MATH399 Course Project Complete Solution
Introduction
The topic that I have chosen is the graduation rates of US high schools for the year 2016-2017 (“

... [Show More] High School Graduation Rates by State 2019”, 2019). The data is found from World Population Review and lists the graduation rates as a percentage for each state. Only public schools are considered within the data as information from private schools are not accessible since they are not run by the government.
Sample Data
89.3 78.2 78 88 82.7 79.1 87.9 86.9 82.3 80.6
82.7 79.7 87 83.8 91 86.5 89.7 78.1 86.9 87.7
88.3 80.2 82.7 83 88.3 85.8 89.1 80.9 88.9 90.5
71.1 81.8 86.6 87.2 84.2 82.6 76.7 86.6 84.1 83.6
83.7 89.8 89.7 86 89.1 86.9 79.4 89.4 88.6 86.2
Table 1. Graduation Rate Data Set in All Fifty States [1]
Problem Computations
Mean
μ=(89.3+78.2+⋯+86.2)/50=84.6
Standard Deviation
σ=√((∑_(i=1)^50▒〖(x_i-84.6)〗^2 )/50)≈4.2697=4.27
Confidence Intervals and Margin of Error
80%
z_c=1.28
E=z_c*σ/√n=1.28*4.27/√50=0.773≈0.77
upper limit=μ+E=84.6+0.77=85.37
lower limit=84.6-0.77=83.83
confidence interval=[83.83,85.37]
95%
z_c=1.96
E=1.96*4.27/√50=1.183≈1.18
upper limit=84.6+1.18=85.78
lower limit=84.6-1.18=83.40
confidence interval=[83.40,85.78]
99%
z_c=2.575
E=2.575*4.27/√50=1.55496≈1.55
upper limit=84.6+1.55=86.15
lower limit=84.6-1.55=83.01
confidence interval=[83.01,86.15]
90% (chosen confidence interval)
z_c=1.645
E=1.645*4.27/√50=0.99
upper limit=84.6+0.99=85.5
lower limit=84.6-0.99=83.5
confidence interval=[83.5,85.5]
Problem Analysis
What trend do you see takes place to the confidence interval as the confidence level rises? Explain mathematically why that takes place.
As the confidence interval rises, the upper limit gets bigger and the lower limit becomes smaller. It is expected to see this observation mostly due to the level of confidence increasing. The bigger the difference between the upper and lower limits, the more confident we are in predicting that the value is within those bounded limits.
Provide a sentence for each confidence interval created in part c) which explains what the confidence interval means in context of topic of your project.
With a confidence interval of 80%, we are 80% confident that the average graduation rate within a sample of US public high schools is between 83.83% and 85.37%.
With a confidence interval of 95%, we are 95% confident that the average graduation rate within a sample of US public high schools is between 83.40% and 85.78%.
With a confidence interval of 99%, we are 99% confident that the average graduation rate within a sample of US public high schools is between 83.01% and 86.15%.
With a confidence interval of 90%, we are 90% confident that the average graduation rate within a sample of US public high schools is between 83.5% and 85.5%.
Explain how Part I of the project has helped you understand confidence intervals better?
This helped me understand the topic of confidence interval much better by giving me the opportunity to do calculations on an example data set. Calculating the upper and lower limits for each confidence interval and comparing each interval allowed me to understand what it means with regards to its context.
How did this project help you understand statistics better?
In this project, I was able to analyze the outcomes of the calculations and understand its overall significance with regards to the topic. The calculations gave me a better understanding of the data set, which is one of the objectives of statistics.
Course Project 2
Chosen Data Set
Volume 58. Number 22. Release Date: July 9, 2009. Column: October 2009
Preliminary Calculations
Summary Table for Live Births
Mean 6754
Median 4643
Standard Deviation 8819.19
Minimum 519
Maximum 46661
Summary Table for Deaths
Mean 4001
Median 2961
Standard Deviation 3986.42
Minimum 278
Maximum 19319
Summary Table for Marriages
Mean 3801
Median 3000
Standard Deviation 3487.41
Minimum 270
Maximum 16447
Summary Table for Divorces
Mean 1582
Median 1274
Standard Deviation 1395.59
Minimum 30
Maximum 7008
Hypothesis Testing
H_o: M≤5000
H_a:M>5000 (claim)
α=0.05
z=(6754-5000)/(8119.19/√52)=1.56
P=1-0.9406=0.0594
P>α
Conclusion: Fail to reject null hypothesis H¬0.
At the 5% significance level, there is not sufficient evidence to conclude that the average amount of live births is over 5000.
H_o: M=6000 (claim)
H_a:M≠6000
α=0.10
z=(4001-6000)/(3986.42/√52)=-3.62
P=2(0.05)=0.1
P≤α
Conclusion: Reject null hypothesis H¬0.
At the 10% significance level, there is not sufficient evidence to conclude that the average amount of deaths is 6000.
H_o: M≥2500(claim)
H_a:M<2500
α=0.05
z=(3801-2500)/(3487.41/√52)=2.69
P=0.9964
P>α
Conclusion: Fail to reject null hypothesis H¬0.
At the 5% significance level, there is sufficient evidence to conclude that the average amount of marriages is greater than or equal to 2500.
H_o: M≤4000 (claim)
H_a:M>4000
α=0.10
z=(1582-4000)/(1395.59/√46)=-11.75
P=1
P>α
Conclusion: Fail to reject null hypothesis H¬0.
At the 10% level of significance, there is sufficient evidence to conclude that the average amount of divorces is less than or equal to 4000.
My own hypothesis: Determine if there is enough evidence to conclude that the average amount of divorces is less than 3000 in the United States and territories at a level of confidence of 0.10.
H_o:M≥3000
H_a:M<3000 (claim)
α=0.10
z=(1582-3000)/(1395.59/√46)=-6.89
P=0
P≤α
Conclusion: Reject null hypothesis H¬0.
At the 10% level of significance, there is sufficient evidence to conclude that the average amount of divorces is less than 3000. [Show Less]