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6-3x=-2x
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6 - 3x = -2x
step 1. Add 3x to both sides
6 - 3x + 3x = -2x + 3x
step 2. simplify both sides
6 = 1x
So x=6
double check by substituting 6 for x
6 - 3 (6) = -2 (6)
6 - 18 = -12
-12 = -12 CHECKS OUT, so x=6 is correct
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West Coast University Data Analysis Descriptive Statistics Project
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West Coast University Data Analysis Descriptive Statistics Project
Survey Information Links https://www.surveymonkey.com/results/SM-RBM2J2FX7/https://www.surveymonkey.com/results/SM-J5BKC2FX7/Research Focus:how creating health programs for children and parents in middle school lead to decreasing the consumption of tobacco in teenagersPurpose of research:Tobacco smoking in the country is the primary source of preventable death. Many smokers (82 percent) began to smoke before 18, and about 3,000 young people continue to smoke every day. School initiatives to discourage the usage of cigarettes may become one of the most successful nicotine approaches implemented. This survey seeks to identify whether creating health programs for children and parents in middle school lead to decreasing the consumption of tobacco in teenagers. The survey will determine whether the reduced use of tobacco among teenagers (dependent variable) is directly challenged by the health programs put in place in middle schools (independent variable). This survey will target teenage students as well as parents in middle school and consequently summarize their feedback regarding the issue. AssignmentBy this week, you should have completed data collection on at least 30 participants who met the qualifications to complete the survey (e.g., at least 18 years or older, have other specific characteristics that make them a member of the population you intended to study).The data for your short survey should be entered into the Excel database that you created last week. With this data set, complete the following:Calculate descriptive statistics for your main independent and other variables you have collected information on (example: demographic variables).Create a bar chart for the participants’ gender and ethnicity.Create a histogram for participants’ age. Calculate descriptive statistics for your dependent variable. Create a summary table of your descriptive statistics.Hint: For categorical variables, you should be calculating frequencies and percentages; for continuous variables, you should be calculating means and standard deviations.Use one hypothesis-testing statistic that examines the relationship between your independent variable and dependent variable.Hint: This test should be one of the following—independent samples t-test, paired t-test, one-way ANOVA, or Pearson’s chi-squared test.Write the null and alternate hypotheses.What is the calculated statistic value?What is the p and critical statistic value?Is the p value significant? Is calculated statistic value greater than critical value?What is your conclusion?
stats homework using R
#Name:
#Student ID:
rm(list=ls())
source('Rallfun-v33.txt')
#PART 1
#A company claims that, when exposed to their to ...
stats homework using R
#Name:
#Student ID:
rm(list=ls())
source('Rallfun-v33.txt')
#PART 1
#A company claims that, when exposed to their toothpaste, 45% of all bacteria related to gingivitis are killed, on average. You run 10 tests and ???nd that the percentages of bacteria killed in each test was:
# 38%, 44%, 62%, 72%, 43%, 40%, 43%, 42%, 39%, 41%
# Assuming normality, you will test the hypothesis that the average percentage of bacteria killed was 45% at alpha=0.05.
#1.1) Write out the Null and Alternative hypotheses
#1.2) Calculate the T-statistic and use Method 1 (we saw in class) to determine if the average bacteria killed was 45%. Do it by "hand".
#Hint: Method 1 is to compare T to a critical value "c".
#1.3) Do you reject or fail to reject the null?
################################################
#PART 2
#Now, let's not assume normality
#2.1) Using the same data as in Part 1, test the hypothesis that the 20% trimmed mean is 45%?
#2.2) Do you reject or fail to reject the null?
#2.3) Assuming your test in 2.1 is the truth, what type of error did you make in #1.3?
################################################
#PART 3
#In a study of court administration, the following times to disposition (in minutes) were determined for twenty cases and found to be:
# 42, 90, 84, 87, 116, 95, 86, 99, 93, 92, 121, 71, 66, 98, 79, 102, 60, 112, 105, 98
#Assuming normality, you will test the hypothesis that the average time to disposition was 99 minutes at alpha=0.05.
#3.1) Write out the Null and Alternative hypotheses
#3.2) Calculate the T-statistic and use Method 2 (we saw in class) to determine if the average time to disposition was 99? Do it by "hand".
#Hint: Method 2 is to evaluate the confidence interval.
#3.3) Do you reject or fail to reject the null?
################################################
#PART 4
#Now, let's not assume normality
#4.1) Using the same data as in Part 3, test the hypothesis that the 20% trimmed mean is 99?
#4.2) Do you reject or fail to reject the null?
#4.3) Assuming your test in 4.1 is the truth, what type of error did you make in #3.3?
################################################
#PART 5
#Suppose you run an experiment, and observe the following values:
# 12, 20, 34, 45, 34, 36, 37, 50, 11, 32, 29
#You will test the hypothesis that the average was 25 at alpha=0.05.
#5.1) Write out the Null and Alternative hypotheses. Conduct the hypothesis test assuming normality. Use the "t.test" function. Do you reject or fail to reject the null?
#5.2) Conduct the hypothesis test without assuming normality. Do you reject or fail to reject the null?
#5.3) Assuming the answer in #5.2 is the truth, what type of error (if any) did you make in #5.1 by assuming normality?
------------------------------------------------------------------------------------------
Lab 7- Lecture Notes (FOR YOUR REFERENCE)
#Lab 7-Contents
#1. Formulating Hypotheses
#2. T-statistics by Hand
#3. Alpha Level
#4. Evaluating Our Results
#5. Using the t.test function
#6. T-tests with Trimmed Means (trimci function)
#7. Type 1 and Type 2 Errors
# Last week we talked about computations for when the Population
#Variance is known and unknown.
# Given that we rarely know the population variance,
#we will use the T-distribution for all of today's lab.
#We will primarily work with the dataset brfss09_lab7.txt:
#########################################################################################################################
#Behavioral Risk Factors Surveilance Survey 2009 (BRFSS09) Data Dictionary:
#------------------------------------------------------------------------------------------------------------------------
#id: "Subject ID"Values[1,998]
#physhlth: "# Days past month phsycial health poor" Values[1,30]
#menthlth: "# Days past month mental health poor"Values[1,30]
#hlthplan: "Have healthcare coverage?"Values 1=Yes, 2=No
#age:"Age in Years"Values[18,99]
#sex:"Biologic Sex"Values 0=Female, 1=Male
#fruit_day: "# of servings of fruit per day"Values[0,20]
#alcgrp: "Alcohol Consumption Groups"Values 1=None, 2= 1-2 drinks/day 3= 3 or more drinks/day
#smoke:"Smoking Status"Values 0=Never, 1=Current EveryDay, 2=Current SomeDays, 3=Former
#bmi:"Body Mass Index"Values[14,70]
#mi:"Myocardial Infarction (heart attack)"Values 0=No, 1=Yes
#------------------------------------------------------------------------------------------------------------------------
# For today's lab, let's start by reading in our datafile
# 'brfss09_lab7.txt' into an object called mydata
mydata=read.table('brfss09_lab7.txt', header=T)
#This file contains:
dim(mydata)#100 Subjects, 11 variables
#With the following variables:
names(mydata)
# We have collected this data and would like to know
#if the values we have found in our sample are different
#from the reported values in the literature.
# For example, it has been reported that the average BMI
# in the population is 27.5. We would like to know if the
#values in our sample are somehow different than this value.
#---------------------------------------------------------------------------------
# 1. Formulating Hypotheses
#---------------------------------------------------------------------------------
#Step 1 of determining if our BMI values differ from the
#national average of 27.5 is to formulate our hypotheses
#We have TWO hypotheses
#1) The Null Hypothesis: H0: mu = 27.5
#2) The Alternative Hypothesis: HA: mu != 27.5
#NOTE: mu=Population Mean
#The above hypotheses are Two-Sided.
#By this I mean that we are looking to see if our sample values of
#BMI are greater than (>) OR less than (<) 27.5.
# A one-sided hypothesis test would look like:
#H0: mu < 27.5
#HA: mu > 27.5
#OR
#H0: mu > 27.5
#HA: mu < 27.5
#We will always use two-sided tests in this class,
#and similarly in the real world two-sided tests dominate.
#Once we have our hypotheses we will evaluate them
#and determine one of two outcomes:
# A) Reject the Null Hypothesis
# B) Fail to Reject the Null Hypothesis
#---------------------------------------------------------------------------------
# 2. T-statistics by Hand (well..with help from the computer)
#---------------------------------------------------------------------------------
#Recall from the last lab, that the formula for a T-statistic is:
# T = (SampleMean - PopMean) / (SampleSD/sqrt(N))
#Another way to write this would be:
# T = (xbar - mu) / (s/sqrt(N))
#In this instance PopMean (mu) is the NULL hypothesis
#value we are testing against.
#We can solve for the other values that we don't yet know:
mu=27.5
xbar=mean(mydata$bmi) #28.22
s=sd(mydata$bmi) #6.32
N=100
T = (xbar - mu) / (s/sqrt(N))
T #1.14
#We end up with a T value of ~ 1.14
#But how does this tell us if our mean is different from 27.5 ???!!!
#Before we move on, I want us to think about why we need
#to evaluate if our mean of 28.22 is different from 27.5.
#Certainly we can see that these are different numbers,
#so what are we really asking here?
#One way to think about it is that we are asking if our
#sample mean of 28.22 is different from 27.5 simply due to chance.
#Think of a coin tossing example:
#Your friend tosses a coin in the air and it lands on heads
#3 times in a row!
#While, kinda cool, seems like that is probably random chance.
#What about if it landed on heads 100 times in a row?!
#You would probably think she was cheating somehow!
#Though it is possible to have 100 heads in a row
#by chance alone, it is very unlikely
#The point at which we say that something is random vs not
#is determined by our alpha level.
#---------------------------------------------------------------------------------
# 3. Alpha Level
#---------------------------------------------------------------------------------
# The alpha level is determined a priori (a head of time)
#and used to set the threshold by which we consider something
#to be random chance
# A common alpha level is 0.05.
# We typically reject the null (think something is not chance)
#when the result we have (eg. 28.22) would only be
#that extreme < 5% of the time by chance.
#Recall from Lab 6, that we use the alpha level
#to help figure out critical values (c)
# c=qt(1-(alpha/2), df)
#---------------------------------------------------------------------------------
# 4. Evaluating our Results
#---------------------------------------------------------------------------------
# There are 3 ways to evaluate if our mean of 28.22
# is different from the null of 27.5
# All three ways will yield the same conclusion.
#1) Compare T to a critical value (c)
#2) Evaluate the Confidence interval
#3) Compare the p value to our alpha level
###########################################################
#1) Compare T to a critical value (c)
#In order to compute the critical value (c),
#we must know the alpha level.
#We will choose a value of 0.05 (which is standard)
alpha=0.05
df=100-1
c=qt(1-(alpha/2), df)
#We can then compare the abosulte value of T (|T|)
#to the critical value c
#A) If |T| > c, then Reject the Null Hypothesis
#B) If |T| < c, then Fail to Reject the Null Hypothesis
#Let's look at T can c
abs(T)
c
#What decision do we make about the Null Hypothesis????
###########################################################
#2) Evaluate the Confidence interval
#Rather than compare T to c,
#we could instead compute the confidence interval.
#Recall the formula for the Confidence interval is:
#LB= xbar - c*(s/sqrt(N))
#UB= xbar + c*(s/sqrt(N))
LB = xbar - c*(s/sqrt(N))
UB= xbar + c*(s/sqrt(N))
#A) If mu is not within the Confidence Interval,
#then Reject the Null Hypothesis
#B) If mu is within the Confidence Interval,
#then Fail to Reject the Null Hypothesis
#Let's look at LB and UB
LB
UB
mu
#What decision do we make about the Null Hypothesis????
###########################################################
#3) Compare the p value to our alpha level
#Lastly, we could find the probability value (or p-value)
#for the T statistic we created.
#We can do this by using the pt() function we learned
#about last week in lab 6.
#There is a forumla for computing P values from T-statitics:
# pval = 2*(1-pt(abs(T), df))
pval = 2*(1-pt(abs(T), df))
#We then compare the p-value to our alpha level
#A) If pval < alpha, then Reject the Null Hypothesis
#B) If pval > alpha, then Fail to Reject the Null Hypothesis
#Let's look at our p-value.
pval
alpha
#What decision do we make about the Null Hypothesis????
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#Exercise 4-1:
#Evaluate if the mean age from our sample (mydata) is different
#than the populatiuin mean age of 56
# A) Write down the Null and Alternative Hypotheses
# B) Calculate the T-statistic by hand
# C) Evaluate the Null hypothesis by using ALL 3 methods that
# we just discussed
# D) Based on the results in C, do you Reject or Fail to Reject
# the Null Hypothesis?
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#A)
#B)
#C)
#Method 1: Compare T to a critical value (c)
#Method 2: Evaluate the Confidence interval
#Method 3: Compare the p value to our alpha level
#D)
#---------------------------------------------------------------------------------
# 5. Using the t.test function
#---------------------------------------------------------------------------------
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
# One Sample T-Test : t.test(data$variable, mu)
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
#It was really awesome that we figured out T by hand!
#And then figured out the confidence intervals and P values!
#From now on, let's just use a program to do all this for us.
#The function t.test will presume an alpha level of 0.05 by default.
t.test(mydata$age, mu=56)
# t.test(mydata$bmi, mu=27.5)
#Much simpler!
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#Exercise 5-1: Use the t.test function to evaluate if
#A) the mean days of physical health (physhlth) is different
# than the population mean of 10? Reject the Null?
#B) the mean fruits per day (fruit_day) is different than
# the populatiuin mean of 4? Reject the Null?
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#A)
#B)
#---------------------------------------------------------------------------------
# 6. T-test with Trimmed Means
#---------------------------------------------------------------------------------
#To use the T-test with trimmed means,
#we will need to load in the source code 'Rallfun-v33.txt'
#The trimmed mean T-test is beneficial in that it does not
#presume a perfect Normal Distribution
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
# Trimmed Mean T-Test:
# trimci(data$variable, tr=0.2, alpha=0.05, null.value=0)
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
#For example, if I wanted to test if the age was equal to 56
#using Trimmed Means I could do:
trimci(mydata$age, null.value=56)
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#Exercise 6-1: Use the trimci function to evaluate if
#A) the 20% trimmed mean of days of physical health (physhlth) is
# different than the populatiuin mean of 10? Reject the Null?
#B) the 20% trimmed mean fruits per day (fruit_day) is different
#than the populatiuin mean of 4? Reject the Null?
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#A)
#B)
#---------------------------------------------------------------------------------
# 7. Type 1 and Type 2 Errors
#---------------------------------------------------------------------------------
#Notice that we had very different answers to the same
#questions in Ex. 5-1 and 6-1
#Depending upon the method that we used.
#This brings us to discussing Type 1 and Type 2 Error
#A Type 1 error is when our test tells us to reject the null,
#but in truth we should not have
#A Type 2 error is when our test tells us to fail to reject the
#null, but in truth we should have rejected the null
#The following 2x2 square might make this easier to see.
# Truth
#------------------------------------
#| H0 | HA |
#-------------- |-------|-----------|
#My Test: H0 | H0 Type 2|
#-------------- |-------|-----------|
#My Test: HA Type 1 | HA |
#------------------------------------
#For the next exercise, let's presume that our test of the trimmed mean is the Truth
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#Exercise 7-1:
#A) What type of error did we make when evaluating the mean
#of physhlth in exercise 5-1?
#B) What type of error did we make when evaluating the mean
#of fruit_day in exercise 5-1?
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#A)
#B)
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Proofs
1. For the following, state what you would assume and what you would want to prove for a In a direct proof, we assume the ...
Proofs
1. For the following, state what you would assume and what you would want to prove for a In a direct proof, we assume the premise (the left part) and ...
West Coast University Data Analysis Descriptive Statistics Project
Survey Information Links https://www.surveymonkey.com/results/SM-RBM2J2FX7/https://www.surveymonkey.com/results/SM-J5BKC2F ...
West Coast University Data Analysis Descriptive Statistics Project
Survey Information Links https://www.surveymonkey.com/results/SM-RBM2J2FX7/https://www.surveymonkey.com/results/SM-J5BKC2FX7/Research Focus:how creating health programs for children and parents in middle school lead to decreasing the consumption of tobacco in teenagersPurpose of research:Tobacco smoking in the country is the primary source of preventable death. Many smokers (82 percent) began to smoke before 18, and about 3,000 young people continue to smoke every day. School initiatives to discourage the usage of cigarettes may become one of the most successful nicotine approaches implemented. This survey seeks to identify whether creating health programs for children and parents in middle school lead to decreasing the consumption of tobacco in teenagers. The survey will determine whether the reduced use of tobacco among teenagers (dependent variable) is directly challenged by the health programs put in place in middle schools (independent variable). This survey will target teenage students as well as parents in middle school and consequently summarize their feedback regarding the issue. AssignmentBy this week, you should have completed data collection on at least 30 participants who met the qualifications to complete the survey (e.g., at least 18 years or older, have other specific characteristics that make them a member of the population you intended to study).The data for your short survey should be entered into the Excel database that you created last week. With this data set, complete the following:Calculate descriptive statistics for your main independent and other variables you have collected information on (example: demographic variables).Create a bar chart for the participants’ gender and ethnicity.Create a histogram for participants’ age. Calculate descriptive statistics for your dependent variable. Create a summary table of your descriptive statistics.Hint: For categorical variables, you should be calculating frequencies and percentages; for continuous variables, you should be calculating means and standard deviations.Use one hypothesis-testing statistic that examines the relationship between your independent variable and dependent variable.Hint: This test should be one of the following—independent samples t-test, paired t-test, one-way ANOVA, or Pearson’s chi-squared test.Write the null and alternate hypotheses.What is the calculated statistic value?What is the p and critical statistic value?Is the p value significant? Is calculated statistic value greater than critical value?What is your conclusion?
stats homework using R
#Name:
#Student ID:
rm(list=ls())
source('Rallfun-v33.txt')
#PART 1
#A company claims that, when exposed to their to ...
stats homework using R
#Name:
#Student ID:
rm(list=ls())
source('Rallfun-v33.txt')
#PART 1
#A company claims that, when exposed to their toothpaste, 45% of all bacteria related to gingivitis are killed, on average. You run 10 tests and ???nd that the percentages of bacteria killed in each test was:
# 38%, 44%, 62%, 72%, 43%, 40%, 43%, 42%, 39%, 41%
# Assuming normality, you will test the hypothesis that the average percentage of bacteria killed was 45% at alpha=0.05.
#1.1) Write out the Null and Alternative hypotheses
#1.2) Calculate the T-statistic and use Method 1 (we saw in class) to determine if the average bacteria killed was 45%. Do it by "hand".
#Hint: Method 1 is to compare T to a critical value "c".
#1.3) Do you reject or fail to reject the null?
################################################
#PART 2
#Now, let's not assume normality
#2.1) Using the same data as in Part 1, test the hypothesis that the 20% trimmed mean is 45%?
#2.2) Do you reject or fail to reject the null?
#2.3) Assuming your test in 2.1 is the truth, what type of error did you make in #1.3?
################################################
#PART 3
#In a study of court administration, the following times to disposition (in minutes) were determined for twenty cases and found to be:
# 42, 90, 84, 87, 116, 95, 86, 99, 93, 92, 121, 71, 66, 98, 79, 102, 60, 112, 105, 98
#Assuming normality, you will test the hypothesis that the average time to disposition was 99 minutes at alpha=0.05.
#3.1) Write out the Null and Alternative hypotheses
#3.2) Calculate the T-statistic and use Method 2 (we saw in class) to determine if the average time to disposition was 99? Do it by "hand".
#Hint: Method 2 is to evaluate the confidence interval.
#3.3) Do you reject or fail to reject the null?
################################################
#PART 4
#Now, let's not assume normality
#4.1) Using the same data as in Part 3, test the hypothesis that the 20% trimmed mean is 99?
#4.2) Do you reject or fail to reject the null?
#4.3) Assuming your test in 4.1 is the truth, what type of error did you make in #3.3?
################################################
#PART 5
#Suppose you run an experiment, and observe the following values:
# 12, 20, 34, 45, 34, 36, 37, 50, 11, 32, 29
#You will test the hypothesis that the average was 25 at alpha=0.05.
#5.1) Write out the Null and Alternative hypotheses. Conduct the hypothesis test assuming normality. Use the "t.test" function. Do you reject or fail to reject the null?
#5.2) Conduct the hypothesis test without assuming normality. Do you reject or fail to reject the null?
#5.3) Assuming the answer in #5.2 is the truth, what type of error (if any) did you make in #5.1 by assuming normality?
------------------------------------------------------------------------------------------
Lab 7- Lecture Notes (FOR YOUR REFERENCE)
#Lab 7-Contents
#1. Formulating Hypotheses
#2. T-statistics by Hand
#3. Alpha Level
#4. Evaluating Our Results
#5. Using the t.test function
#6. T-tests with Trimmed Means (trimci function)
#7. Type 1 and Type 2 Errors
# Last week we talked about computations for when the Population
#Variance is known and unknown.
# Given that we rarely know the population variance,
#we will use the T-distribution for all of today's lab.
#We will primarily work with the dataset brfss09_lab7.txt:
#########################################################################################################################
#Behavioral Risk Factors Surveilance Survey 2009 (BRFSS09) Data Dictionary:
#------------------------------------------------------------------------------------------------------------------------
#id: "Subject ID"Values[1,998]
#physhlth: "# Days past month phsycial health poor" Values[1,30]
#menthlth: "# Days past month mental health poor"Values[1,30]
#hlthplan: "Have healthcare coverage?"Values 1=Yes, 2=No
#age:"Age in Years"Values[18,99]
#sex:"Biologic Sex"Values 0=Female, 1=Male
#fruit_day: "# of servings of fruit per day"Values[0,20]
#alcgrp: "Alcohol Consumption Groups"Values 1=None, 2= 1-2 drinks/day 3= 3 or more drinks/day
#smoke:"Smoking Status"Values 0=Never, 1=Current EveryDay, 2=Current SomeDays, 3=Former
#bmi:"Body Mass Index"Values[14,70]
#mi:"Myocardial Infarction (heart attack)"Values 0=No, 1=Yes
#------------------------------------------------------------------------------------------------------------------------
# For today's lab, let's start by reading in our datafile
# 'brfss09_lab7.txt' into an object called mydata
mydata=read.table('brfss09_lab7.txt', header=T)
#This file contains:
dim(mydata)#100 Subjects, 11 variables
#With the following variables:
names(mydata)
# We have collected this data and would like to know
#if the values we have found in our sample are different
#from the reported values in the literature.
# For example, it has been reported that the average BMI
# in the population is 27.5. We would like to know if the
#values in our sample are somehow different than this value.
#---------------------------------------------------------------------------------
# 1. Formulating Hypotheses
#---------------------------------------------------------------------------------
#Step 1 of determining if our BMI values differ from the
#national average of 27.5 is to formulate our hypotheses
#We have TWO hypotheses
#1) The Null Hypothesis: H0: mu = 27.5
#2) The Alternative Hypothesis: HA: mu != 27.5
#NOTE: mu=Population Mean
#The above hypotheses are Two-Sided.
#By this I mean that we are looking to see if our sample values of
#BMI are greater than (>) OR less than (<) 27.5.
# A one-sided hypothesis test would look like:
#H0: mu < 27.5
#HA: mu > 27.5
#OR
#H0: mu > 27.5
#HA: mu < 27.5
#We will always use two-sided tests in this class,
#and similarly in the real world two-sided tests dominate.
#Once we have our hypotheses we will evaluate them
#and determine one of two outcomes:
# A) Reject the Null Hypothesis
# B) Fail to Reject the Null Hypothesis
#---------------------------------------------------------------------------------
# 2. T-statistics by Hand (well..with help from the computer)
#---------------------------------------------------------------------------------
#Recall from the last lab, that the formula for a T-statistic is:
# T = (SampleMean - PopMean) / (SampleSD/sqrt(N))
#Another way to write this would be:
# T = (xbar - mu) / (s/sqrt(N))
#In this instance PopMean (mu) is the NULL hypothesis
#value we are testing against.
#We can solve for the other values that we don't yet know:
mu=27.5
xbar=mean(mydata$bmi) #28.22
s=sd(mydata$bmi) #6.32
N=100
T = (xbar - mu) / (s/sqrt(N))
T #1.14
#We end up with a T value of ~ 1.14
#But how does this tell us if our mean is different from 27.5 ???!!!
#Before we move on, I want us to think about why we need
#to evaluate if our mean of 28.22 is different from 27.5.
#Certainly we can see that these are different numbers,
#so what are we really asking here?
#One way to think about it is that we are asking if our
#sample mean of 28.22 is different from 27.5 simply due to chance.
#Think of a coin tossing example:
#Your friend tosses a coin in the air and it lands on heads
#3 times in a row!
#While, kinda cool, seems like that is probably random chance.
#What about if it landed on heads 100 times in a row?!
#You would probably think she was cheating somehow!
#Though it is possible to have 100 heads in a row
#by chance alone, it is very unlikely
#The point at which we say that something is random vs not
#is determined by our alpha level.
#---------------------------------------------------------------------------------
# 3. Alpha Level
#---------------------------------------------------------------------------------
# The alpha level is determined a priori (a head of time)
#and used to set the threshold by which we consider something
#to be random chance
# A common alpha level is 0.05.
# We typically reject the null (think something is not chance)
#when the result we have (eg. 28.22) would only be
#that extreme < 5% of the time by chance.
#Recall from Lab 6, that we use the alpha level
#to help figure out critical values (c)
# c=qt(1-(alpha/2), df)
#---------------------------------------------------------------------------------
# 4. Evaluating our Results
#---------------------------------------------------------------------------------
# There are 3 ways to evaluate if our mean of 28.22
# is different from the null of 27.5
# All three ways will yield the same conclusion.
#1) Compare T to a critical value (c)
#2) Evaluate the Confidence interval
#3) Compare the p value to our alpha level
###########################################################
#1) Compare T to a critical value (c)
#In order to compute the critical value (c),
#we must know the alpha level.
#We will choose a value of 0.05 (which is standard)
alpha=0.05
df=100-1
c=qt(1-(alpha/2), df)
#We can then compare the abosulte value of T (|T|)
#to the critical value c
#A) If |T| > c, then Reject the Null Hypothesis
#B) If |T| < c, then Fail to Reject the Null Hypothesis
#Let's look at T can c
abs(T)
c
#What decision do we make about the Null Hypothesis????
###########################################################
#2) Evaluate the Confidence interval
#Rather than compare T to c,
#we could instead compute the confidence interval.
#Recall the formula for the Confidence interval is:
#LB= xbar - c*(s/sqrt(N))
#UB= xbar + c*(s/sqrt(N))
LB = xbar - c*(s/sqrt(N))
UB= xbar + c*(s/sqrt(N))
#A) If mu is not within the Confidence Interval,
#then Reject the Null Hypothesis
#B) If mu is within the Confidence Interval,
#then Fail to Reject the Null Hypothesis
#Let's look at LB and UB
LB
UB
mu
#What decision do we make about the Null Hypothesis????
###########################################################
#3) Compare the p value to our alpha level
#Lastly, we could find the probability value (or p-value)
#for the T statistic we created.
#We can do this by using the pt() function we learned
#about last week in lab 6.
#There is a forumla for computing P values from T-statitics:
# pval = 2*(1-pt(abs(T), df))
pval = 2*(1-pt(abs(T), df))
#We then compare the p-value to our alpha level
#A) If pval < alpha, then Reject the Null Hypothesis
#B) If pval > alpha, then Fail to Reject the Null Hypothesis
#Let's look at our p-value.
pval
alpha
#What decision do we make about the Null Hypothesis????
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#Exercise 4-1:
#Evaluate if the mean age from our sample (mydata) is different
#than the populatiuin mean age of 56
# A) Write down the Null and Alternative Hypotheses
# B) Calculate the T-statistic by hand
# C) Evaluate the Null hypothesis by using ALL 3 methods that
# we just discussed
# D) Based on the results in C, do you Reject or Fail to Reject
# the Null Hypothesis?
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#A)
#B)
#C)
#Method 1: Compare T to a critical value (c)
#Method 2: Evaluate the Confidence interval
#Method 3: Compare the p value to our alpha level
#D)
#---------------------------------------------------------------------------------
# 5. Using the t.test function
#---------------------------------------------------------------------------------
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
# One Sample T-Test : t.test(data$variable, mu)
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
#It was really awesome that we figured out T by hand!
#And then figured out the confidence intervals and P values!
#From now on, let's just use a program to do all this for us.
#The function t.test will presume an alpha level of 0.05 by default.
t.test(mydata$age, mu=56)
# t.test(mydata$bmi, mu=27.5)
#Much simpler!
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#Exercise 5-1: Use the t.test function to evaluate if
#A) the mean days of physical health (physhlth) is different
# than the population mean of 10? Reject the Null?
#B) the mean fruits per day (fruit_day) is different than
# the populatiuin mean of 4? Reject the Null?
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#A)
#B)
#---------------------------------------------------------------------------------
# 6. T-test with Trimmed Means
#---------------------------------------------------------------------------------
#To use the T-test with trimmed means,
#we will need to load in the source code 'Rallfun-v33.txt'
#The trimmed mean T-test is beneficial in that it does not
#presume a perfect Normal Distribution
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
# Trimmed Mean T-Test:
# trimci(data$variable, tr=0.2, alpha=0.05, null.value=0)
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
#For example, if I wanted to test if the age was equal to 56
#using Trimmed Means I could do:
trimci(mydata$age, null.value=56)
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#Exercise 6-1: Use the trimci function to evaluate if
#A) the 20% trimmed mean of days of physical health (physhlth) is
# different than the populatiuin mean of 10? Reject the Null?
#B) the 20% trimmed mean fruits per day (fruit_day) is different
#than the populatiuin mean of 4? Reject the Null?
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#A)
#B)
#---------------------------------------------------------------------------------
# 7. Type 1 and Type 2 Errors
#---------------------------------------------------------------------------------
#Notice that we had very different answers to the same
#questions in Ex. 5-1 and 6-1
#Depending upon the method that we used.
#This brings us to discussing Type 1 and Type 2 Error
#A Type 1 error is when our test tells us to reject the null,
#but in truth we should not have
#A Type 2 error is when our test tells us to fail to reject the
#null, but in truth we should have rejected the null
#The following 2x2 square might make this easier to see.
# Truth
#------------------------------------
#| H0 | HA |
#-------------- |-------|-----------|
#My Test: H0 | H0 Type 2|
#-------------- |-------|-----------|
#My Test: HA Type 1 | HA |
#------------------------------------
#For the next exercise, let's presume that our test of the trimmed mean is the Truth
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#Exercise 7-1:
#A) What type of error did we make when evaluating the mean
#of physhlth in exercise 5-1?
#B) What type of error did we make when evaluating the mean
#of fruit_day in exercise 5-1?
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#A)
#B)
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