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# Final Exam XXXXXXXXXXMathematics and Statistics for Social Sciences Author: XXXXXXXXXXVanda Salari LL.B 2019/2020 year student Student Number: B019080 Introduction The definition of success in the...

Final Exam
XXXXXXXXXXMathematics and Statistics for Social Sciences
Author: XXXXXXXXXXVanda Salari
LL.B 2019/2020 year student
Student Number: B019080
Introduction
The definition of success in the world of sports can be interpreted in a number of ways.
Success in basketball can be defined using the value of the player. The National
Basketball Association (NBA) uses different factors to determine its most successful
player. It is important to look into characteristics which could be attributed to a playerâ€™s
success.
From College to the NBA, Evans examines the characteristics the NBA considers to
determine a playerâ€™s success. The study gathered data from 2006 to 2013 draftees.
Evans looked into how the duration of the playerâ€™s career prior to being drafted, as well
as their qualification. The variables included statistics from the playersâ€™ games. Othe
variables which were considered were their age, height, and BMI. The result of the
esearch showed that height and in game statistics were co
elated to success.
Meanwhile, number of fouls, age, and late start were co
elated to higher draft positions
written by NBAstuffer themself to help their fans and was last updated on june 14, 2019
Moxley and Towne also focus on the early career success of NBA players. They studied
multiple factors and found the top variable to be the playersâ€™ performance history. In
addition, height, arm span, vertical leap, and agility are also taken into consideration.
The results of the study showed that although physical attributes are taken into account,
they are not distinguishing factors to the playerâ€™s success. The factors which were found
to be useful were age, win shares, and college quality.
Measuring Greatness in the NBA looks into the question of how greatness and success
is measured in the NBA. Whitmoyer believes that a single factor is insufficient when
determining the success of the player, noting that other factors should also be taken into
account. The playerâ€™s compensation is often used as a reflection of their value,
however, it is not always reliable. The perception of the player is also key to a playerâ€™s
value.
The articles took into account qualitative and quantitative factors to justify their thesis.
All three journals look into qualitative and quantitative profiles to find what determines
success. They found physical attributes to not be as important of a factor as the
performance and other ratings. Age was also found to be rather important, as well as
the history of the player. How long they have been playing professionally is also a big
determining factor. Evans takes into account more in game statistical data as compared
to the other two, especially Whitmoyer. Whitmoyer looks into the measurement of
success more than the actual statistics. Physical attributes arenâ€™t much mentioned in the
3rd journal, but are taken into account in Evans and Moxleyâ€™s findings. The articles have
used more specific and thorough data to explore the topic.
There are a number of factors which need to be considered to find what indicates
success. Different studies have been conducted to find which factors are most
co
elated to NBA Career success. Although the data acquired has been useful in
finding possible factors to success, it is important to gather more information to na
ow
Research Questions
What is the minimum wage for an NBA player?
How many NBA players out of 202 have an average yearly rating?
During which are the most successful and earn most?
Methodology
A. Study of NBA players and their values
The Data used for the research is taken for the NBA PlayOffs between the year of 2018-
2019 from the official NBA official webpage; https:
www.nbastuffer.com/ XXXXXXXXXXnba-
player-stats/. The data summarized the NBA players yearly value and wage according to
their performance throughout the year which affect their yearly ratings data Analysis.
Due to each NBA player not having the same exact characteristics the variables were
esearched separately such as each NBA player's net worth and income depending on
their yearly ratings which is calculated at the end of each playoffs.
B. Data Analysis
The results are achieved and will be analyzed in the Microsoft Excel system. It can be used to
test outcomes, to verify hypotheses, degrees of trust and association. Three theories will be
tested out.
Hypothesi: 1
Ho: p=>0.77
Ha: pâ‰  0.77
It is a one-sided theory test of 0.06 alpha and 90 per cent of the Confidence level. Before the
hypothesis is tested, the data must be modeled and the modeling parameters and assumptions
followed.
Independence:
The values of this database are independent from each NBA player.
Sample Size:
The Sample is large enough due to 157/202 students being only american nationals.
https:
www.nbastuffer.com/ XXXXXXXXXXnba-player-stats
https:
www.nbastuffer.com/ XXXXXXXXXXnba-player-stats
Use
Sticky Note
Inco
ect null hypothesis.
Use
Sticky Note
How can alpha be 0.06 when your confidence level is 90%?
Use
Sticky Note
These hypotheses don`t match the table below in the results section.

Randomization:
There is an application of randomization to the applicant of the database.
10% Condition:
The data includes more than half of the population of the American players therefore this
satisfies the 10% condition. The conditions are satisfied, so I will use the Usual Template to
consider a z-interval in one proportion.
Hypothesis: 2
H0: Î¼ = 45
HA: Î¼ > 45
Hypothesis 2 is a one-sided hypothesis with an alpha of 0.06 and a confidence interval of 95 per
cent, with data on average NBA players ranked between 75-77 percent on an annual basis. To
un an independence, randomization and 10 per cent t-test.To ca
y out a t-test of equality,
andomization and 10 percent criteria must be met, and all of these conditions and assumptions
have been met, as stated in the Hypothesis 1 test. Hence we can ca
y on the work.
Hypothesis: 3
H0: Î¼ = 0.6
Ha: Î¼ = 0.6
As mentioned before, all the assumptions and parameters for this data set are satisfied, and a
two-sided hypothesis test for the null hypothesis and alternate hypothesis was ca
ied out as
follows: this data was analyzed with a confidence interval of 95% and an alpha of 0.06
Results
Hypothesis 1
Sample proportion XXXXXXXXXXZ-stat XXXXXXXXXX
Use
Sticky Note
How can alpha be 0.06 when your confidence level is 95%?
Why are you changing confidence levels during the paper?
Use
Sticky Note
Inco
ect alternative hypothesis.

sample size n 202 P(Z<=z) one tail XXXXXXXXXX
Hypothesized proportion 0.6. z Critical one-tail XXXXXXXXXX
Alpha 0.06 P(Z<=z) Two tail XXXXXXXXXX
90% Confidence
Interval
The study results suggest that the p value is = 0.77, because the p value is higher than the 0.05
alpha value, we reject the null hypothesis and accept the alternative hypothesis. Therefore, our
study allowed us to assume with 90% accuracy, to assume that this is not a fair benefit, and to
dismiss the null hypothesis.
Hypothesis 2
Sample mean 201 t-Stat XXXXXXXXXX
Sample standard deviation 123.2 P(T<=t) one-tail XXXXXXXXXX
Sample Size 22 t Critical lower one-tail XXXXXXXXXX
Hypothesized mean 45 t Critical upper one-tail XXXXXXXXXX
Alpha 0.05 P(T<=t) two-tail XXXXXXXXXX
95% Confidence
Interval
The observational results show that the mean of the sample is 0.2 as the median is greater than
the prediction of 0.05 alpha, we reject the null hypothesis and accept the alternative hypothesis.
Our study therefore allowed us to predict an accuracy of 95 per cent, to infer that this is not a
fair advantage, and to dismiss the null hypothesis.
Hypothesis 3
Mean XXXXXXXXXX
Variance XXXXXXXXXX
StdDev XXXXXXXXXX
Observations 77
Hypothesized Mean difference 800
Use
Sticky Note
Completely wrong interpretation of the results. Also, I can't understand what is written here.
Use
Sticky Note
Completely wrong interpretation of the results. Also, I can't understand what is written here.
Use
Sticky Note
Which two groups are you comparing?

df 88
t-Stat 0
P(T<=t) one-tail XXXXXXXXXX
t Critical lower one-tail XXXXXXXXXX
t Critical upper one-tail XXXXXXXXXX
The table indicates certain predictive values for the study's two samples, the analysis is a one-
sided conclusion since we were interested in checking whether a large difference in mean exists
and that one group received better outcomes than another. We refuse to accept the argument
null as the significance p is smaller than the significance alpha XXXXXXXXXXSo we can tell with 95
percent trust.
Co
elation and Regression
Use
Sticky Note
Completely wrong interpretation of the results. Also, I can't understand what is written here.
Regression Statistics
Multiple R XXXXXXXXXX
R square XXXXXXXXXX
Standard E
or -
Observations 202
The present model of linear relationship was examined and it was found that there was a strong
positive co
elation (R=0.674) between the total number of applicants and the quantity of
applicants permitted. The R squared value was 0.454, which indicated that this model was very
ealistic and compensated for 45 per cent of the tests, with 55 percent remaining out of the
esiduals. The scatter plot shows a straight upward axis, with a 0.90x equation. The intercept y
is a negative integer, and is thus excluded.
Conclusion:
In conclusion, 77% of the NBA players are American nationals leaving only 23% of the NBA
players international, the analization of all these factors to render figures on which club pays
higher salaries, at which age players have the highest value and earnings, which average
anking and job performance is required for higher value and higher wages, which ethnicity
football players are most productive and which foot they usually prefer; as well sorting the
columns using value and wages from highest to lowest, along with age, yearly rating, nationality,
working rates, positions, and prefered clubs.

Use
Sticky Note
Did we only discuss one type of charts during this course?
Use
Sticky Note
What estimates did you find for
Answered Same Day Jun 15, 2021

## Solution

Biswajit answered on Jun 15 2021
Data Analysis:
Hypothesis test 1: Z proportion test nationality
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proportion test
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H0
proportion of American players is 0.5
0.5
Assumed proportion
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Ha
proportion of American players is more than 0.5
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sample size
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202
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P
proportion of American players in sample
0.77
Proportion as per data
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level of significance
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0.05
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Z statistic
9.118642058
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p value
0.0000
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As the p value is lower than level of significance (0.05), we reject the null hypothesis.so we say that there are substantially more no of American players in comparison to others
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conclusion

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Assumptions
np >10
156
yes
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np(1-p)>10
36
yes
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samples are independent
Â
yes
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Data points more than 30 to use normality assumption as per central limit theorem
202
yes
Hypothesis test 2: T test for difference of ratings between American & rest of others players
F-Test Two-Sample for Variances
Â
Â
Â
Â
Â
Â
American yearly ratings
Rest of others yearly ratings
Mean
77.24516129
78.36170213
Variance
35.77067449
49.14893617
Observations
155
47
df
154
46
F
0.727801602
Â
P(F<=f) one-tail
0.078827566
Â
F Critical one-tail
0.690790631
Â
Level of significance
Â
0.05
Â
variances are equal
as p value > .05
t-Test: Two-Sample Assuming Equal Variances
Â
Â
Â
Â
Â
Â
American yearly ratings
Rest of others yearly ratings
Mean
77.24516129
78.36170213
Variance
35.77067449
49.14893617
Observations
155
47
Pooled Variance
38.84767467
Â
Hypothesized Mean Difference
0
Â
df
200
Â
t Stat
-1.075799136
Â
P(T<=t) one-tail
0.141656571
Â
t Critical one-tail
1.652508101
Â
P(T<=t) two-tail
0.283313141
Â
t Critical two-tail
1.971896224
Â
Level of significance
Â
0.05
conclusion
American yearly ratings are not statistically different from rest pf others players
As p value is greater than level of significance 0.05
Hypothesis test 3:t Test for difference of Age between American players & others
F-Test Two-Sample for Variances
Â
Â
Â
Â
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American Age
Others Age
Mean
26.86451613
27.19148936
Variance
15.10490155
17.81036078
Observations
155
47
df
154
46
F
0.848096327
Â
P(F<=f) one-tail
0.228354436
Â
F Critical one-tail
0.690790631
Â
Â
Â
Â
Â
variances are equal
as p value > .05
Â
Â
Â
t-Test: Two-Sample Assuming Equal Variances
Â
Â
Â
Â
Â
Â
American Age
Others Age
Mean
26.86451613
27.19148936
Variance
15.10490155
17.81036078
Observations
155
47
Pooled Variance
15.72715717
Â
Hypothesized Mean Difference
0
Â
df
200
Â
t Stat
-0.495137912
Â
P(T<=t) one-tail
0.310523156
Â
t Critical one-tail
1.652508101
Â
P(T<=t) two-tail
0.621046312
Â
t Critical two-tail
1.971896224
Â
Â
Â
Â
conclusion
American Ages are not statistically different from rest of others players
As p value is greater than level of significance 0.05
Hypothesis test 4: t test for comparison of wages of American players & others
F-Test Two-Sample for...
SOLUTION.PDF