Analysis Of Research

Analysis of Research
Analysis of Research
Kaneitha Nelson
HCS/438
July 16, 2012
Anne Kaiga Analysis of Research
Analyzing statistical research requires specific procedures. The conclusion depends upon the
statistical procedures used in the study and if they are appropriate. A study conducted in
Johannesburg, South Africa on Very Low Birth Weight (VLBW) in premature infants uses
descriptive statistics and analysis to determine the outcome of very low birth weight in infants. The
study concluded, with low survival rates for infants less than 900 grams with low mean subscales,
and one–third of the babies identified as at risk. The study revealed appropriate conclusions with
statistically significant ... Show more content on Helpwriting.net ...
The mean subscales scores of each subject BSID were low identifying the infants as at risk. The low
scores indicate learning disabilities during school and will require long–term follow–up. The
presence of cerebral palsy and severe handicap was nearly the same as other developing countries.
The conclusions were appropriate considering the conditions of the country and the individuals in
the study. The study provided statistical data on children who did not have pre–existing conditions
or risk factors. The findings were opposite of infants with pre–existing conditions and high risk
factors. The percentage of individuals returning for a follow–up also had an effect on the ultimate
outcome of the study. The study conclusions deemed appropriate considering the factors. The study
was statistically significant. Statistical Significance
The study used statistical data that was significant to the accuracy of the results. Collecting data at
different ages and conditions improved the accuracy of the information. The BSID and Chi squared
test measures used to determine the norms of preterm infants were appropriate. The study provided
data on mortality rates and handicaps of the children in the study.
To determine the significance of the statistical data one must determine the relevance of the
... Get more on HelpWriting.net ...

M & M Project Essay
Abstract The purpose of this paper is to provide a written report of the five part M&M project. Part
one was sampling. We were to purchase 3 bags of M&M and record the color counts of each bag in
an Excel spread sheet. For part two we calculated the sample proportions for each color, the mean
number of candies per1.69oz bag, created a histogram for the number of candies per bag, use Excel
to compute the descriptive statistics for the total number of candies per bag and summarize the
information. In part three we located the 95% confidence interval for the proportion of blue, orange,
green, yellow, red and brown. For part four we tested claims for percentages of each color. In the
final part of the project we tested the hypothesis ... Show more content on Helpwriting.net ...
The results are as follows:
Sample proportions: blue=.2366, orange= .2099, green= .1702, yellow= .1428, red= .1156, brown=
.1249; the histogram is skewed left
Mean= 55.5667; standard deviation= 2.0003; total # of candies=5001; # of bags= 90
The standard deviation indicates that the difference between the actual number of M&Ms per bag
and the mean number of M&Ms per bag is 2.0003.
Part 3: Method, Analysis, Results The objective of this part of the project is was to construct a 95%
confidence interval for the proportions of blue, orange, green, yellow, red and brown m&ms. The
results were:
Blue (.22477, .24833) Orange (.19867, .22125) Green (.15975, .18058) Yellow (.13308, .15247)
Red (.10672, .12444) Brown (.11581, .13414) Mean (55.153338, 55.979862 Bonus 438
Part 4: Method, Analysis, Results For this portion we were to tests Masterfoods' claims about which
proportion of colors the greatest number of people found attractive.
Blue; H0: p=.24 claim, H1: p[pic] .24; Z= –.5630, Fail to reject
There is insufficient evidence to suggest the true proportion is not .24.
Orange; H0: p=.20 claim, H1: p[pic].20; Z= 1.75; Fail to reject
There is insufficient evidence to suggest the true proportion is not .20.
Green; H0: p=.16 claim, H1: p[pic] .16; Z= 1.9676; reject
There is sufficient evidence to suggest the true proportion is not .16.
Yellow; H0: p=.14 claim, H1: p[pic] .14;

Qnt561 One and Two Samples Sets Week 5
One– and Two–Sample Tests of Hypothesis, Variance, and Chi–squared Analysis Problem Sets
University of Phoenix
Applied Business Research and Statistics
QNT 561
August 5, 2011
Chapter 10
Exercise Question 31: A new weight–watching company, Weight Reducers International, advertises
that those who join will lose, on the average, 10 pounds the first two weeks with a standard
deviation of 2.8 pounds. A random sample of 50 people who joined the new weight reduction
program revealed the mean loss to be 9 pounds. At the .05 level of significance, can we conclude
that those joining Weight Reducers on average will lose less than 10 pounds? Determine the p–value
To calculate the test statistics:
Z=(9–10)/(2.8/sqrt(50))= –2.525 ... Show more content on Helpwriting.net ...
He selects a sample of 15 families, some with only a single insured driver, others with several
teenage drivers, and pays each family a stipend to contact the two companies and ask for a price
quote. To make the data comparable, certain features, such as the deductible amount and limits of
liability, are standardized. The sample information is reported below. At the .10 significance level,
can we conclude that there is a difference in the amounts quoted?
Geico Mutual Insurance
Family | Progressive Car Insurance | GEICO Mutual Insurance | Becker | 2,090 | 1,610 | Berry |
1,683 | 1,247 | Cobb | 1,402 | 2,327 | Debuck | 1,830 | 1,367 | DuBrul | 930 | 1,461 | Eckroate | 697 |
1,789 | German | 1,741 | 1,621 | Glasson | 1,129 | 1,914 | King | 1,018 | 1,956 | Kucic | 1,881 | 1,772 |
Meredith | 1,571 | 1,375 | Obeid | 874 | 1,527 | Price | 1,579 | 1,767 | Phillips | 1,577 | 1,636 | Tresize |
860 | 1,188 |
Chapter 12
Exercise Question 23: A real estate agent in the coastal area of Georgia wants to compare the
variation in the selling price of homes on the oceanfront with those one to three blocks from the
ocean. A sample of 21 oceanfront homes sold within the last year revealed the standard deviation of
the selling process was $45,600. A sample of 18 homes, also sold within the last year, that were one

to three blocks from the ocean revealed that the standard deviation was $21,330. As the .01
significance level, can

The 7 Ecosystem Functions Between Two Habitats Types
In order to analyze the data for the 7 ecosystem functions between two habitat types (forest vs.
pasture, for example) to address the questions, I am going to use the following statistical analyses: 1.
Nested ANOVA, which will tell us the if there is a significant difference of the ecosystem functions
between two habitat types. 2. Correlation matrix, which will tell us if there is an association between
the microbial respiration and biological nitrogen fixation with the expression and activity of 5
extracellular enzymes. 3. Analysis of beta–diversity patterns of ecosystem functions by principle
coordinate analysis (PCoA). 1. Nested ANOVA: Since the question says that "You have collected
data on 7 ecosystem functions", I am assuming I do not need to describe how I have got the data. As
far as I understood I just need analyze the data by appropriate statistical method(s). The soil
sampling protocol was designed to provide sufficient replication within and across land use types
such that statistically rigorous techniques could be used for comparative analysis of the data for7
ecosystem functions. If I tabulate the data for statistical analysis, it should look like the following:
Data on each of 7 ecosystem functions (nitrogen fixation rate/respiration rate/ alpha–
amylase/lipase/protease/xylanase/pectinase) of two contrasting habitat types Forest Pasture Site–A
Site–B Site–C Site–D Site–E Site–F Site–G Site–H Site–I Site–J X1 X4 X7 X10 X13 Y1 Y4 Y7
Y10 Y13 X2 X5 X8 X11 X14

Essay on Aj Davis Department Store Part B
AJ Davis Department Store Part B AJ Davis Department Store Introduction The following
information will show whether or not the manager's speculations are correct. He wants to know the
following information: Is the average mean greater than $45,000? Does the true population
proportion of customers who live in an urban area exceed 45%? Is the average number of years
lived in the current home less than 8 years? Is the credit balance for suburban customers more than
$3200? Hypothesis testing and confidence intervals for situations A–D are calculated. A. THE
AVERAGE (MEAN) ANNUAL INCOME WAS LESS THAN $45,000. Solution: Step 1: Null
Hypothesis: The average (mean) annual income was equal to $45,000. H_0: μ=45,0000 Step2:
Alternate ... Show more content on Helpwriting.net ...
Solution: Step 1: Null Hypothesis: The average (mean) number of years lived in the current home is
equal to 8 years. H_0: μ=8 Step 2: Alternate Hypothesis: The average (mean) number of years lived
in the current home is less than 8 years. H_a: μ 50 requires that the z–test for mean be used to test
the given hypothesis. The alternative hypothesis is Ha:μ 3200 Step 3: Test Statistic: z= Following
the provided information, the Significance Level is α=0.05. The alternative hypothesis is Ha:
μ>3200; therefore, the given test is a one–tailed (upper–tailed) z–test. Step 4: Critical Value and
Rejection Region: The critical value for significance level α=0.05 for an upper–tailed z–test is given
as 1.645. Rejection Region: Reject H_0,if z–statistic>1.645. Step 5: Assumptions: The sample size
in this speculation is greater than 30, therefore, The Central Limit Theorem (CLT) will apply, and no
assumptions need to be made. Step 6: Calculation of test statistic: One–Sample Z: Credit Balance
($) Test of mu = 3200 vs > 3200 The assumed standard deviation = 742.365 95% Lower Variable N
Mean StDev SE Mean Bound Z P Credit Balance ($) 15 4675 742 192 4360 1.96 0.025 Step 7:
Interpretation: According to the above results from MINITAB, the p–value of 0.038 is smaller than
the significance level of 0.05; consequently, the null hypothesis will be rejected. There is sufficient
evidence to support the

Understanding the Unbiased Estimator
Terms: 1estimator, estimate (noun), parameter, bias, variance, sufficient statistics, best unbiased
estimator. The Department of Finance and Actuarial Science have recently introduced a new way to
help actuarial science students by hiring tutors. All tutors were selectively picked by the Dean of the
department based on their overall performance. Any student that faces any problem regarding
actuarial science subject can visit the tutor. The tutor room is available every Monday till Friday
8AM to 5PM. The tutor room is open to make sure that students get help for their actuarial science
subject. 2However, is it reasonable that the tutor room is open from 8AM to 5PM? How many
students will actually want to visit the tutor room at the opening hours? The answers to these
questions would help the Department of Finance and Actuarial Science to reduce its hiring expenses
by determining the number of tutor they should hire. By estimating the average number of students
who will visit the tutor room during the opening hours, we can determine whether the Department of
Finance and Actuarial Science can close down the tutor room during some specific hours so that
they can reduce the hiring expenses. Firstly, we need to identify the distribution needed for this test.
3As the numbers of students who will visit the tutor room during the opening hours are subject to
the timing students usually study and all this and would not affect one another, this random variable
is independently and

Advantages And Disadvantages Of Hypothesis Significance...
Null hypothesis significance testing (NHST)
Sir Ronald Fisher introduced NHST in 1925 and since then it has been a keystone foundation in the
inferential statistical analysis of research data (Tryon 2001). Inferential statistics is a line of statistics
that is widely used by researchers in social sciences to help obtain a value for the population under
study (Argyrous 2011). The NHST is predominantly known as the experimenter's pillar for
constructing inductive inferences from populations under study (Kreueger 2010).
The null hypothesis is defined by Frick (1996) as a statistical calculation constructed in such a way
that "there is no difference between two variables" and assuming that the null is always true, then a
p–value is calculated which is predictably defined as the probability of obtaining a pattern of results
that are statistically insignificant and to indicate if a relationship between two variables exists or not
(Argyrous 2011). However, the calculation of NHST ... Show more content on Helpwriting.net ...
Despite the great deal of controversies about the null hypothesis yet it still dominates the field of
social sciences and it is still being used in research that involves scientific psychology (Krueger
2010).
Another prominent weakness of NHST is that it does not give any indication of the magnitude of the
statistical relationships between the two variables under study. However, the NHST reduces the
statistical inference to a process of a binary decision making and that is making a choice between
two alternatives (Sim and Reid 1999). Furthermore, it seems useless and senseless to test a
hypothesis that is almost certainly false any way and that might encourage an undesired, binary
focus on data outcomes (Killeen

Unit 4 Agression Analysis Paper
In this case, we are dealing with a between subject experiment where one of the groups is tested in
condition A while another group is tested with condition B. We can see from the set of data that we
got outliers. To get rid of the outliers, we can calculate the interquartile mean by find the
interquartile range first. This method can help to trim the outliers in the data.
To run a randomization test to determine if there is a difference in the interquartile means between
the two groups, I have to collect the observed data first. The observed data is the participants' data
that we get from two different groups (condition A and condition B). After I did this, I have to
decide on the test statistic or observed statistic that I would use to compare with the simulation of
the null model. If I get test statistics that is very unlikely to happen under the null model, it means
that the null model is possibly wrong. In this case, I used interquartile mean as the test statistic.
Then, I had to come up with a null model, which I assumed that the ... Show more content on
Helpwriting.net ...
I did this simulation by ripping paper into 32 small pieces of papers (because there are 32
participants that join this study) and then write each participant's data on each paper. By doing this
way, I can easily re–assign and randomize each data to a group. Then, I repeated this randomization
processes 20 times/simulations (as the question asked me to do 20 times). I calculated the
interquartile range difference for each of the randomization process/simulation that I did. Image 1.0
shows the difference of interquartile range in condition A and B that I got from the simulation that I
did. Then, I calculated the interquartile means based on the difference I got in the table 1.0, and I get
3.12685 for the interquartile

Keller Math 533 Project Part B
Math 533 Project Part B In regards to the dataset from AJ Department store, your manager has
speculated the following: the average (mean) annual income is less than $50,000, the true population
proportion of customers who live in an urban area exceeds 40%, the average (mean) number of
years lived in the current home is less than 13 years, the average (mean) credit balance for suburban
customers is more than $4300. Part 1. Using the sample data, perform the hypothesis test for each of
the above situations in order to see if there is evidence to support your manager's belief in each case
a.–d. In each case use the Seven Elements of a Test of Hypothesis, in Section 6.2 of your text book
with α = ... Show more content on Helpwriting.net ...
Therefore, we will reject 〖the null hypothesis,H〗_0,if z>1.645 Element 5: Assumptions Element
6:Interpretation of Results Since the P–value (0.386) is greater than the significance level (0.05), we
fail to reject the null hypothesis. The p–value implies the probability of rejecting a true null
hypothesis. Element 7: Conclusion At a significance level of 0.05, there is no sufficient evidence to
support the claim that the true population proportion of customers who live in an urban area is
greater than 40%. The average (mean) number of years lived in the current home is less than 13
years Element 1. Null hypothesis (H0): The average (mean) number of years lived in the current
home is greater than or equal to 13 years H_0: μ≥13 Element 2. Alternative (research) hypothesis
(Ha): The average (mean) number of years lived in the current home is less than 13 years. H_a: μ 30
we will use a z–test for the mean to test the given hypothesis. As the alternative hypothesis is Ha:μ
30 we will use z–test for mean to test the given hypothesis. As the alternative hypothesis is
Ha:μ>4300 , the given test is a one–tailed (upper–tailed) z–test. One–Sample Z: Credit Balance ($)
_1 Test of mu = 4300 vs > 4300 The assumed standard deviation = 770.339

A Research Study On The Field Of Psychology
The field of psychology is truly fascinating, as the discovery of new and remarkable ideas arise with
every question a researcher chooses to explore. Where questions can develop into a research study,
which can either solidify, contribute, support, inspire, or provide answers in the field of psychology.
In a sense, research is a trial and error approach as researchers are not always cognizant of the
results that a study will provide. However, with sufficient knowledge the researcher can conduct a
study, with less error, should the researcher be aware of a few simple ingredients.
First, as simple as it sounds, knowing whether if a question is sufficient is a good place to start.
Once that has been determined, then the researcher can ... Show more content on Helpwriting.net ...
In conclusion, this writer will be discussing how the statistical model created in outline form can be
useful, its limitations, along with what this writer learned in the process of creating a unique
statistical model of this writer's liking and understanding.
Decision's Model in Outline Form Is your question adequate and sufficient? This is determined by
figuring out whether if you have put enough thought into your question, if only you can answer the
following 4 questions (Sukal, 2013).
(1). What is your overarching research question? (Sukal, 2013)
Note: According to Sukal (2013), this can be determined by figuring out whether if there are
keywords within the question, such as "effects" to refer to cause and effect (Sukal, 2013). The word
"association", indicates figuring out if there is a relationship amongst the variables (Sukal, 2013).
The word "prediction" can signify regression (Sukal, 2013).
(2). How many Independent Variables (IV), Dependent Variables (DV), and covariate variables are
in the study? (Sukal, 2013).
Note: The dependent variable is thought of as the criterion or consequence variable, while
independent variable is thought to be the predictor or the variable that brings out an effect on the
dependent variable (Sukal, 2013). Further, a variable is something a person plans to observe,
manipulate, test, record, or evaluate

Do Students Visit The Holman Library Information Desk?
Do Students Visit the Holman Library Information Desk? Introduction Making appropriate staffing
decisions is one of the most important tasks for a manager or of any company. The decision a
company makes about staffing their business can have dramatic effects on the quality of their work,
their retention rates and level of customer service. Because of this, we would like to investigate
whether a sufficient number of staff already exists at the Holman Library of Green River College. It
is important to note that we cannot just look at one factor in order to conclude whether a company
has sufficient staffing. In fact, the knowledge and skill of the staff, the type of problem that the staff
has to solve, and the peak hours of the library, are ... Show more content on Helpwriting.net ...
To ensure randomness on our research, TI–84 calculator was also used to generate random numbers
from 1 to 7 that are assigned to every 30 minute time intervals during the day. Procedure A number
from one to seven was assigned to seven different 30 minute time intervals during the day. To avoid
bias, a TI–84 calculator was used to select the timeframe of observations for each day from Monday
to Friday. The first five numbers generated was the timeframes that we used for the observation. The
randomly selected time(s) are summarized on Table 1.0 on the right. During the observation, an
observer from the group stood in front of the information desk on the first floor of the Holman
Library to count how many people came into the library using a tally counter, and how many people
first went to the information desk for help by drawing tally on a sheet of paper at the same time. As
for the data analysis, the criterion for significance for all data was set at α=0.01. Results The
observation that were conducted from October 19th to October 23rd went very well. Sometimes a
group of future students who enters the library for a tour would first visit the information desk to get
more info about the library. We thought that since this does not happen at a daily basis, their visit
would keep us away from getting the accurate result. We were glad that it did not happen during the
observation. The bar chart above (bar chart 1.0) summarizes the data that were collected.

Math 300 MM Project
Strayer University
Math 300
MM Project PT 4
August 14, 2011
Solution: We want to test the following null and alternative hypotheses
We need to use the z–statistic, which is calculated using
Observe that the sample proportion is
This corresponds to a two–tailed z–test for proportions. The z–statistics is computed by the
following formula:
The critical value for for this two–tailed test is. The rejection region is given by
Since, then we reject the null hypothesis H0.
Hence, we have enough evidence to reject the claim that the true proportion of blue M&Ms®
candies is 0.24.
3 pts. Test their claim that the true proportion of orange M&Ms® candies is 0.20 at the ... Show
more content on Helpwriting.net ...
3 pts. Test their claim that the true proportion of yellow M&Ms® candies is 0.14 at the 0.05
significance level.
Solution: We need to test the following hypotheses
We need to use the z–statistic, which is computed using
This corresponds to a two–tailed z–test for proportions. The z–statistics is given by the following

formula:
Since, then we reject the null hypothesis H0.
Hence, we have enough evidence to reject the claim that the true proportion of yellow M&Ms
candies is 0.14.
3 pts. Test their claim that the true proportion of red M&Ms® candies is 0.13 at the 0.05
significance level. Solution: We are interested in testing the following hypotheses
We need to use the z–statistic, which is calculated using
This corresponds to a two–tailed z–test for proportions. The z–statistics is given by the following
formula:
Since, then we fail to reject the null hypothesis H0.
Hence, we don't have enough evidence to reject the claim the true proportion of red M&Ms
candies is 0.13.
3 pts. Test their claim that the true proportion of brown

Statistics Purpose Statement
Statement of Purpose
Junru Xia
I first discovered my interest in statistics in my sophomore year at the University of California,
Berkeley. In a statistics lecture, the professor explained how statistics can be applied in real life by
giving an example that Amazon often promotes products by emailing its customers and
recommending products according to sophisticated analysis of customers' search records, as well as
their transaction data. Previously having been immersed in the world of abstract statistics, I realized
that in addition to being an essential tool to confirm theories already proposed, data itself can lead to
entirely new ideas. The data patterns can be interpreted from many different perspectives, which
triggered my eagerness to explore the stories hidden in data and improve my understanding of the
world through data analysis.
With a strong urge to enhance my quantitative foundation, I finished all my courses in mathematics
with high scores, and learned Python and Data Scientists' Toolbox through Codecademy and ...
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My job in the personal loan center was to assist in maintaining the database and generating
evaluations. One project which I was involved in was using personal credit scoring model to make a
risk forecast and detect bank fraud based on loan documentation in the database. During the project,
I realized that rather than just about complicated methods, statistics was about simplifying and
making sense. Sufficient data and in–depth analysis can not only help companies make valuable
decisions, but also enable us to understand real life in a more direct and precise way. As the trend of
information explosion is inevitable, I am sure that the power of data will become increasingly
overwhelming, and the role of data analysis will be indispensable. Therefore, along with my goal to
pursue graduate studies in statistics, my desire to pursue a career in the field of data analysis was

Statistical Analysis And Quantitative Methods
Looking back, this course was very helpful in expanding my knowledge pertaining to statistical
analysis and quantitative methods. Although I had some knowledge going into this semester from
prior classes, I felt uneasy about statistical analysis and its applications. There were so many
elements I had questions about and a variety of factors I wished to understand but that were unclear.
Specifically, advanced statistical methods seemed to be shrouded in mystery – complex,
indecipherable, abstract, and inaccessible thought constructs. Indeed, completing the weekly
assignments and conducting the various statistical tests has aided in expanding my understanding of
some of the advanced statistical tests, making them less incomprehensible. Working on the
applications has also reemphasized the importance of wholly understanding the statistical analysis to
be performed, its underlying assumptions, and possible limitations of the analysis to arrive at a
plausible and cogent interpretation of the data. Furthermore, the brief exposure to statistical research
designs at the beginning of the course has caused me to think critically about the type of analysis I
will be conducting. Despite having been unable to settle on a specific design for my intended study,
the review has provided me with the tools to understand the advantages and limitations, benefits and
drawbacks of the various approaches. What's more, gaining a better understanding of the designs
and methods will enable me to

Purpose. The Overall Purpose/Hypothesis Of The Article,
Purpose The overall purpose/hypothesis of the article, was it to find out how physical activity is
linked to the preservation of neurocognition, but more specifically to address the fundamental
question–– "does exercise improve cognition?" (Masley et al., 2009). The secondary reason for this
study is to use a technological method to determine the field of cognitive domains, in relation to
memory, mental speed, reaction time, attention, and cognitive flexibility.
Subjects
The subjects from this study were volunteers from the Carillon Wellness Center, St Anthony's
Hospital, St. Petersburg, Florida. There were 91 male and female subjects who were between the
ages of 18–70. No subjects have a history of elevated heart levels or any major ... Show more
content on Helpwriting.net ...
Findings and Statistics The comparison between demographic and physiological data was found on
Table 1. The statistics showed that there was no significant difference between both the control
group and the combined intervention group in relation it age, BMI, years of education, and VO2max
fitness. The only shown difference was gender, a higher percentage of women were added onto the
study. The study showed that there was no statistical significant change in VO2max levels.
"VO2max levels had increased by 6.5% in the control group, 12% in the moderate intervention
group, 17.3 % in the combined intervention group, and 21.3 % in the high exercise intervention
group" (Masley et al., 2009). The independent t–test showed the comparison between the control
group and the high exercise group was rather significant, with a p value of 0.53. Also, the Spearman
rank correlation coefficient compared the change from the control, to the moderate, to the high
exercise groups, with a coefficient of 0.23 and a VO2 max p value of 0.06. Neurocognitive data that
was analyzed by MANOVA, was collected in Table 2. For this study, there was a high significant
difference is the observes psychomotor speed, attention and cognitive flexibility, with a p value of
0.03. When t–test compared the change in the combined intervention group with the controlled
group, only cognitive flexibility

Aj Davis Course Project Parts a and B
The following report presents a detailed statistical analysis of AJ DAVIS department store
customers. Data was collected from a sample of 50 AJ DAVIS credit customers for the purpose of
learning more about the customers of AJ DAVIS.
The first variable considered is Location, a categorical variable. The three subcategories are Urban,
Suburban and Rural. The frequency distribution and pie chart are included. Measures of central
tendency and descriptive statistics are not calculated due to the categorical nature of the variable.
Frequency Distribution: LOCATION | FREQUENCY | Urban | 22 | Suburban | 15 | Rural | 13 |
The largest number of customers belong to the Urban Location category (44%), followed by those in
the ... Show more content on Helpwriting.net ...
There is a linear positive relationship between Income and Credit Balance variables. Where income
increases, credit balance also increases.
The relationship between the variables Years and Credit Balance is illustrated in the following
scatter plot:
There is no clear relationship between the two variables in the scatter plot. The points are in no
specific pattern, suggesting that there is no significant correlation between the variables years and
credit balance.
It can be concluded that some variables such as Income are strongly related to the credit balance of
AJ DAVIS department store customers. Several other variables appear to be unrelated.
PART B: Hypothesis Testing and Confidence Intervals
1.a– The average (mean) annual income was less than $50,000 I found the average (mean) income
to be $43,740, with a standard deviation of $14,640. According to the hypothesis test (see appendix),
the calculated test statistic of –3.0236 does fall in the rejection region of z<–1.645 therefore I can
reject the null hypothesis and say there is sufficient evidence to indicate µ<50 or $50,000. The p–
value of 0.001 (see appendix for data), supports the rejection of the null hypothesis since the p value
is less than α=0.05. Based on the confidence interval, we can be 95% certain that the average
income lies between $39,680–$47,800.
1.b– The true population proportion of customers who live in an urban area exceeds 40% Of those
surveyed, 22 out of 50

Data Analysis Assignment Essay
Amanda Mueller
MBA6018 – Data Analysis
Unit 3 Activity 1
January 23, 2013
Practical Application Scenario 1
In 2010, Playbill Magazine contracted Boos Allen to conduct a survey aimed at determining the
average annual household income of Playbill readers. 300 readers were randomly pulled and
sampled from the list of customers provided by Playbill Magazine. From that sampling effort, Boos
Allen was confident that the population average household income is $119,155 and that the
population sample household income standard deviation is $30,000.
Two Playbill executives recently hypothesized that the average annual household income of its
readership has increased and so believe that the magazine price should also increase. From a ...
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Using the table provided on page 242 of the text, one can determine that the p–value is equal to 1 –
0.99886 or 0.00114.
Step 5: Reject H0 if the p–value is less than α. Interpret the statistical results
The p–value of 0.00114 is less than the α value of 0.05. Therefore, H0 would be rejected at the level
of significance 0.05 and the magazine price could be raised.
Inputs –– | | Hypothesized Population Mean: | 119155 | Population Standard Deviation (sigma): |
30000 | Sample size (n): | 300 | Sample Mean (X–bar) | 124450 | | | Intermediate Calculations –– | |
Standard Error of the Estimate: | 1732.0508 | Test Statistic (z): | 3.0570697 | | | Results –– | | | | One
tailed, H0: Mu =>119155, p= | 0.9989 | | | One tailed, H0: Mu <=119155, p= | 0.0011 | | | Two–
tailed, H0: Mu = 119155, p = | 0.0022 |
| For the Alpha level given, H0 should be | | Alpha: | 0.01 | 0.05 | 0.1 | | not rejected | not rejected | not
rejected | | | | | | Rejected | Rejected | Rejected | | | | | | Rejected | Rejected | Rejected |
One can be statistically confident in the conclusion to reject the null hypothesis H0 by performing
calculations to derive the confidence intervals. Using the Excel function NORMSINV and inputting
the necessary variables (see highlighted figures below), one can determine both the upper and lower

Part AAj Davis Department Store
Course Project Part A
September 15, 2013
Applied Managerial Statistics
Professor Mayers
Brief Introduction
The following report presents a detailed statistical analysis of AJ Davis Department Store credit
customers. Data was collected from a sample of 50 AJ Davis credit customers on five variables
which are Location, Size, Income, Years, and Credit Balance. Out of the five variables,
Location,Size, and Income is emphasize more in this analysis. AJ Davis Department Store is very
determined to find out more information about their credit customers. So by doing a in–depth
analysis of the variables and their relationships through graphical, numerical summary and
interpretation should give a detailed summary of their ... Show more content on Helpwriting.net ...
Descriptive Statistics: Years
Variable Location Mean StDev Variance Median Range IQR Mode
Years Rural 12.46 4.94 24.44 13.00 16.00 7.00 13, 15, 18 Suburban 6.467 2.949 8.695 6.000 9.000
5.000 10 Urban 10.045 3.982 15.855 10.000 17.000 5.000 10
The 2nd pairing of variables I combined together is Location and Years. I demonstrated the variables
in a dot plot to illustrate the number of years the customer has been living in that location. The most
years was more than 18 years and the location was in an urban area. The highest amount of dots was
10 years. The shape of the distribution is symmetric.
The last pairing of variables I combined together is Income and Size and it demonstrated in a scatter
plot. The household size of 7 or 8 has the highest income is with over $69,000 and more. The shape
of distribution is positive linear relationship.
Regression Analysis: Income ($1000) versus Size
The regression equation is
Income ($1000) = 33.5 + 2.78 Size
Predictor Coef SE Coef T P
Constant 33.499 3.523 9.51 0.000
Size 2.7824 0.6844 4.07 0.000

S = 12.0983 R–Sq = 25.6% R–Sq(adj) = 24.1%
Conclusion
As the result shows, the urban location is where most of AJ Davis Department Store comes from
with 44%. The urban location also has a higher credit

Hypothesis Testing The Boiler Room Stats
Hypothesis Testing the Boiler Room Stats According to the manager four aspects require evaluation
from the statistics gathered for the week of calls made in the boiler room. The four aspects are the
basis for the hypothesis testing; the manager's suspicions are a) the null which are as follows:
1. Hypothesis Test One
a) Null Hypothesis Ho=u > Mean sales exceeds 41.5 sales per telemarketer
b) Alternative Ha=u< less than 55% receiving online training
b) Alternative Ha=u>or= 55% receiving online training
3. Hypothesis Test Three
a) Null Hypothesis Ho=u < Mean calls among no training is less than 145
b) Alternative Ha=u >or= Mean calls among no training is 145
4. Hypothesis Test Four
a) Null Hypothesis Ho=u > Mean time per call ... Show more content on Helpwriting.net ...
For hypothesis test two the manager suspects less than 55% of the 100 telemarketers received online
training. Looking at the type of training data; two types of training were offered and none at all,
provides the three options. Therefore, if those receiving either group training or none at all total
more than 45 telemarketers of the 100 sales people then the alternative hypothesis is accepted
because there is sufficient evidence to support it. If the manager's suspicions are true: the null
hypothesis that states there are less than 55% of the sales force who received online training is true
then there must be sufficient evidence to support that statement. 20 of 100 callers have no training
and 30 telemarketers received group training while 50 sales people received online training
according to the Excel data sheet provided. Therefore, 50% either received no training or group
training and 50% received online training. The manager's suspicions are supported. There is
sufficient evidence to support the null hypothesis and therefore the alternative is false. With
hypothesis test three the manager states the mean amount of calls made by those with no training is
less than 145 calls per person for the week of data recorded.
To prove the null; calls are less than 145 per person for those who are listed as no training or none,
the total amount of calls for the week made by the none group are divided by the 20 telemarketers to
compute the mean

Data Analysis Notes
Lecture 7. Sampling Distributions.
Statistical Inference: Using statistics calculated from samples to estimate the values of population
parameters. Select Random Sample Sample for (statistic) Calculate to estimate Becomes Population
Parameter.
BASIC Example: Soft Drink Bottler μ=600, σ=10. Normal Distribution. What is P(X>598)?
p(x<598) .
Sampling Dist.of the Mean – Distribution of all Possible Sample Means if you select a sample of a
certain size. μX= μ. μ = i=1NXiN (formula for mean) . σ = i=1N(Xi–μ)2N Although you do not
know how close the sample mean of any particular sample selected comes to the pop mean, you
know that the mean of all possible sample means that could have been selected = the pop mean.
Standard ... Show more content on Helpwriting.net ...
OR conduct a pilot study and estimate σ with S.
Determining Sample Size for π where e =
To determine sample size for proportion you must know the desired level of confidence (1 –∝),
which determines the critical Z value, The acceptable sampling error (e), and the true proportion of
'successes',(π) . π can be estimated with past data, a pilot sample, or conservatively use π = 0.5
HYPOTHESIS TESTING TOPIC 9
In the inferential method of hypothesis testing, you consider the evidence (sample statistic) to see
whether the evidence better supports the statement (null hypothesis) or the mutually exclusive
alternative. Hypothesis testing is based on sample information. Methodology enables you to make
inferences about a population parameter by analysing differences between the results observed
(sample stat) and the results you expect to get if some underlying hypothesis is actually true. The
hypothesis that assumes the status quo – that the old theory, method or standard is still true; the
complement of the alternative hypothesis.
NULL HYPOTHESIS: Always contains '=' , '≤' or '≥' sign, May or may not be rejected, Is always
about a population parameter, μ, not about a sample statistic ,Similar to the notion of innocent until
proven guilty
ALTERNATE HYPOTHESIS: The hypothesis that complements the null hypothesis. Usually it is
the hypothesis that the researcher is interested in proving. They are mutually exclusive and null is
assumed to be true. Burden

Math Week 4 Case Study
Week 4 Practice Problems
1. Two sets of data both have a mean of 17. Set A has a standard deviation of 3.5. Set B has a
standard deviation of 6.8. Explain specifically what the different standard deviation measurements
tell you as a researcher about the two data sets?
A= mean: 17 SD: 3.5 B= mean: 17 SD: 6.8
2. Which of the two sets of basketball players were more consistent in their foul shots? Explain how
you know.
Team A: Mean number of foul shots made = 20, standard deviation = 4.
Team B: Mean number of foul shots made = 21, standard deviation = 2.1.
Team B is more consistent because there is less difference
3. You are interested in how many Scioto County 3rd – grade students can improve their reading
proficiency using a ... Show more content on Helpwriting.net ...
Alternate Hypothesis: The new lightbulb claims that it has average life of more than 1000 hours.
Ho = p=1000 hours HA : p > 1000 hours.
7. Write the null and alternate hypotheses for this situation:
A cereal manufacturer uses a filling process designed to add exactly 18 ounces of cereal to each box.
State the null and alternative hypotheses that would be used to verify this claim.
Null Hypothesis: It may be the cereal manufacturer filled by adding less than 18 ounces of cereal.
Alternate Hypothesis: manufacturer claims that designed to add exactly 18 ounces of cereal to each
box.
Ho = p < 18 HA = p=18
8. Write the null and alternate hypotheses for this situation:
In the last census, taken five years ago, it was determined that 6% of school–aged children in a
certain state lived with their grandparent(s). To support a bill on tax breaks for seniors, a
congressional member plans to take a random sample of school–aged children to determine if that
percent has increased.
Null Hypothesis: 6% or even less than that in a certain state lived with their grandparents.
Alternative Hypothesis: The percent of children in the state who live with their grandparents has
increased greater than 6%. Ho: p= 0.6 HA: p > 0.6

9. Let's suppose you have completed a statistical analysis. The null/research hypotheses are listed
below, along with the p–value that you obtained from your testing. Explain whether you have
significant evidence to "reject the null" or

The Effect Of Fdi Inflow On Exports
Data
This paper uses time series data on Bangladesh from the WDI databank to find the impact of FDI
inflow on exports. We've considered the time period from 1976. Bangladesh became independent in
1971. Just after birth, the nation adopted an import substitution trade policy. And consequently,
discouraged export oriented industrialization. And FDI was forbidden till 1976. In 1976, Bangladesh
took newer policies and thereby trade openness began. FDI also started to arrive. That's why, we've
excluded the years before 1976.
Dependent Variable: Export, measured in US dollar.
Independent Variable: Foreign Direct Investment, FDI inflow measured in US dollar.
Methodology
Before explaining the methodology of data analysis it is important to ... Show more content on
Helpwriting.net ...
Granger–casualty tests the hypothesis if one variable is useful in forecasting the change in another
variable. Here, the time series variable 'FDIBD' would Granger–cause time series 'X' (exports) if it
could be statistically established.
Results of the study suggests causality in both directions (i.e. 'XBD' depends on 'FDIBD' and
'FDIBD' depends on 'XBD'). It may be mentioned that the main objective of this analysis is related
to first one only. However, second causation also provides worthwhile information for further
studies. Results of the analysis are given in Table–1. The results reject the null hypothesis (of no
causality) and statistically prove causality between FDI and exports in both directions.
Cointegration Relationship Test
In spite of statistically significant Granger–causality between FDI and Exports, this test alone is not
that strong to fully establish dependence of one variable on other. It, in fact, only establishes that
explanatory variable leads the dependent variables– a necessary (but not sufficient) condition for an
explanatory variable. For unbiased estimation of regression parameters another necessary condition
is that the explanatory and dependent variables should have mutual cointegration. Granger–causality
and cointegration together are thought to be

The Net Inflow Of Fdi Into The Developing Countries
Sample Data
Figure 1 shows the net inflow of FDI into the developing countries. There is a fall into the amount
of FDI going to the developing countries from late 1980 to early 1990 and in the late 2000. Overall,
there is an upward trend of amount of FDI going to the developing countries. The same trend with
ODA shown in Figure 2. The amount of Net ODA received by developing countries from 1990 to
mid–1990 is fluctuating then continued to fall until 2000. From year 2000 onwards, there is a steady
increase of ODA received by the low–income countries. Compared with FDI and Net ODA,
personal remittances has a steady upward trend (shown in Figure 3), noting a huge increase in 2008
of US$4.5 billion from US$15.2 billion in 2007. With the ... Show more content on Helpwriting.net
...
The independent variables (all are current U.S. dollars) are FDI defined as sum of equity capital,
reinvestment of earnings, other long–term capital, and short–term capital as shown in the balance of
payments less disinvestment; Net ODA consists of net repayments of principal and grants by
Development Assistance Committee members agencies and by non–DAC countries to promote
economic development; and personal remittances is consists of personal transfers (all current
transfers between resident and non–resident individuals) and compensation of employees (income of
workers who are employed in foreign country).
Descriptive statistics in Table 3 employed the logarithmic function of the variables as suggested by
the reference literatures and thus will also use in the regression analysis. It shows that among
variables, there are different observations due to in some years, data is not given. This also somehow
affect the deviation of variables from the mean. LogFDI and LogRem have the highest deviations
from the mean which are 2.59201 and 2.393887 respectively. The average amount of FDI low–
income countries received increases by 17.00671% annually, and the lowest percentage increase
2.374347%. For Net ODA, countries have an average of 19.69657% increase in the amount they
received annually and the percentage increase could be high as 22.43425%. In the case of personal
remittances, low–income countries received an average increase of 17.47987% and the highest
increase could

Quantitative Statistics Answers
SMCU002: ADVANCED STATISTICS
LESSON ONE.
Names & Reg.no: Eric Wagobera Junior (MIRD/43004/2016)
Answers
1 a). Data is defined as the collected details from which numerical information is derived. The
collected data is processed and organized into vital statistical information which can be in the form
of facts or figures. This data is then interpreted or analyzed through statistical methods for the
purposes of gaining knowledge about a phenomenon, which can help us to make informed
decisions.
b). Information is data that has been recorded, classified, and structured within a statistical
framework with an aim of making it useful for decision–making purposes. This information is
processed from the collected raw data to come up with meaningful conclusions.
c). Population refers to a set of unit with common characteristics from which statistical sample is
drawn for purposes of obtaining the needed data. The information collected from a sample
population can be used by statisticians to develop generalized observations which can be applied to
the whole population.
2 a). The difference between continuous and discrete data is that variables in discrete data are given
in whole number values which are countable. Such figures are distinct and aren't either in decimal or
fractional formats, i.e. positions or rankings. On the other hand, continuous data includes variables
whose number values are either in decimal or fractional format which are measurable on a scale.
This implies that

Stat 200 Final Exam 100% Correct Answers Essay
STAT 200 Final Exam 100% Correct Answers https://homeworklance.com/downloads/stat–200–
final–exam–100–correct–answers/ To Get this Tutorial Copy & Paste above URL Into Your Browser
Hit Us Email for Any Inquiry at: Lancehomework@gmail.com Visit our Site for More Tutorials: (
http://homeworklance.com/ ) STAT 200 Final Exam 100% Correct Answers 1. True or False. Justify
for full credit. (25 pts) (a) The normal distribution curve is always symmetric to its mean. (b) If the
variance from a data set is zero, then all the observations in this data set are identical. (c) . of
complement the is where, 1) AND(AAAAPc c (d) In a hypothesis testing, if the p–value is less than
the significance level α, we do not have ... Show more content on Helpwriting.net ...
Show all work. Just the answer, without supporting work, will receive no credit. A fair coin is tossed
4 times. 7. How many outcomes are there in the sample space? (5 pts) 8. What is the probability that
the third toss is heads, given that the first toss is heads? (10 pts) 9. Let A be the event that the first
toss is heads, and B be the event that the third toss is heads. Are A and B independent? Why or why
not? ––––––––––––––––––––––––––––––––––––––––– Refer to the following situation for
Questions 10, 11, and 12. The boxplots below show the real estate values of single family homes in
two neighboring cities, in thousands of dollars. For each question, give your answer as one of the
following: (a) Tinytown; (b) BigBurg; (c) Both cities have the same value requested; (d) It is
impossible to tell using only the given information. Then explain your answer in each case. 10.
Which city has greater variability in real estate values? 11. Which city has the greater percentage of
households with values $85,000 and over? 12. Which city has a greater percentage of homes with
real estate values between $55,000 and $85,000? ––––––––––––––––––––––––––––––––––––––
Refer to the following information for Questions 13 and 14. Show all work. Just the answer,

Making Appropriate Staffing Decision At The Holman Library...
Making appropriate staffing decision is one of the most important tasks for a manager or of any
company. The decision a company makes about staffing their business can have dramatic effects on
the quality of their work, their retention rates and level of customer service. Because of this, we
would like to investigate whether a sufficient number of staff already exists at the Holman Library
of Green River College. It is important to note that we cannot just look at one factor in order to
conclude whether a company has sufficient staffing. In fact, the knowledge and skill of the staff, the
type of problem that the staff has to solve, and the peak hours of the library, are some other factors
that one may want to look at in order to see whether staffing has been efficient. However, since we
only have limited resources to conduct this research, we only looked at the peak hours of a day. Our
hypothesis for this research was that 25% of visitors to the library would first visit the information
desk. It is important for the librarians to know the peak hours of the library so that students can get
what they need without having to wait in line for too long. Method Participants In this research, all
Holman Library visitors will be the population and the sample size will be the number of visitors
within a specified time during the day. Materials A hand–held tally counter was used to count the
number of people entering the library. To ensure randomness on our research, TI–84

Wentworth Medical Center
Results for the Good Health Survey Florida: Confidence Level at 95% is 1.001191951; Upper Limit
is 6.551; Lower Limit is 4.548. The expected value has a 95% chance of being in the confidence
range. In the case of the above problem, it has a 95% chance of being between 6.551 and 4.548.
New York: Confidence Level at 95% is 1.0298559; Upper Limit is 9.0299; Lower Limit is 7.0299.
The expected value has a 95% chance of being in the confidence range. In the case of the above
problem, it has a 95% chance of being between 9.0299 and 7.0299. California: Confidence Level at
95% is 1.3278753; Upper Limit is 8.3779; Lower Limit is 5.7221. The expected value has a 95%
chance of being in the confidence range. In the case of the above problem, ... Show more content on
Helpwriting.net ...
Some body that has heart ailment will probably be more depressed than somebody with hyper
tension. Two–way ANOVA: 1–What is the response variable? The levels of Depression 2–What are
the factors? Graphical Location and Health. 3–How many levels do the factors have? Factor1 has
two levels (Good Health and Chronic Health), and Factor2 has three levels (Florida, North Carolina,
and New York) 4–How many treatments are there, and what are they? Six treatments: GH–F, GH–
NY, GH–NC, CH–F, CH–NY–CH–NC. These are sometimes referred to as factor–levels. 5–How
many replicates are there in each treatment? 19 replicate each treatment. Null hypothesis: H0: All
treatment means are equal Alternative hypothesis: Ha: at least one treatment mean is not equal
Conclusion, at the .05 significance level, there is a difference between Good Health and Bad Health,
between Florida, North Carolina and New York, and between Heath and States. Tukey Analysis
First, we would reject the null hypothesis of no interraction, p = .2620 suggests that there is no
interraction between health and States. Second, we can reject the null hypothesis concerning no
differences between health, p = .7.11E–27 is Very strong evidence of a difference between Good
Healty and Chronic Health. Third, we can reject the null hypothesis concerning states, p = .05 is
evidence that there is differences

Understanding Business Research Terms and Concepts: Part 2...
Understanding Business Research Terms and Concepts: Part 2
Ming D. Lee
RES/351
April 18, 2016
Dr. Linda F Florence
Understanding Business Research Terms and Concepts: Part 2
Descriptive statistics
Descriptive statistics suggests a straightforward quantitative outline of a data–set which has been
gathered. It helps us comprehend the experimentation or data–set in–detail and tells people
concerning the mandatory details that help show the data perceptively. Descriptive statistics, we just
convey exactly what the data reveals and tell us.
Most of the statistical averages and numbers we estimate are essentially illustrative averages. For
instance the Dow Jones Industrial tells us about the typical performance of select firms. The ... Show
more content on Helpwriting.net ...
The research can use strategies including surveys, observation, area experiments, interviews and
quantity analysis. As it is indeed diverse in technique and utilization, some researchers favor its use
in numerous disciplines including advertising, medical–health, and psychology.
On the possible negative aspect, "Descriptive research" can at times be used to match the
requirements of the researcher. Example: when preparing a survey, one may load the survey
questions to direct the reader to respond in a specific manner. Or in another instance, a comparison
between two merchandise or product, it is possible to give one commodity an unfair edge to get the
wanted outcome. There are two principal limits to the usage of "Inferential Statistics". The primary
and most critical limitation, which will be present in all "inferential statistics", is that if you are
supplying data of a population which you haven't entirely quantified, and thus, cannot actually be
entirely sure the values/numbers you compute are right. Keep in mind, "inferential statistics" are
based in the notion of utilizing the values measured in an example to estimate/infer the values that
might be quantified in a population; there will be a level of doubt in this way. The 2nd restriction is
linked with the primary restriction. Again, there will likely be

A Study On The Rate Of Exercise Is Greater
Question 2 It would be a one–tailed test. Hypothesis for an one–tail test: Ho:μHo≥μHa Compute
Zobt: The Z critical Value = ±1.645 Ho: should be rejected, the researcher should conclude that the
new toothpaste prefix better at helping prevent cavities. 95% confident that the population mean is
within the range: Question 4 This is a type two error Question 6 Yes, this is a two–tailed
Ho:μo≠μ1,μo<μ1 95% Confidence interval: 56.80479 < µ < 61.19521 2.160 No, The researcher
should fail to reject Question 8 X^(2 )=2.55 df=1 X^(2 )=3.84 X^2 (1,N=120)=2.55,p>0.05 Reject
the Null Hypothesis and concluded the national rate of exercise is greater. Jackson even–numbered
Chapter exercises (pp. 273–275) Question 2 He should use an independent t–Test Ho: μo music=μa
no music | Ho: μo music ≤ μa no music 6–7.6 √(2.88/9+4.56/9=–1.76 ) Yes, he should reject; he
should conclude that studies without music produce higher test grades. yes; it is significant Test
Statistic, t: –2.6515, Critical t: ±2.146376, P–Value: 0.0191, Degrees of freedom: 14, 95%
Confidence interval: –2.895174 < µ1–µ2 < –0.3048258 Degrees of freedom Degrees of freedom are
a collection of sample data comprised of a number of sample values that varies according to
dynamic restrictions that may be present. They are the number of data points (N) that can change
without affecting the mean of the data set. Jackson p. 210 without affecting the mean. Additionally,
they are

Wildcatozine Testing
Introduction
This paper contains the report of the analysis carried out on a sample size of 30 to determine the
extent to which Wildcatozine drug helps in the reduction of depression in people. Based on the data
collected from the sample population, descriptive statistics, as well as the analysis of drug with the
mean comparison, was done. The analysis involved the use of student t–test since a sample
population of 30 was chosen. At 5% level of significance, the null hypothesis was then tested, and
the overall results and discussion of the results were as presented below. Precisely, to test the
effectiveness of the drug, the test was carried out at three levels: using descriptive statistics, using
the z–score transformation statistics and ... Show more content on Helpwriting.net ...
Therefore, there is sufficient evidence that there is a significant difference at the top 2.5% and the
bottom 2.5% in the probability distribution of the given data set. Additionally, the resulting p–value
at a level of significance of 0.05 is 2.76, which is greater than our hypothesized value. We, therefore,
reject the null hypothesis and conclude that the sample mean is different from 5.63 as
desired(McGowan & Vaughan, 2011, p. 64). These conclusions are based on the assumptions that no
Type I or Type II errors were made in the course o hypothesis

Example Practice Problems For Stat 509 Final Exam Essay
Example Practice Problems for STAT 509 Final Exam 1. The newest Boeing 787–9 Dreamliner is
designed to be 206 feet in length. The manufacturing process for assembling the 787–9 follows a
normal distribution with mean 205.8 feet and standard deviation 0.5 feet. a) If the specification for
the 787–9 Dreamliner is 206 feet plus or minus 1 foot, or (205' , 207'), what percentage of the
assembled airplanes are beyond the specifications? Draw a Normal Graph of the data. (5) b) Find
the 90% percentile of all 787–9 airplanes in terms of length? Show on a Normal Graph (5) c) Find
the probability that a random sample of 5 airplanes exceeds a mean of 206.5 feet. Draw a Normal
Graph of the data. (5) 2. A group of 300 people were surveyed. Their marital status was recorded
along with answers to several questions. One question asked was whether "Friends and social life"
or "Job or primary activity" contributes most to their general well–being. The results from this
question appear in the table below. Single (never married) Married Widowed or divorced Friends
and social life 47 59 56 Job or primary activity 33 61 44 Find the probability that a randomly
selected person chose "Friends and social life". Find the probability that a randomly selected person
is "Married" and chose "Friends and social life". Find the probability that a randomly selected
person chose "Friends and social life" given the person is "Married". Are the events chose "Friends
and

Description Of The Manager Of Snowpea Restaurant Collected...
Background
Snowpea is a Restaurant specialized in Chinese delivery and carryout. Recently it hired four new
employees for delivery staff. However, after hiring the new delivery people, Snowpea starts getting
complaints from customers about the excessively long delivery times.
Descriptive Statistics
The manager of Snowpea Restaurant collected data on the new four delivery people to find the
possible cause of the complaints. Figure two shows the descriptive statistics for the preparation time,
in minutes, from when the orders are placed until they are ready to be delivered to the customers.
Based on the descriptive statistics of the sample data for the preparation time, the deliverers two and
four have a mean preparation time slightly higher than the total mean of the four people, which
could contribute to a longer delivery time of the orders. However, the mean of those two delivery
people are not significantly higher, so the preparation time could not be the cause of the delays in
delivery.
On the other hand, figure one shows descriptive statistics for the speed, measured in kilometers per
hour, for the four deliverers and also the total speed for the four people. Deliverers one and two,
with average speeds of 42.46 and 48.15 respectively, have a mean speed higher than the mean speed
of the total average of 38.79 of the four deliverers with two having the highest average speed.
However, with a mean speed of 33.51 and 28.14, deliverers three and four have significant lower

Part B Course Project Math 533 Essay
Course Project Part B
a. the average (mean) annual income was less than $50,000
Null and Alternative Hypothesis
H0: mu= 50 (in thousands)
Ha: mu<50 (in thousands)
Level of Significance
Level of Significance = .05
Test Statistic, Critical Value, and Decision Rule
Since alpha = .05, z<–1.645, which is lower tailed
Rejection region is, z<–1.645
Calculate test statistic, x–bar=43.74 and s=14.64
Z=(43.74–50)/2.070=–3.024 2.070 is calculated by: s/sq–root of n
Decision Rule: The calculated test statistic of –3.024 does fall in the rejection region of z<–1.645,
therefore I would reject the null and say there is sufficient evidence to indicate mu<50.
Interpretation of Results and Conclusion ... Show more content on Helpwriting.net ...
the average (mean) credit balance for suburban customers is more than $4300.
15 people of 50 surveyed live in a suburban community, so I will be conducting a t–test because
15<30.
Null and Alternative Hypothesis
H0: mu=$4,300
Ha: mu>$4,300
Level of Significance
Level of Significance= .05

Test Statistic, Critical Value, and Decision Rule
Since alpha= .05, t>1.761, which is upper tailed.
Rejection region is t>1.761
Calculate the test statistic, x–bar=4675 and s=742
T=(4675–4300)/742sqrt15=1.957
Decision Rule: 1.957 is greater than 1.761, which means it does fall in the rejection region, so I
would reject H0. Because I am rejecting H0, this means that there is sufficient evidence to conclude
that the average credit balance for the suburban customers is greater than $4300
Interpretation of Results and Conclusion
p–value=.035
.035<.05
Because the p–value of .035 is less than the significance level of .05, I will reject the null hypothesis
at 5% level.
95%=(4264, 5086)– I am 95% confident that the average credit balance for suburban customers falls
between $4264 and $5,086.
Minitab Output
One–Sample T
Test of mu = 4300 vs > 4300
95% Lower N Mean StDev SE Mean Bound T P
15 4675 742 192 4338 1.96 0.035
Final Report of Results
Before parts a–d are broken down, provided below are a the

Data Analysis And Evaluation Of Findings
Data Analysis and Evaluation of Findings
Brian A. Biacan
TUI University Data Analysis and Evaluation of Findings Students at the Trident University
MAE504 course are taught to develop principles and techniques to conduct research in education.
Understanding the different research methods gives greater appreciation for the importance of
research. This paper will take a research question and go through the process of identifying the
aspects of research that will help meaningfully answer that question to generalize the results to
students and educators alike. Among the topics will include the research methodology, the variables,
the sample population, and the data collection and analysis.
Background.
The research question is, "Is there a significant difference in final grades of students who are team–
taught and students who are not team–taught in the MAE504 course at Trident University." There
are a number of articles that discuss the pros and cons of team–teaching (Kete, 2011, and Money &
Coughlan, 2016), but do not address the statistical data of better or worse grades in comparison to
courses taught by a single instructor. This question narrows the study down to specifically students
of the MAE504 course at Trident University. When conducting research on this question, despite the
specificity of the course, there still are multiple variables to consider to come to an accurate and
useful result.
Research Methodology. The question calls for a difference in

Best Preforming Drug Within The Groups
In the following exploratory data analysis, of the relevant variables, it is our hope to identify the best
preforming drug within the groups (groups A, group B, group C). The exploratory analysis provides
a view of how each group performed over the course of 12 weeks. In figure 1, below is the
descriptive table. The descriptive table provides a view of all variables associated with the
HDL/LDL drug study. The output with the descriptive table contains the overall Norman clashers of
the study, there is a section identified as group, statistics, and standard error. The section labeled
group; it identifies the group of participants in the study. In addition, statistics are provided in the
form of the mean scores for each group as well as the number of participants and error associated
with each group (some data isn't visible due to pivot table activation).
Figure 1
In the LDL group, the overall mean total score of 97.98 (SD = 17.165).
Control group (M = 101.10) Drug A (M = 86.20), drug B (121.40) drug C (M=83.20), and scores
82.25, 94.00, and 109.75 represented the 25th, 50th, and 75th percentiles, respectively.
However, it is necessary to evaluate the associated graphs to aid in the development of
understanding of the data (Field, 2013). Perhaps, one graph that is of importance is the box graph, it
gives the researcher a better view of how the mean scores compare. Using the graphs and
descriptive information in tables gives the researcher enough information to formulate a

Interest In Politics
30050 Applications for Economics Management and Finance
Applied research project and report
"Man is by nature a Political Animal"
FACTORS SHAPING INDIVIDUAL'S INTEREST IN POLITICS
INDEX:
I. ABSTRACT
II. INTRODUCTION
III. DATA DESCRIPTION: Dependent variable and Independent variables
IV. MULTIVARIATE ANALYSIS: Factor, Regression and ANOVAAnalysis
V. CONCLUSIONS
VI. REFERENCES
I. ABSTRACT
The purpose of this project is to analyze what stimulates individual's "Interest in politics". Indeed
politics, either directly or indirectly, shapes community's every day life. It is of capital importance as
it affects individual's welfare and well–being. After gathering the required data from ... Show more
content on Helpwriting.net ...
Inexpensive
2. Quickly obtained
3. Generally high data quality
1. Format
2. Data definitions (scope)
3. Uncertainties about data quality
We should also remember to take into consideration social desirability bias: the tendency of
respondents to answer questions in a way they think will be viewed favorably by others, which may
influence the reliability of our findings.
i. Dependent variable

The dependent variable chosen is "Interest in Politics", a numeric ordinal variable. We had to invert
the scale so that larger values of Y imply higher interest in politics, in line with the other variable's
scale. It now ranges from 1 ("not at all interested) to 4 ("very interested"). With respect to the initial
data screening, we checked the presence of missing values in our dependent variable. As shown in
the table, the percentage of missing data is not significant in relation to our threshold of 5%
considered to be the general rule of thumb.
ii. Independent variable
While describing the independent variables chosen, we have to consider that some are latent nature,
meaning that the underlying concepts cannot be measured directly. In order to create a meaningful
value for said variables, we carried out a factor analysis with the aim of extracting two factors. In
our analysis we also included 3 dummy variables, one single variable and the usual control
variables. A brief

Quantitative Research Paper On Multitasking
Introduction
Multitasking refers to the ability to handle more than one task simultaneously. A popular example of
multitasking is answering a call or text message while driving, which has been identified as a major
cause of road accidents. While some scholars argue that multitasking results in time saving, others
contend that multitasking causes an undivided attention, thereby resulting in a waste of time. It,
however, is worth noting that some tasks require less brain–power than others do; therefore, the
products of multitasking may vary according to the amount of brain–power the tasks require. For
example, chewing while walking, which most people are good at, is an example of multitasking.
Nevertheless, both activities require less brain–power. ... Show more content on Helpwriting.net ...
A quantitative methodology is utilized because it promotes representation, deduction, and data
analysis. According to the Information Resources Management Association (2011), quantitative
research seeks to capture the essence of reality, hence objective. Moreover, quantitative research is
representative, and facilitates the generalization of sample characteristics to the population.
Quantitative research seeks to generate answers to questions such as how much and how many. The
current study seeks to investigate how much efficient multitasking is, in the completion of jobs;
hence the suitability of quantitative research. A random sample of 60 students from whom the data
for this study is sourced, is used. The random sampling technique is preferred because it is
representative of the population, thereby facilitates generalization of findings. Moreover, the use of
this sampling technique promotes the internal and external validity of the study. The selected sample
is equally sufficient and representative of the population. The determination of the sample size is
based on the expected margin of error in gathering information and the confidence level upon which
the statistical tests are based. Each sample had an equal number of male and female

Math 221 Homework Week 7
Math 221 Homework Week 7 – LATEST
IF You Want To Purcahse A+ Work then Click The Link Below For Instant Down Load
http://www.acehomework.net/wp–admin/post.php?post=3168&action=edit IF You Face Any
Problem Then E Mail Us At JOHNMATE1122@GMAIL.COM
1. Math 221 Homework Week 7 – LATEST
2.
3.
4. Use the given statement to represent a claim. Write it's complement and state which is Ho and
which is Ha. u> 635
Find the complement of the claim. u < 635
2. A null and alternative hypothesis are given. Determine whether the hypothesis test is left–tailed,
right tailed, or two–tailed.
What type of test is being conducted in this problem?
Answer: Right–tailed test
3. Write the null and alternative hypotheses. Identify which is ... Show more content on
Helpwriting.net ...
Z = 1.40
Area = 0.919
P Value = 0.081
Reject Ho
At the 3% significance level, there is not enough evidence to support the administrator's claim that
the mean score for the state's 8th graders on the exam is more than 270.
13. A company that makes cola drinks states that the mean caffeine content per 12–ounce bottle of
cola is 45 milligrams. You want to test this claim. During your tests, you find that a random sample
of thirty 12–ounce bottles of the cola has a mean caffeine content of 45.5 milligrams with a standard
deviation of 6.1 milligrams. At a = 0.08, can you reject the company's claim?
The critical values are = 1.75 z = 0.45
Since z is not in the rejection region, fail to reject the null hypothesis.
At the 8% significance level, there is not enough evidence to reject the company's claim that the
mean caffeine content per 12–ounce bottle of cola is equal to 45 milligrams.
14. A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least

975 hours. A random sample of 72 light bulbs has a mean life of 954 hours with a standard deviation
of 85 hours. Do you have enough evidence to reject the manufacturer's claim? Use a = 0.04.
Zo = –1.75
Z = –2.10
Reject Ho. There is sufficient evidence to reject the claim that mean bulb life is at least 975 hours.
15. An environmentalist estimates that the mean waste recycled by adults in the country is more than
1 pound per person per day. You want to this test

Hcs/438 Dq's Essay
HCS/438 DQ's
Week 1:
DQ1: What are the differences between descriptive and inferential statistics?
According to Bennett (2009), the biggest difference between descriptive and inferential statistics is
that descriptive statistics "deals with describing raw data in the form of graphics and sample of
statistics" and inferential statistics "deals with estimating population parameters from sample data."
This means that inferential statistics would be an estimate because the data would be estimated from
sample data rather than using specific data whereas descriptive statistics would be more accurate.
An example of descriptive statistics would be trying to find an average of something such as a
G.P.A. or your overall grade in a class. ... Show more content on Helpwriting.net ...
We come up with the number 7. 7 would be the mean in this case.
"The median is the middle value of the data set. To find a median we arrange the values in ascending
(or descending) order, repeating data values that appear more than once. If the number of values is
odd, there is exactly one value in the middle of the list, and this value is the median. If the number
of values is even, there are two values in the middle of the list, and the median is the number that
lies halfway between them. For an example the list 3, 4, 6, 6, 10. The median number is 6 because 6
is the middle number in the list." (Bennett, Briggs, & Triola, 2009, p. 146).
"The mode is the most common value or group of values in a data set. For an example the mode in
the number set 3, 4, 6, 6, 10 is 6 because this value occurs twice in the data set."(Bennett, Briggs,
& Triola, 2009, p. 146).
We would use mean, median and mode in healthcare to find the average of many things such as how
effective a new medication would be on the average population. Or how far into a new treatment
patients start seeing results or improvements. Bennett, J.O., Briggs, W.L., & Triola, M.F.
(2009). Statistical Reasoning for Everyday Life (3rd ed.). Boston, MA: Pearson Education Inc
DQ2: Measures of variability are range, Interquartile range, variance, and standard deviation. Range
is the difference between the highest and the lowest

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