Lab7
Lance Nuique
11/1/2021
The Survey
Exercise 1
In the first paragraph, several key findings are reported. Do these percentages appear to be sample statistics (derived from the data sample) or population parameters?
The data is from a sample because it comes from a poll survey.
Exercise 2
The title of the report is “Global Index of Religiosity and Atheism”. To generalize the report’s findings to the global human population, what must we assume about the sampling method? Does that seem like a reasonable assumption?
The sampling method has to be independent and random. It does seem like a reasonable assumption because it is going to compare differnt countries and those countries will have different populations.
The data
##downloads the file from the database and puts it into R
download.file("http://www.openintro.org/stat/data/atheism.RData", destfile = "atheism.RData")
load("atheism.RData")Exercise 3
What does each row of Table 6 correspond to? What does each row of atheism correspond to?
The row represents each country. Each row of atheism corresponds to each person’s answer
Exercise 4
Using the command below, create a new dataframe called us12 that contains only the rows in atheism associated with respondents to the 2012 survey from the United States. Next, calculate the proportion of atheist responses. Does it agree with the percentage in Table 6? If not, why?
us12 <- subset(atheism, nationality == "United States" & year == "2012")
##This line looks for the amount of atheist responses in the subset
atheism1<-subset(us12, response=="atheist")
##Showing how many people answered the question and it shows 1002 people were surveyed
table(us12)## , , year = 2012
##
## response
## nationality atheist non-atheist
## Afghanistan 0 0
## Argentina 0 0
## Armenia 0 0
## Australia 0 0
## Austria 0 0
## Azerbaijan 0 0
## Belgium 0 0
## Bosnia and Herzegovina 0 0
## Brazil 0 0
## Bulgaria 0 0
## Cameroon 0 0
## Canada 0 0
## China 0 0
## Colombia 0 0
## Czech Republic 0 0
## Ecuador 0 0
## Fiji 0 0
## Finland 0 0
## France 0 0
## Georgia 0 0
## Germany 0 0
## Ghana 0 0
## Hong Kong 0 0
## Iceland 0 0
## India 0 0
## Iraq 0 0
## Ireland 0 0
## Italy 0 0
## Japan 0 0
## Kenya 0 0
## Korea, Rep (South) 0 0
## Lebanon 0 0
## Lithuania 0 0
## Macedonia 0 0
## Malaysia 0 0
## Moldova 0 0
## Netherlands 0 0
## Nigeria 0 0
## Pakistan 0 0
## Palestinian territories (West Bank and Gaza) 0 0
## Peru 0 0
## Poland 0 0
## Romania 0 0
## Russian Federation 0 0
## Saudi Arabia 0 0
## Serbia 0 0
## South Africa 0 0
## South Sudan 0 0
## Spain 0 0
## Sweden 0 0
## Switzerland 0 0
## Tunisia 0 0
## Turkey 0 0
## Ukraine 0 0
## United States 50 952
## Uzbekistan 0 0
## Vietnam 0 050/1002## [1] 0.0499002Based on the tables there were 50 people who said atheist as a response compared to the total of 1002 people surveyed which beomes around 5%. The table says it is also 5 percent so both numbers agree with each other.
Inference on Proportions
Exercise 5
Write out the conditions for inference to construct a 95% confidence interval for the proportion of atheists in the United States in 2012. Are you confident all conditions are met?
The observations must be independent and the individuals were selected randomly. The sample is also less than 10% of total population. I am confident that the conditions are met.
##This does the CI interval
inference(us12$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.0499 ; n = 1002
## Check conditions: number of successes = 50 ; number of failures = 952
## Standard error = 0.0069
## 95 % Confidence interval = ( 0.0364 , 0.0634 )1.96*.0069## [1] 0.013524Exercise 6
Based on the R output, what is the margin of error for the estimate of the proportion of the proportion of atheists in US in 2012?
The margin of error is 1.96SE= 1.96(0.0069)= 0.013524
Exercise 7
Using the inference function, calculate confidence intervals for the proportion of atheists in 2012 in two other countries of your choice, and report the associated margins of error. Be sure to note whether the conditions for inference are met. It may be helpful to create new data sets for each of the two countries first, and then use these data sets in the inference function to construct the confidence intervals.
##This code is basically what was did earlier but now picks different countries from the list
jpn2012 <- subset(atheism, nationality == "Japan" & year == "2012")
inference(jpn2012$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.3069 ; n = 1212
## Check conditions: number of successes = 372 ; number of failures = 840
## Standard error = 0.0132
## 95 % Confidence interval = ( 0.281 , 0.3329 )##Another line for another country
cnd2012 <- subset(atheism, nationality == "Canada" & year == "2012")
inference(cnd2012$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.0898 ; n = 1002
## Check conditions: number of successes = 90 ; number of failures = 912
## Standard error = 0.009
## 95 % Confidence interval = ( 0.0721 , 0.1075 )The conditions are going to be similar to the US sample which means that the conditions are going to be met.
How does the proportion affect the margin of error?
Exercise 8
Describe the relationship between p and me.
Based on the graph there is a quadratic relationship between them because it shows a half circle. As the p increases to .5 the highest margin of error is reached and then goes back down to being less as the p gets further away from .5 closer to 1.
Success-failure condition
n <- 1000
p <- seq(0, 1, 0.01)
me <- 2 * sqrt(p * (1 - p)/n)
plot(me ~ p, ylab = "Margin of Error", xlab = "Population Proportion")
Exercise 9
Describe the sampling distribution of sample proportions at n=1040 and p=0.1. Be sure to note the center, spread, and shape.
Hint: Remember that R has functions such as mean to calculate summary statistics.
The mean is 0.1 and the center is .1 the spread is from 0.06731 to .13654. The shape is relatively.
p <- 0.1
n <- 1040
p_hats <- rep(0, 5000)
for(i in 1:5000){
samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
p_hats[i] <- sum(samp == "atheist")/n
}
hist(p_hats, main = "p = 0.1, n = 1040", xlim = c(0, 0.18))
summary(p_hats)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.07019 0.09327 0.09904 0.09969 0.10577 0.12981Exercise 10
Repeat the above simulation three more times but with modified sample sizes and proportions: for n=400 and p=0.1, n=1040 and p=0.02, and n=400 and p=0.02. Plot all four histograms together by running the par(mfrow = c(2, 2)) command before creating the histograms. You may need to expand the plot window to accommodate the larger two-by-two plot. Describe the three new sampling distributions. Based on these limited plots, how does n appear to affect the distribution of p^? How does p affect the sampling distribution?
All the distributions are unimodal and symmetrical. Having a larger n makes the graphs have a tighter spread around the p value and if n is smaller they are wider.
par(mfrow = c(2, 2))
##Adds all the different situations to a table
p2<-.01
n2<-400
phat2<-rep(0,5000)
for(i in 1:5000){
samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p2, 1-p2))
phat2[i] <- sum(samp == "atheist")/n
}
hist(phat2, main = "p = 0.1, n = 400", xlim = c(0, 0.18))
##another scenario
p3<-.02
n3<-1040
phat3<-rep(0,5000)
for(i in 1:5000){
samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p3, 1-p3))
phat3[i] <- sum(samp == "atheist")/n
}
hist(phat3, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))
##another scenario
p4<-.02
n4<-400
phat4<-rep(0,5000)
for(i in 1:5000){
samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p4, 1-p4))
phat4[i] <- sum(samp == "atheist")/n
}
hist(phat4, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))