Statistics in R and Python
This section showcases my work from the Statistics course at the University of South Florida (USF), where I applied statistical techniques using R and Python.
- Language & Tools: R, Python
- Topics Covered:
- Descriptive statistics
- Inferential statistics
- Regression analysis
- Data visualization
You can view my course projects and related works below:
Final Project
RStudio Code
RStudio Console
This is related to our classwork, as we learned anova, paired t-tests, and TukeyHSD methods of analysis in class.
First R Studio Assignment LIS 4372
1 assignment = c(6,18,14,22,27,17,22,20,22)
2 myMean = function(assignment) {
3 return(sum(assignment)/length(assignment))
4 }
5 myMean(assignment)
So, what happens in the code above?
1: we created a container "assignment" that holds the data values of 6,18,14,22,27,17,22,20,22
2-4: we created a function "myMean" that returns the mean of a given container
5: we called "myMean" with a parameter of (assignment) which returned 18.66667 as the mean of assignment
Module 3 R Studio Assignment LIS 4372
s1 = c(10,2,3,2,4,2,5)
s2 = c(20,12,13,12,14,12,15)
# for each set
# compute mean, median, mode under Central Tendency
# mean
s1mean = mean(s1)
s2mean = mean(s2)
# median
s1median = median(s1)
s2median = median(s2)
# create mode function
getMode = function(data) {
uniqv = unique(data)
uniqv[which.max(tabulate(match(data, uniqv)))]
}
# mode
s1mode = getMode(s1)
s2mode = getMode(s2)
# compute range, interquartile, variance, standard deviation under Variation
# range
s1range = range(s1)
s2range = range(s2)
# interquartile
s1IQR = quantile(s1)
s2IQR = quantile(s2)
# variance
s1variance = var(s1)
s2variance = var(s2)
# standard deviation under variation
s1SD = sd(s1)
s2SD = sd(s2)
# results
summary(s1)
summary(s2)
Comparing Datasets s1 and s2
The values in the dataset s2 are equivalent to the values of dataset s1 + 10. The interquartile range, mean, median, mode, and range of s2 are equivalent to the values in s1 + 10. The variance of s1 and s2 is the same at 8.3 repeating. The standard deviation of s1 and s2 is the same at ~2.887. The variance and standard deviation of the datasets are the same because the values deviated from the mean the same amount.
Module 4 Assignment
A.
a1. P(A) = 10/90 + 20/90 = 30/90 = 1/3
a2. P(B) = 10/90 + 20/90 = 30/90 = 1/3
a3. P(A | B) = P(A + B) = P(A) + P(B) - P(A n B)
P(A | B) = 1/3 + 1/3 - 1/9 = 5/9
a4. P(A | B) = P(A) + P(B)?
P(A | B) = 5/9
P(A) + P(B) = 1/3 + 1/3 = 2/3
5/9 != 2/3
P(A | B) != P(A) + P(B)? no
B.
b1. The answer of 11% is true.
b2. We know the answer is true. Bayes’ Theorum allows us to evaluate the probability of the weatherman’s successful prediction of rain.
This value, 11.11%, tells us that it will not rain more often than not when the weatherman predicts rain.
Therefore, there is about a 89% chance it will not rain on Jane’s wedding day tomorrow.
C.
* using RStudio we get
C1. dbinom(10, 10, 0.2) = 1.024e-07
The probability of 10 successes in a row given a probability of success of 0.2 (20%) is 1.024e-07.
Module 5 Assignment : P-val and Correlation Analysis
First Question:
The director of manufacturing at a cookies needs to determine whether a new machine is production a particular type of cookies according to the manufacturer’s specifications, which indicate that cookies should have a mean of 70 and standard deviation of 3.5 pounds. A sample pf 49 of cookies reveals a sample mean breaking strength of 69.1 pounds.
A. State the null and alternative hypothesis ___
Null Hypothesis (H_o) : u ≥ 70
The cookies have a breaking strength of 70 or greater.
Alternate Hypothesis (H_a) : u < 70
The cookies have a breaking strength of less than 70.
B. Is there evidence that the machine is not meeting the manufacturer’s specifications for average strength? Use a 0.05 level of significance ___
C. Compute the p value and interpret its meaning ___
1. Specify the level of significance
level of significance (a) = 0.05
2. Compute the Z-value
z = (mean - u)/(sd / sqrt(n))
mean = 70
u = 69.1
sd = 3.5
n = 49
z = (70 - 69.1) / (3.5 / sqrt(49)) = 1.8
3. Compute p-val
for z = 1.8, cumulative probability = 0.964070
p-val = 1 - cumulative probability = 1 - 0.964070 = 0.03593
4. Determine whether to reject H_o
Because the p-value of 0.03593 < 0.05, there is sufficient statistical evidence to infer the cookies do not meet the manufacturer’s specification of an average breaking strength of 70 lbs.
D. What would be your answer in (B) if the standard deviation were specified as 1.75 pounds?__
if sd = 1.75, z = ?
z = (70 - 69.1) / (1.75 / sqrt(49)) = 3.6
cumulative probability = 0.99984
p-val = 1 - 0.99984 = 0.00016
Because the p-value of 0.00016 < 0.01, there is a significant amount of statistical evidence to infer the cookies do not meet the manufacturer’s specification of an average breaking strength of 70.
E. What would be your answer in (B) if the sample mean were 69 pounds and the standard deviation is 3.5 pounds? __
if mean = 69, sd = 3.5, z = ?
z = (70 - 69) / (3.5 / sqrt(49)) = 2
cumulative probability = 0.97725
p-val = 1 - 0.97725 = 0.02275
Because the p-value of 0.02275 > 0.05, there is sufficient statistical evidence to infer the cookies do not meet the manufacturer’s specification of an average breaking strength of 70.
Second Question:
If x̅ = 85, σ = standard deviation = 8, and n=64, set up 95% confidence interval estimate of the population mean μ.
given : mean = 85, sd = 8, n=64
set up a 95% confidence interval estimate of the population mean u
z = 1.96 for 95% confidence
Confidence Interval Formula
mean +- z(sd / sqrt(n))
85 + 1.96(8/sqrt(64)) = 86.96
85 - 1.96(8/sqrt(64)) = 83.04
With 95% confidence the population mean is between 83.04 and 86.96, based on 64 samples.
Third Question
using the following dataset
a. Calculate the correlation coefficient for this data set _____
The correlation coefficient comparing the total values for girls and boys is 0.9999681; This indicates a strong positive correlation of 0.99.
More Correlations can be found below.
b. Pearson correlation coefficient _____
The Pearson correlation coefficient comparing the total values for girls and boys is 0.9999681; this indicates a strong positive correlation of 0.99.
More Correlations can be found below.
c. Create plot of the correlation
Completed in R below
# Module 5 Assignment
# Correlation Analysis
setwd("C:/Users/justi/Desktop/r")
# x = girls, y = boys
# vars = goals, grades, popular, time
# create data
# create girls
goals = c(4, 5, 6)
grades = c(49, 50, 69)
popular = c(24, 36, 38)
timeSpent = c(19, 22, 28)
total = c(92, 108, 135)
girls = data.frame(goals, grades, popular, timeSpent, total)
girls
# create boys
# goals is the same
grades = c(46.1, 54.2, 67.7)
popular = c(26.9, 31.6, 39.5)
timeSpent = c(18.9, 22.2, 27.8)
total = c(95.9, 113, 141)
boys = data.frame(goals, grades, popular, timeSpent, total)
boys
cor(girls, boys)
data = data.frame(girls, boys)
data
# correlation
cor(data, method="pearson")
cor(data, method="spearman")
# plot
corrgram(data)
Module 6 Assignment
A
// Module 6 Assignment
// Justin Stutler
// A
// consider a population: 8,14,16,10,11
p = c(8,14,16,10,11)
// compute mean and standard deviation
pmean = mean(p)
pmean
psd = sd(p)
psd
// select a random sample of size 2
sample = c(16, 11)
// compute mean and standard deviation for sample
smean = mean(sample)
smean
ssd = sd(sample)
ssd
// compare sample mean and sd to population
smean
pmean
ssd
psd
// smean of 13.5 is larger than pmean of 11.8
// ssd of 3.536 is larger than psd of 3.194
B
// B
// n = 100, p =0.95
pn = 0.95 * 100
pn
// pn = 95 >= 10
qn = 0.05 * 100
// qn = 5 ! >= 10
// therefore not normal distribution
// 10 = 0.05n
n = 10 / 0.05
n
// smallest value is n = 200
C
explain how rbinorm is probably more suited for coin flip simulation than sample.
rbinorm is suited for something with 2 mutually exclusive outcomes with consistent probability such as flipping a coin. The 2 mutually exclusive outcomes are landing heads or tails with each of the options having a consistent probability of 0.5.
Sampling can be used with replacement or without. Sampling without replacement would not work in the case of flipping a coin. Sampling with replacement would work if the sample contained an equal amount of heads and tails options, but the concept of flipping a single object with a binary outcome just comes more naturally froma binomial distribution.
Module 7 Assignment
R Code
// Module 7 Assignment
// Regression
// Justin Stutler
// 1.1 Define the relationship between the predictor and response variable
// Predictor: X
// Response: y
// Describe x by y
d.data = data.frame(
x = c(16, 17, 13, 18, 12, 14, 19, 11, 11, 10),
y = c(63, 81, 56, 91, 47, 57, 76, 72, 62, 48)
)
d.data
// 1.2 calculate the coefficients
d.regression = lm(x ~ y, data = d.data)
d.regression
// 2
// create data
visit.data = data.frame(
discharge = c(3.600, 1.800, 3.333, 2.283, 4.533, 2.883),
waiting = c(79, 54, 74, 62, 85, 55)
)
visit.data
// 2.1
// Predictor: discharge
// Response: waiting
// Describe discharge by waiting
visit.regression = lm(discharge ~ waiting, data = visit.data)
visit.regression
// 2.2 extract coefficients
visit.coeffs = coefficients(visit.regression)
visit.coeffs
// 2.3 fit discharge duration
wait = 80
duration = visit.coeffs[1] + visit.coeffs[2]*wait
duration
// 3
// load mtcars data set
car.data = mtcars[,c("mpg","disp","hp","wt")]
print(head(car.data))
// 3.1 Examine the relationship Multi Regression Model as stated above and its Coefficients using 4 different variables from mtcars (mpg, disp, hp and wt).
// what does the multi regression model and coefficients tells about the data?
car.regression = lm(formula = mpg ~ disp + hp + wt, data = car.data)
car.regression
// (Intercept) gives the y intercept
// disp, hp, wt display the change in these variables per 1 mpg change
// 4
// required library
library(ISwR)
// 4.1 According to the fitted model, what is the predicted metabolic rate for a body weight of 70 kg?
plot(metabolic.rate~body.weight,data=rmr)
r = lm(metabolic.rate ~ body.weight, data = rmr)
r
coeffs = coefficients(r)
coeffs
body_weight = 70
dur = coeffs[1] + coeffs[2]*body_weight
dur
// Based on the simple linear regression model, if the body weight is 70 kg, we expect the metabolic rate to be 1305.394
R Studio Console
Assignment 8 : ANOVA
R Studio Code
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Assignment 9
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Assignment 10
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Assignment 11
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Assignment 12
R Studio Code
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