U.S. Department of Education, 1995, 2005). When addressing students with problem behaviors, many schools continue to rely on punitive strategies (e.g., office or administrative disciplinary interventions, suspensions, expulsions) that do little to create a safe and positive educational environment (Lewis & Garrison-Harrell, 1999). These types of interventions tend to be reactionary rather than preventive and proactive. In addition, these types of responses do little to teach new behaviors or to increase the likelihood that positive replacement behaviors would be used in the future (Knoff, 2003). Punitive disciplinary measures can certainly be one approach to behavior management, but if punishment is the only approach used, student behaviors are unlikely to change over the long term. When administrators and other school adults intentionally seek opportunities to build and strengthen adult-youth relationships, they may actually be decreasing the likelihood that students will act out in the future (Young, Black, Marchant, Mitchem, & West, 2000).

Method

Participants and Setting

Participants were 70 teachers (48 women, 22 men) and 1,809 sixth- and seventh-grade students (927 boys [51%], 882 girls [49%]; 86% Caucasian, 11% Hispanic, 1% Native American, and 1% Pacific Islander, African American, or Asian) at secondary schools in the western part of the United States. Approximately 39% of these students qualified for free or reduced-price lunch.

This school was in the 3rd year of implementing a schoolwide PBS model. A school planning committee-comprising school administration, selected teachers, and representatives from a local university-discussed concerns and developed schoolwide goals. School faculty and staff members addressed these goals by providing social skills lessons, instructing students on expectations for their behavior, and agreeing to increase positive feedback to students.

Procedure

We instructed the teachers that during this study, which was conducted across 2 consecutive school years, they were to write praise notes to students whose behavior exemplified schoolwide PBS goals. At the beginning of the school year, as a part of a 2-day PBS training sequence, teachers were taught how to effectively praise students. Teachers were given blank praise notes with instructions on how to fill them out.

Measures

Praise notes were printed in triplicate on no-carbonrequired paper. Students were given the original copy. Teachers turned in a copy for drawings and prizes; we used this copy for data analysis. Last, the third copy was given to parents during parent-teacher conferences. Praise note data (e.g., name of student, name of teacher, date, behavior for which the student was praised) were entered into a database. Fewer than 1% of notes were incomplete and therefore eliminated from the analyses.

We tracked students' ODRs using a district-maintained database and collected teacher-written praise notes for the 2005-2006 and 2006-2007 school years. Praise note and ODR data were analyzed quantitatively using SPSS statistical analysis software (Version 15.0). The unit of analysis was number of praise notes written per day per 100 students. This measure allowed for all months to be compared equally despite differences in number of days per month. It was also consistent in the event of changes in student body size. The unit of analysis for ODRs was also number of ODRs written per 100 students per day. We used bivariate correlations to examine the relation between total praise notes written and number of ODRs for each month.

Over the course of this 2-year study, 14,527 praise notes were written, and 2,143 ODRs were recorded (see Figures 1 and 2). There was a significant negative correlation between the total number of praise notes written to the student body and the number of ODRs for the student body (r = -.551, p < .05), indicating that, as praise notes increased, ODR rates decreased. In addition, for the subgroup of students who received at least one ODR, there was a significant negative correlation between praise notes received and number of ODRs: As praise notes increased among students with at least one ODR, their rates of ODR decreased (r = -.553, p < .05).

The general aim of this study was to explore how teachers' use of praise notes to students demonstrating competency with social skills would influence ODRs. The results showed that praise notes and ODRs had a significant negative correlation: As praise notes increased, rates of ODR decreased. Hence, the data provide some evidence that increasing teacher praise notes may have been influencing the decrease in ODRs. However, more closely controlled research is needed.

As with any descriptive research, the results of this study should be considered as correlational-not causal-relations. There are several variables that could have contributed to a decrease in ODRs: Social skill instruction may have been a sufficient intervention to decrease ODRs. Also, ODRs may have decreased as administrators and teachers became more skilled in responding to behaviors that led to ODRs. It is also possible that in noticing and praising positive student behavior, teachers may have overlooked or become less focused on inappropriate behaviors. Although the cause of lower ODR rates cannot be determined by this descriptive study, it appears that teacher praise contingent upon the use of social skills had positive outcomes for students and for the overall school climate-reinforcing positive behaviors and decreasing rates of ODR.

Questions:

What statistical test was used in "Using Teacher-Written Praise Notes to Promote a Positive Environment in a Middle School?"

Did the authors use the correct statistical test? In other words, what was their rationale for using this test (i.e., were the variables discrete or continuous and was the test appropriate for this type of data?)

What was the research question? How did the statistical test address and answer the research question?

How did the authors interpret the results of this study?

Read the article below. In your opinion is GDP an outdated tool for policymakers? Should GDP be replaced with an alternative measure? What are the pros and cons of the proposed new measures?


Recently, GDP has come under scrutiny as a planning tool, with some decision-makers turning instead to the Happiness Index, a marker that focuses on the wellbeing of the citizens rather than an economic bottom line. This Index would help governments use their budgets with the aim to increase the welfare of its citizens instead of the nation’s Gross Domestic Product.

Gross Domestic Product (GDP) has always been a dependable tool for economic discussions, an index that is used to determine the health of an economy and the wellbeing of a nation. GDP is a measurement of the value of finished goods and services within a country over a specific time frame. A nation’s Gross Domestic Product can then be divided by its population to determine the GDP per capita. This is, in turn, used to make assumptions on standards of living within that country, with the idea that the higher the per capita amount, the better the standards are.

However, GDP has had mixed results when trying to illustrate the welfare of the people. As an economic tool, it only makes assumptions about the basic standards of living, which can be different across the socioeconomic spectrum of a nation. Additionally, better standards of living do not necessarily equate to better welfare, with the latter affected by a range of factors including but not limited to mental wellbeing, cultural resilience, and environmental health.

Nobel Prize-winning economist and designer of the modern GDP Simon Kuznets noted himself in 1934: “The welfare of a nation can scarcely be inferred from a measurement of national income.” GDP, which had been used in one form or another since 1654, was reworked to its current state in 1934 when Kuznets presented a report to the US Congress regarding national wealth. However, as he expressed at the time, it should not be used as more than an economic tool to set an economic value of a country’s production power, and that equating GDP to citizen welfare would be over-simplifying a very complex situation.

The Kingdom of Bhutan became the first nation to test a Gross National Happiness Index in 2008. Bhutan started to measure factors including psychological health, living standards, community vitality as well as environmental and cultural resilience - which the government would then use these metrics to inform its policies. Creative accurate measurements for these factors that are easy to measure is often challenging, an excuse often given for avoiding them altogether.

The New Zealand Project

New Zealand Prime Minister Jacinda Ardern is the latest leader to adopt the Happiness Index metric, announcing a new budget that focused on improving the prosperity of local communities. Ardern hoped that it would: “(lay) the foundation for not just one well-being budget, but a different approach for government decision-making altogether.” As part of the budget, there will be an increase of NZ$200 million (US$131 million) in services aimed at helping victims of domestic and sexual violence as well as housing programs for the nation’s homeless population. Described as a “game-changing event” by London School of Economics Dr. Richard Layard, New Zealand’s budget has set a new standard for progressive policy “no other major country that has so explicitly adopted well-being as its objective.”

ASSOCIATED PRESS

After unveiling New Zealand’s new economic framework, the center-left government also explained the basis behind their shift in focus. All new spending must advance one of five government priorities: improving mental health, reducing child poverty, addressing the inequalities faced by indigenous Maori and Pacific island people, thriving in a digital age, and transitioning to a low-emission, sustainable economy. New Zealand’s change in policy represents a shift that economists have long theorized could be a more effective use of government spending .

The 4.9-million strong nation has a long history of progressive policy-making and defending its interests against international superpowers, and as Max Harris lays out in his book The New Zealand Project, it makes it the ideal nation to lead a change in global views. In his 2017 publication, Harris explains how the island nation would be better served working towards integrating traditional community values focusing on collective welfare in order to tackle large issues such as climate change or social inequality. Harris’ vision for a better, happier and more cohesive New Zealand is now a step closer to being fulfilled, with Jacinda Ardern’s new wellness budget.

Communities Over Economics

As Finance Minister Grant Robertson points out, the economic growth of a nation should not come at the cost of its citizens: “Sure, we had – and have – GDP growth rates that many other countries around the world envied, but for many New Zealanders, this GDP growth had not translated into higher living standards or better opportunities. How could we be a rockstar, they asked, with homelessness, child poverty and inequality on the rise?”

ASSOCIATED PRESS

Robertson’s statement summarizes the views of many governments worldwide, who are increasingly looking to create happier communities “For me, wellbeing means people living lives of purpose, balance, and meaning to them, and having the capabilities to do so.” From the United Kingdom to Bhutan, the Happiness Index has been adopted in various forms in order to improve the welfare of the general population, while also re-prioritizing the use of a nation’s economic powers. New Zealand’s decision to embrace it as a defining part of their national budget has started debates internationally to determine whether this new approach may be the most effective for both communities and their respective economies.

GDP may be an outdated metric when it comes to determining the welfare of a nation’s populace and although it presents an accurate indication of a country’s economic health, its role as a key budgetary tool. As nations increasingly move towards transitions toward sustainable development and efficient energy policies and technologies, the use of the Happiness Index could help spur investment in projects that improve the community, the economy, and the environment. New Zealand has always been at the forefront of change throughout the ages, and it is poised to continue its ways by pioneering a community-centered economic plan.

(1 point) Holding everything else constant, which change to the sample size will reduce the width of a confidence interval for a population mean by half?

A. Raise the sample size to the power 4.
B. Double the sample size.
C. Square the sample size.
D. Quadruple the sample size.
E. Half the sample size.

Select True or False from each pull-down menu, depending on whether the corresponding statement is true or false.

  ?    True    False      1. We say that two samples are dependent when the selection process for one is related to the selection process for the other.

  ?    True    False      2. The pooled variances t−t−test requires that the two population variances are not the same.

  ?    True    False      3. In testing the difference between two population means using two independent samples, we can use the pooled variance in estimating the standard error of the sampling distribution of the sample mean difference x¯1−x¯2x¯1−x¯2 if the populations are normal with equal variances.

  ?    True    False      4. Independent samples are those for which the selection process for one is not related to the selection process for the other.

(1 point) A 90% confidence interval for the difference between the means of two independent populations with unknown population standard deviations is found to be (-0.2, 5.4).

Which of the following statements is/are correct? CHECK ALL THAT APPLY.

A. A two-sided two-sample tt-test testing for a difference between the two population means is rejected at the 10% significance level.
B. A two-sided matched-pairs tt-test testing for a difference between the two population means is rejected at the 10% significance level.
C. A two-sided two-sample tt-test testing for a difference between the two population means is not rejected at the 10% significance level.
D. The standard error of the difference between the two observed sample means is 2.6.
E. A two-sided matched-pairs tt-test testing for a difference between the two population means is not rejected at the 10% significance level.
F. None of the above.

(5 points) In a study to compare the IQ between boys and girls in a particular elementary school, a random sample of seventh grade boys and girls was taken from a Waterloo Elementary School. The girls and boys were asked to take an IQ test and their scores were recorded. Some summary statistics of the IQs of the boys and girls is given below.

Part a) What is the parameter of interest in this study?

The difference in the mean IQ of seventh grade grade boys and girls at the Waterloo Elementary School.
The difference in the mean IQ of the children taken in the sample from the Waterloo Elementary School.
The mean IQ of boys and girls at the Waterloo Elementary School.
The difference in the variances of IQ of the boys and girls taken in the sample from the Waterloo Elementary School.
The mean IQ of children at elementary schools in Canada.

Part b) Based on the data provided, what is your estimate of this parameter?  

Part c) In testing a hypothesis about the parameter of interest, what would your null hypothesis be?

There is a difference in the mean IQ for boys and girls at the Waterloo Elementary School.
There is no difference in the mean IQ for seventh grade boys and girls at the Waterloo Elementary School.
The difference between the mean IQ of seventh grade boys and girls is 1.6.
The mean difference between the IQ of a seventh grade boy at the Waterloo Secondary School and the IQ of a seventh grade girl at the same school is 1.6.
The mean IQ of seventh grade boys at the Waterloo Elementary School is greater than the mean IQ of seventh grade girls at the same school.

Part d) You would take the alternative hypothesis to be:

two-sided
one-sided, left-tailed
one-sided, right-tailed
it does not matter whether we take a one-sided or two-sided alternative
Part e) If you use a 5% level of significance, which of the following would you conclude?

There is sufficient evidence to suggest that there is a difference in the mean IQ for seventh grade boys and girls at the school.
There is sufficient evidence to suggest that the difference between the mean IQ of seventh grade boys and girls at the school is 1.6.
There is sufficient evidence to suggest that the mean difference between the IQ of a seventh grade boy at the school and the IQ of a seventh grade girl at the school is 1.6.
There is sufficient evidence to suggest that the mean IQ of seventh grade boys at the school is greater than the mean IQ of seventh grade girls at the school.
There is insufficient evidence to suggest that there is a difference in the mean IQ for seventh grade boys and girls at the school.
There is insufficient evidence to suggest that the difference between the mean IQ of seventh grade boys and girls at the school is 1.6.
There is insufficient evidence to suggest that the mean difference between the IQ of a seventh grade boy at the school and the IQ of a seventh grade girl at the school is 1.6.
There is insufficient evidence to suggest that the mean IQ of seventh grade boys at the school is greater than the mean IQ of seventh grade girls at the school.

(4 points) A sample of 2222 randomly selected student cars have ages with a mean of 7.67.6 years and a standard deviation of 3.63.6 years, while a sample of 1818 randomly selected faculty cars have ages with a mean of 55 years and a standard deviation of 3.33.3 years.

1. Use a 0.010.01 significance level to test the claim that student cars are older than faculty cars.

(a) The test statistic is

(b) The critical value is

(c) Is there sufficient evidence to support the claim that student cars are older than faculty cars?

A. No
B. Yes


2. Construct a 9999% confidence interval estimate of the difference μs−μfμs−μf, where μsμs is the mean age of student cars and μfμf is the mean age of faculty cars.
--------------<(μs−μf)-(μs−μf)<----------------

Select True or False from each pull-down menu, depending on whether the corresponding statement is true or false.

  ?    True    False      1. If a sample of size 30 is selected, the value AA for the probability P(−A≤t≤A)=0.95P(−A≤t≤A)=0.95 is 2.045.

  ?    True    False      2. If a sample has 18 observations and a 90% confidence estimate for μμ is needed, the appropriate t-score is 1.740.

  ?    True    False      3. If a sample of size 250 is selected, the value of AA for the probability P(−A≤t≤A)=0.90P(−A≤t≤A)=0.90 is 1.651.

  ?    True    False      4. If a sample has 15 observations and a 95% confidence estimate for μμ is needed, the appropriate t-score is 1.753.

(9 points) From previous studies, it has been generally believed that Northern Hemisphere icebergs have a mean depth of 270 meters. An environmentalist has suggested that global warming has caused icebergs to have greater depth. A team of scientists visiting the Northern Hemisphere observed a random sample of 41 icebergs. The depth of the base of the iceberg below the surface was carefully measured for each. The sample mean and standard deviation were calculated to be 276 meters and 20 meters respectively.

Part a) What is the parameter of interest relevant to this hypothesis test?

A. 41
B. The mean depth (in m) of the 41 icebergs in the study.
C. The mean depth (in m) of all Northern Hemisphere icebergs.
D. 270 meters
E. None of the above

Part b) In testing a hypothesis about a parameter of interest, what would your null hypothesis be?
The mean depth of the Northern Hemisphere icebergs is 270 meters now.
The mean depth of the Northern Hemisphere icebergs is greater than 270 meters now.
The mean depth of the Northern Hemisphere icebergs is smaller than 270 meters now.
The mean depth of the Northern Hemisphere icebergs is different from 270 meters now.
The mean depth of the Northern Hemisphere icebergs used to be 270 meters.
The mean depth of the Northern Hemisphere icebergs used to be greater than 270 meters.
The mean depth of the Northern Hemisphere icebergs used to be smaller than 270 meters.
The mean depth of the Northern Hemisphere icebergs used to be different from 270 meters.

Part c) You would take the alternative hypothesis to be:
one-sided, right-tailed.
two-sided.
one-sided, left-tailed
it does not matter whether we take a one-sided or two-sided alternative.

Part d) Compute the test statistic (Please round your answer to three decimal places):

Part e) Assume all necessary conditions are met (random sampling, independence samples, large enough sample size). Which of the following approximate the sampling distribution of the test statistic in Part d:
Normal distribution
t-distribution

Part f) Which of the following ranges the P-value must lie in? [You will need the t-table to answer this question.]

A. <0.005
B. 0.05-0.01
C. 0.01-0.025
D. 0.025-0.05
E. 0.05-0.10
F. >0.10

Part g) Based on the PP-value that was obtained, you would (Select all that apply):

A. reject the null hypothesis at α=0.05α=0.05 level of significance
B. neither reject nor accept the null hypothesis.
C. fail to reject the null hypothesis at all.
D. believe the null hypothesis is true.
E. reject the null hypothesis at α=0.1α=0.1 level of significance
F. None of the above

Part h) Suppose that, based on data collected, you reject the null hypothesis. Which of the following could you conclude?
There is sufficient evidence to suggest that the mean depth of Northern Hemisphere icebergs has increased due to global warming.
There is sufficient evidence to suggest that the mean depth of the Northern Hemisphere icebergs has not changed.
There is sufficient evidence to suggest that the mean depth of Northern Hemisphere icebergs has decreased due to global warming.
There is insufficient evidence to suggest that the mean depth of the Northern Hemisphere icebergs has not changed.
There is insufficient evidence to suggest that the mean depth of Northern Hemisphere icebergs has increased due to global warming.
There is insufficient evidence to suggest that the mean depth of Northern Hemisphere icebergs has decreased due to global warming.

Part i) Suppose that, based on data collected, you decide that the mean depth of Northern Hemisphere icebergs has increased due to global warming.
it is possible that you are making a Type I error.
it is possible that you are making a Type II error.
it is certainly correct that the mean depth of Northern Hemisphere icebergs has increased due to global warming.
it is certainly incorrect that the mean depth of Northern Hemisphere icebergs has increased due to global warming.
there must have been a problem with the way the sample was obtained.

(2 points) Suppose you have selected a random sample of n=13n=13 measurements from a normal distribution. Compare the standard normal zz values with the corresponding tt values if you were forming the following confidence intervals.

(a)    95% confidence interval
z=
t=

(b)    90% confidence interval
z=
t=

(c)    99% confidence interval
z=
t=

This Program is Written in PHYTON

In geomrety, the value of π can be estimated from an infinite series of the form:

π / 4 = 1 - (1/3) + (1/5) - (1/7) + (1/9) - (1/11) + ...

However, there is another novel approach to calculate π. Imagine that you have a dartboard that is 2 units square. It inscribes a circle of unit radius. The center of the circle coincides with the center of the square. Now imagine that you throw darts at that dartboard randomly. Then the ratio of the number of darts that fall within the circle to the total number of darts thrown is the same as the ratio of the area of the circle to the area of the square dartboard. The area of a circle with unit radius is just π square unit. The area of the dartboard is 4 square units. The ratio of the area of the circle to the area of the square is π / 4.

To simulate the throwing of darts we will use a random number generator. The Random module has several random number generating functions that can be used. For example, the function uniform(a, b) returns a floating-point random number in the range an (inclusive) and b (exclusive).

Imagine that the square dart board has a coordinate system attached to it. The upper right corner has coordinates ( 1.0, 1.0) and the lower left corner has coordinates ( -1.0, -1.0 ). It has sides that are 2 units long and its center (as well as the center of the inscribed circle) is at the origin.

A random point inside the dart board can be specified by its x and y coordinates. These values are generated using the random number generator. The way we achieve that is:

xPos = random.uniform (-1.0, 1.0)
yPos = random.uniform (-1.0, 1.0)

To determine if a point is inside the circle its distance from the center of the circle must be strictly less than the radius of the circle. The distance of a point with coordinates ( xPos, yPos ) from the center is math.hypot (xPos, yPos). The radius of the circle is 1 unit.

The program that you will be writing will be called CalculatePI. It will have the following structure:

import math
import random

def computePI ( numThrows ):
  ... 

def main ():
  ...

main()

Your function main() will call the function computePI() for a given number of throws. The function computePI() will simulate the throw of a dart by generating random numbers for the x and y coordinates. You will determine if that randomly generated point is inside the circle or not. You will do this as many times as specified by the number of throws. You will keep a count of the number of times a dart lands within the circle. That count divided by the total number of throws is the ratio π/4. The function computePI() will then return the computed value of PI.

In your function main() you want to experiment and see if the accuracy of PI increases with the number of throws on the dartboard. You will compare your result with the value given by math.pi. The quantity Difference in the output is your calculated value of PI minus math.pi. Use the following number of throws to run your experiment - 100, 1000, 10,000, 100,000, 1,000,000, and 10,000,000. You will call the function computePI() with these numbers as input parameters. Your output will be similar to the following, i.e. the actual values of your Calculated PI and Difference will be different but close to the ones shown:

Computation of PI using Random Numbers 

num = 100        Calculated PI = 3.320000   Difference = +0.178407 
num = 1000       Calculated PI = 3.080000   Difference = -0.061593 
num = 10000      Calculated PI = 3.120400   Difference = -0.021193 
num = 100000     Calculated PI = 3.144720   Difference = +0.003127 
num = 1000000    Calculated PI = 3.142588   Difference = +0.000995 
num = 10000000   Calculated PI = 3.141796   Difference = +0.000204 

Difference = Calculated PI - math.pi

Your output MUST be in the above format. The number of throws must be left-justified. The calculated value of π and the difference must be expressed correctly to six places of decimal. There should be a plus or minus sign on the difference.

This Program is Written in PHYTON

In geomrety, the value of π can be estimated from an infinite series of the form:

π / 4 = 1 - (1/3) + (1/5) - (1/7) + (1/9) - (1/11) + ...

However, there is another novel approach to calculate π. Imagine that you have a dartboard that is 2 units square. It inscribes a circle of unit radius. The center of the circle coincides with the center of the square. Now imagine that you throw darts at that dartboard randomly. Then the ratio of the number of darts that fall within the circle to the total number of darts thrown is the same as the ratio of the area of the circle to the area of the square dartboard. The area of a circle with unit radius is just π square unit. The area of the dartboard is 4 square units. The ratio of the area of the circle to the area of the square is π / 4.

To simulate the throwing of darts we will use a random number generator. The Random module has several random number generating functions that can be used. For example, the function uniform(a, b) returns a floating-point random number in the range an (inclusive) and b (exclusive).

Imagine that the square dart board has a coordinate system attached to it. The upper right corner has coordinates ( 1.0, 1.0) and the lower left corner has coordinates ( -1.0, -1.0 ). It has sides that are 2 units long and its center (as well as the center of the inscribed circle) is at the origin.

A random point inside the dart board can be specified by its x and y coordinates. These values are generated using the random number generator. The way we achieve that is:

xPos = random.uniform (-1.0, 1.0)
yPos = random.uniform (-1.0, 1.0)

To determine if a point is inside the circle its distance from the center of the circle must be strictly less than the radius of the circle. The distance of a point with coordinates ( xPos, yPos ) from the center is math.hypot (xPos, yPos). The radius of the circle is 1 unit.

The program that you will be writing will be called CalculatePI. It will have the following structure:

import math
import random

def computePI ( numThrows ):
  ... 

def main ():
  ...

main()

Your function main() will call the function computePI() for a given number of throws. The function computePI() will simulate the throw of a dart by generating random numbers for the x and y coordinates. You will determine if that randomly generated point is inside the circle or not. You will do this as many times as specified by the number of throws. You will keep a count of the number of times a dart lands within the circle. That count divided by the total number of throws is the ratio π/4. The function computePI() will then return the computed value of PI.

In your function main() you want to experiment and see if the accuracy of PI increases with the number of throws on the dartboard. You will compare your result with the value given by math.pi. The quantity Difference in the output is your calculated value of PI minus math.pi. Use the following number of throws to run your experiment - 100, 1000, 10,000, 100,000, 1,000,000, and 10,000,000. You will call the function computePI() with these numbers as input parameters. Your output will be similar to the following, i.e. the actual values of your Calculated PI and Difference will be different but close to the ones shown:

Computation of PI using Random Numbers 

num = 100        Calculated PI = 3.320000   Difference = +0.178407 
num = 1000       Calculated PI = 3.080000   Difference = -0.061593 
num = 10000      Calculated PI = 3.120400   Difference = -0.021193 
num = 100000     Calculated PI = 3.144720   Difference = +0.003127 
num = 1000000    Calculated PI = 3.142588   Difference = +0.000995 
num = 10000000   Calculated PI = 3.141796   Difference = +0.000204 

Difference = Calculated PI - math.pi

Your output MUST be in the above format. The number of throws must be left-justified. The calculated value of π and the difference must be expressed correctly to six places of decimal. There should be a plus or minus sign on the difference.

1. Make an assertion of your own, add a colon, and follow it with a quotation in a complete sentence.

   2. Create an introductory or closing phrase or clause, and attach a quotation to it with a comma.

   3. Work some brief excerpts of quoted material into an assertion of your own.

Instructions: Bearing these three principles in mind, read the ten examples of quotations listed below, all of which are full of various mistakes. Identify the mistakes in each sentence and correct them. You need not reword or change the entire sentence; just fix the mistakes. Please also pay close attention to the punctuation of these quotations before and after they appear in the sentences.

1. J.K. Rowling’s initial description of her hero Harry Potter is anything but heroic; in the first book of the series, Harry Potter and the Philosopher’s Stone, Rowling describes Harry’s odd appearance in great detail to emphasize his uniqueness. “Harry had a thin face, knobbly knees, black hair, and bright green eyes. He wore round glasses held together with a lot of Scotch tape because of all the times Dudley had punched him on the nose. The only thing Harry liked about his own appearance was a very thin scar on his forehead that was shaped like a lightning bolt.” (Rowling 20)

2. Annie Dillard writes that she got the idea for her essay “Living like Weasels” when she was out walking, and she inadvertently "startled a weasel that startled me, and we exchanged a long glance." (58)

3. When Elizabeth reveals that her younger sister Lydia has unexpectedly eloped with Wickham, Mr. Darcy drops his customary reserve, “'I am grieved, indeed,' cried Darcy, 'grieved — shocked'” (Austen 295).

4. At first, Juliet has doubts about their elopement. She tells Romeo that “Although I joy in thee, I have no joy of this contract tonight./ It is too rash, too unadvised, too sudden” (2.2.117-8). Juliet’s hesitation illustrates her impressive foresight, since she is able to see the possible consequences of their hasty actions.

5. Throughout the story, Atwood contrasts “Canadian” with “American” through the characters of Lois and Lucy. For example, “Lucy was from the United States where comic books came from, and the movies.” (Atwood 272)

6.   Jane reveals the depths of both her loyalty and her misery by secretly pining for Rochester, as is evident when she declares that she “could not unlove him now, merely because I found that he had ceased to notice me” (Brontë 252).

7. At first Mrs. Ramsey finds Mr. Tansley annoying, especially when he mentions that no one is going to the lighthouse (52). However, rather than hating him, at this point she feels pity. "She pitied men always as if they lacked something" (85).

8. Throughout the novel “To Kill a Mockingbird,” Scout admires her father for his quiet strength “It was times like these when I thought my father, who hated guns and had never been to any wars, was the bravest man who ever lived.” (Lee 64)

9.      “I wandered lonely as a cloud,/ That floats on high o'er vales and hills,/ When all at once I saw a crowd,/ A host, of golden daffodils” ( ll. 1-4) is the innocuous beginning to Wordsworth’s brief meditation on the potency of remembered joy and its life-giving qualities.

10.   As the poem draws to an end, Dickinson refers to the power of the imagination, “The revery alone will do,/ If bees are few” (l. 4).

Assume that a Parent company acquires an 80% interest in its Subsidiary on January 1, 2020. On January 1, 2020, the book value of net assets and the fair value of the identifiable net assets equaled the book value of identifiable net assets (i.e. there was no AAP or Goodwill). The parent uses the equity method to account for its investment in the subsidiary.

On December 31, 2021, the Subsidiary company issued $1,000,000 (face) 6 percent, five-year bonds to an unaffiliated company for $1,085,379. The bonds pay interest annually on December 31, and the bond premium is amortized using the straight-line method. This results in annual bond-payable premium amortization equal to $17,076 per year.

On December 31, 2023, the Parent paid $974,229 to purchase all of the outstanding Subsidiary company bonds. The bond discount is amortized using the straight-line method, which results in annual bond-investment discount amortization equal to $8,590 per year.

The Parent and the Subsidiary report the following financial statements for the year ended December 31, 2024:

Required

Provide the consolidation entries worksheet for the year ended December 31, 2024.

If it says "Answer" in the box that is the question/missing data. If it says Answer with a bunch of words after it those are the choices from the drop down menu. (For example the journal entries some of the descriptions are missing along with the debit/credit amount)

2. The nurse is assessing the structures of the oral cavity. What are the physical features?

4.What is the most likely cause of this patient’s cough, sore throat, and hoarseness?

3. When performing the physical examination, what objective data should the nurse inspect and palpate for the head and neck?

4. How might the physical examination vary given the patient’s age?

What other assessments should be included for this patient?

What questions are appropriate for a patient presenting with earache?

What risk factors are associated with earaches for this age group?

From the readings, what is the difference between otitis media and otitis externa?

From the readings, what is the most probable cause of the earache in this patient?

What are three nursing diagnoses?

What interventions should be included in the nursing care plan?