In August of 2012, news outlets ranging from the Washington Post to the Huffington Post ran a story about the rise of atheism in America. The source for the story was a poll that asked people, “Irrespective of whether you attend a place of worship or not, would you say you are a religious person, not a religious person or a convinced atheist?” This type of question, which asks people to classify themselves in one way or another, is common in polling and generates categorical data. In this lab we take a look at the atheism survey and explore what’s at play when making inference about population proportions using categorical data.

The survey

To access the press release for the poll, conducted by WIN-Gallup International, click on the following link:

https://github.com/jbryer/DATA606/blob/master/inst/labs/Lab6/more/Global_INDEX_of_Religiosity_and_Atheism_PR__6.pdf

Take a moment to review the report then address the following questions.

  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?

Answer: The percentages appear to be sample Statistics

  1. 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?

Answer: We must assume the sample is random and independent. We also expect to see at least 10 successes and 10 failures in our sample. The assumption is reasonable.

The data

Turn your attention to Table 6 (pages 15 and 16), which reports the sample size and response percentages for all 57 countries. While this is a useful format to summarize the data, we will base our analysis on the original data set of individual responses to the survey. Load this data set into R with the following command.

download.file("http://www.openintro.org/stat/data/atheism.RData", destfile = "atheism.RData")
load("atheism.RData")
  1. What does each row of Table 6 correspond to? What does each row of atheism correspond to?

Answer: Each row of Table 6 correspond to each country’s sample, and percentage of religious, not religious, atheist and did not respond. Each row of atheism correspond to individual person in the country and the year the poll was taken.

To investigate the link between these two ways of organizing this data, take a look at the estimated proportion of atheists in the United States. Towards the bottom of Table 6, we see that this is 5%. We should be able to come to the same number using the atheism data.

  1. 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?
library(plyr)
library(knitr)
us12 <- subset(atheism, nationality == "United States" & year == "2012")
number_non<-count(us12$response=="non-atheist")
number_atheist <- count(us12$response == 'atheist')
proportion_atheist<-number_atheist/(number_atheist+number_non)
proportion_atheist
##     x      freq
## 1 NaN 0.9500998
## 2 0.5 0.0499002

It matches with the percentage of table 6.

Inference on proportions

As was hinted at in Exercise 1, Table 6 provides statistics, that is, calculations made from the sample of 51,927 people. What we’d like, though, is insight into the population parameters. You answer the question, “What proportion of people in your sample reported being atheists?” with a statistic; while the question “What proportion of people on earth would report being atheists” is answered with an estimate of the parameter.

The inferential tools for estimating population proportion are analogous to those used for means in the last chapter: the confidence interval and the hypothesis test.

  1. 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?

Answer: We need to satisfy two condition. First, sample’s observation are independent. Base on how the data was collected, we know that the observation is independent. Second, we expected to see at least 10 successes and 10 failures in the sample. np>=10 and n(1-p)>=10. The observation satisfy the success-failure condition. (952+50)*0.05~=50. Since both values are least 10, we meet the requirement.

If the conditions for inference are reasonable, we can either calculate the standard error and construct the interval by hand, or allow the inference function to do it for us.

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 )

Note that since the goal is to construct an interval estimate for a proportion, it’s necessary to specify what constitutes a “success”, which here is a response of "atheist".

Although formal confidence intervals and hypothesis tests don’t show up in the report, suggestions of inference appear at the bottom of page 7: “In general, the error margin for surveys of this kind is \(\pm\) 3-5% at 95% confidence”.

  1. 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?
me<-1.96*0.0069
me
## [1] 0.013524
  1. 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.

China

china12<-subset(atheism, nationality == "China" & year == "2012")
count(china12)
##   nationality    response year freq
## 1       China     atheist 2012  235
## 2       China non-atheist 2012  265
inference(china12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.47 ;  n = 500 
## Check conditions: number of successes = 235 ; number of failures = 265 
## Standard error = 0.0223 
## 95 % Confidence interval = ( 0.4263 , 0.5137 )
#margin_of_error
me_china<-1.96*0.0223
me_china
## [1] 0.043708

The sample from China is independent and also meet the success-failure condition. Total 500 observations and number of atheist is 235, which greater than np=5000.47=235>=10. Number of non-atheist is 265, greater than n (1-p) =500(1-0.47) =265>=10. In addition, 95 % Confidence interval is between 0.4263 and 0.5137. And associated margins of error is 0.043708.

Japan

Japan12<-subset(atheism, nationality == "Japan" & year == "2012")
count(Japan12)
##   nationality    response year freq
## 1       Japan     atheist 2012  372
## 2       Japan non-atheist 2012  840
inference(Japan12$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 )
#margin_of_error
me_Japan<-1.96*0.0132
me_Japan
## [1] 0.025872

The sample from Japan is independent and also meet the success-failure condition. Total 1212 observations and number of atheist is 372, which greater than np=12120.3069=372>=10. Number of non-atheist is 840, greater than n (1-p) =1212(1-0.306) =841>=10. In addition, 95 % Confidence interval is between 0.281 and 0.3329. And associated margins of error is 0.025872.

How does the proportion affect the margin of error?

Imagine you’ve set out to survey 1000 people on two questions: are you female? and are you left-handed? Since both of these sample proportions were calculated from the same sample size, they should have the same margin of error, right? Wrong! While the margin of error does change with sample size, it is also affected by the proportion.

Think back to the formula for the standard error: \(SE = \sqrt{p(1-p)/n}\). This is then used in the formula for the margin of error for a 95% confidence interval: \(ME = 1.96\times SE = 1.96\times\sqrt{p(1-p)/n}\). Since the population proportion \(p\) is in this \(ME\) formula, it should make sense that the margin of error is in some way dependent on the population proportion. We can visualize this relationship by creating a plot of \(ME\) vs. \(p\).

The first step is to make a vector p that is a sequence from 0 to 1 with each number separated by 0.01. We can then create a vector of the margin of error (me) associated with each of these values of p using the familiar approximate formula (\(ME = 2 \times SE\)). Lastly, we plot the two vectors against each other to reveal their relationship.

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")

  1. Describe the relationship between p and me.

Answer: Margin of error increase when the population proportion increase. We can see the max of margin of error when the population proportion reach 0.5 and then it decrease.

Success-failure condition

The textbook emphasizes that you must always check conditions before making inference. For inference on proportions, the sample proportion can be assumed to be nearly normal if it is based upon a random sample of independent observations and if both \(np \geq 10\) and \(n(1 - p) \geq 10\). This rule of thumb is easy enough to follow, but it makes one wonder: what’s so special about the number 10?

The short answer is: nothing. You could argue that we would be fine with 9 or that we really should be using 11. What is the “best” value for such a rule of thumb is, at least to some degree, arbitrary. However, when \(np\) and \(n(1-p)\) reaches 10 the sampling distribution is sufficiently normal to use confidence intervals and hypothesis tests that are based on that approximation.

We can investigate the interplay between \(n\) and \(p\) and the shape of the sampling distribution by using simulations. To start off, we simulate the process of drawing 5000 samples of size 1040 from a population with a true atheist proportion of 0.1. For each of the 5000 samples we compute \(\hat{p}\) and then plot a histogram to visualize their distribution.

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))

These commands build up the sampling distribution of \(\hat{p}\) using the familiar for loop. You can read the sampling procedure for the first line of code inside the for loop as, “take a sample of size \(n\) with replacement from the choices of atheist and non-atheist with probabilities \(p\) and \(1 - p\), respectively.” The second line in the loop says, “calculate the proportion of atheists in this sample and record this value.” The loop allows us to repeat this process 5,000 times to build a good representation of the sampling distribution.

  1. 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.
summary(p_hats)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## 0.07019 0.09327 0.09904 0.09969 0.10577 0.12981
mean(p_hats)
## [1] 0.09969

Answer: The sample is normal distribution with mean about 0.01.

  1. 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 \(\hat{p}\)? How does \(p\) affect the sampling distribution?
par(mfrow = c(2, 2))

#first histogram
p <- 0.1
n <- 400
p_hats400 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats400[i] <- sum(samp == "atheist")/n
}

#second histogram
p <- 0.02
n <- 1040
p_hats1040 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats1040[i] <- sum(samp == "atheist")/n
}

#third histogram
p <- 0.02
n <- 400
p_hats400_2<- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats400_2[i] <- sum(samp == "atheist")/n
}


#fourth histogram
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))
hist(p_hats400, main = "p = 0.1, n = 400", xlim = c(0, 0.18))
hist(p_hats1040, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))
hist(p_hats400_2, main = "p = 0.02, n = 400", xlim = c(0, 0.18))

par(mfrow = c(1, 1))

Answer: When the p is the same, the larger of the size, the narrower of the spread and smaller of the margin of the error. When we compare the p with the same size, the histograms show the opposite. The larger of the p, the narrow of the spread.

Once you’re done, you can reset the layout of the plotting window by using the command par(mfrow = c(1, 1)) command or clicking on “Clear All” above the plotting window (if using RStudio). Note that the latter will get rid of all your previous plots.

  1. If you refer to Table 6, you’ll find that Australia has a sample proportion of 0.1 on a sample size of 1040, and that Ecuador has a sample proportion of 0.02 on 400 subjects. Let’s suppose for this exercise that these point estimates are actually the truth. Then given the shape of their respective sampling distributions, do you think it is sensible to proceed with inference and report margin of errors, as the reports does?

Answer: Australia should satisfy with the normal distribution condition with sample proportion of 0.1 on a sample size of 1040, but Ecuador didn’t meet the condition because 400*0.02=8<10.

On your own

The question of atheism was asked by WIN-Gallup International in a similar survey that was conducted in 2005. (We assume here that sample sizes have remained the same.) Table 4 on page 13 of the report summarizes survey results from 2005 and 2012 for 39 countries.

Spain05 <- subset(atheism, nationality == "United States" & year == "2005")
Spain12 <- subset(atheism, nationality == "United States" & year == "2012")

inference(Spain05$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.01 ;  n = 1002 
## Check conditions: number of successes = 10 ; number of failures = 992 
## Standard error = 0.0031 
## 95 % Confidence interval = ( 0.0038 , 0.0161 )
inference(Spain12$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 )

Answer: The 95% confidence intervals for the true proportion of atheists are (0.083, 0.1177) for 2005 and (0.0734 and 0.1065) for 2012. We can see a big overlap, so there is not convincing evidence that Spain change in its atheism index between 2005 and 2012.

**b.** Is there convincing evidence that the United States has seen a
change in its atheism index between 2005 and 2012?
us05 <- subset(atheism, nationality == "Spain" & year == "2005")
us12 <- subset(atheism, nationality == "Spain" & year == "2012")

inference(Spain05$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.01 ;  n = 1002 
## Check conditions: number of successes = 10 ; number of failures = 992 
## Standard error = 0.0031 
## 95 % Confidence interval = ( 0.0038 , 0.0161 )
inference(Spain12$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 )

Answer: Answer: The 95% confidence intervals for the true proportion of atheists are (0.0038, 0.0161) for 2005 and (0.0364 and 0.0634) for 2012.Since they do not overlap and we have a strong evidence that United States change in its atheism index between 2005 and 2012.

Answer: To get this percent, we would simply multiply .05 by the total number of countries (39). We would expect to detect a change about 1.95 of those countries.

n<-0.5^2/(0.01/1.96)^2
n
## [1] 9604