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:

http://www.wingia.com/web/files/richeditor/filemanager/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?

    I would say that those percentages appear to be derived from samples because the study analyses data that was taken from global poll.

  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 following assuptions should be made:

    1. Samples are random. Selection of any individual should happen by chance, rather than by choice. Random sampling reduces the chance that differences in materials or conditions strongly bias results.
    2. Observations are independend. Value of one observation does not influence or affect the value of other observations

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")
head(atheism)
##   nationality    response year
## 1 Afghanistan non-atheist 2012
## 2 Afghanistan non-atheist 2012
## 3 Afghanistan non-atheist 2012
## 4 Afghanistan non-atheist 2012
## 5 Afghanistan non-atheist 2012
## 6 Afghanistan non-atheist 2012

  1. What does each row of Table 6 correspond to? What does each row of atheism correspond to?

    Each row of Table 6 correspond to countries while each row of atheism correspond to individuals.

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?
us12 <- subset(atheism, nationality == "United States" & year == "2012")
nrow(subset(us12,response=="atheist"))/nrow(us12)
## [1] 0.0499002

The proportion of atheist responses in US is 4.99% (almost 5%).

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?

    Let’s check samples sizes.

by(us12$nationality,us12$response,length)
## us12$response: atheist
## [1] 50
## -------------------------------------------------------- 
## us12$response: non-atheist
## [1] 952

First, we’ve already assumed that the samples that were used in the study are independent (see question 2).

Second, we’ve already assumed that the samples are independent. Moreover, both samples are seem to be less than 10% of the population (I would assume that the number of athists in US in 2012 is much greater than 50 people and the number of non-athists in US in 2012 is much greater than 952 people).

Third, both random samples sizes are above 30.

So that, the conditions necessary for inference are satisfied.

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 )

The average proportion of atheists in US in 2012 falls in the range of (0.0364 , 0.0634) 95% of the time.

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?

    Let’s calculate margin of error based on the formula showd below:

    ME=SE*Z

    Z-score for 95 % Confidence interval is 1.96 Standard error was calculated by inference function.

#method 1
Z <- 1.96
SE <- 0.0069 
ME <- Z*SE
ME
## [1] 0.013524
#method 2
lower_bound <- 0.0364 
upper_bound <- 0.0634
ME <- (upper_bound - lower_bound)/2
ME
## [1] 0.0135

  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.

    Let’s create data sets called ch12 and ja12that contain only the rows in atheism associated with respondents to the 2012 survey from China ad Japan respectfully.

ch12 <- subset(atheism, nationality == "China" & year == "2012")
ja12 <- subset(atheism, nationality == "Japan" & year == "2012")

Let’s find samples sizes.

by(ch12$nationality,ch12$response,length)
## ch12$response: atheist
## [1] 235
## -------------------------------------------------------- 
## ch12$response: non-atheist
## [1] 265
by(ja12$nationality,ja12$response,length)
## ja12$response: atheist
## [1] 372
## -------------------------------------------------------- 
## ja12$response: non-atheist
## [1] 840

First, we’ve already assumed that the samples that were used in the study are independent (see question 2).

Second, we’ve already assumed that the samples are independent. Moreover, both samples are seem to be less than 10% of the population.

Third, both random samples sizes are above 30.

So that, the conditions necessary for inference are satisfied.

Now, let’s use inference function.

inference(ch12$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 )
inference(ja12$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 )

The average proportion of atheists in China in 2012 falls in the range of (0.4263 , 0.5137) 95% of the time whereas the average proportion of atheists in China in 2012 falls in the range of (0.4263 , 0.5137) 95% of the time (0.281 , 0.3329).

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.

The plot is bell-curved. The margin of error reaches its maximum when proportion reaches 0.5. When the proportions gets closer 0.00 or 1.00 the margin of error decreases. p and me are positevly correlated from 0 to 0.5 (pincreases asme increases) and negatively correlated from 0.5 to 1 (p increases as me decreases).

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
hist(p_hats)

The distribution of this population is unimodel since it has one clear peak. The distribution of area is bell-shaped and close to symmetric. It’s a little bit right skewed as the mean is a little bit greater than the median.

  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?
#build sample distribution for n=400 and p=0.1
p <- 0.1
n <- 400
p_hats1 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats1[i] <- sum(samp == "atheist")/n
}


#build sample distribution for n=1040 and p=0.02
p <- 0.02
n <- 1040
p_hats2 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats2[i] <- sum(samp == "atheist")/n
}


#build sample distribution for n=400 and p=0.02
p <- 0.02
n <- 400
p_hats3 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats3[i] <- sum(samp == "atheist")/n
}


par(mfrow = c(2, 2))
hist(p_hats1, main = "p = 0.1, n = 400", xlim = c(0, 0.18))
hist(p_hats2, main = "p = 0.02, n = 400", xlim = c(0, 0.18))
hist(p_hats3, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))

Based on the histograms above we can conclude that p affects the spread of a distribution while n affects the shape of a distribution. The spread decreases as p decreases. Two distributions of n=400 are closer to bimodal(with two peaks) distribution than unimodal distribution (with one clear peak) wheres the distribution of n=1040 is clearly unimodal. As the sample sizes get larger the shape of the distribution goes closer to the shape of normal distribution.

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?

    Let’s check success-failure condition before making inference. According to success-failure condition he sample proportion can be assumed to be nearly normal if it is
    1. based upon a random sample
    2. observations are independent
    3. both np≥10 and n(1−p)≥10.

    We’ve already assumed that all samples used in the study are random and all observations are independent. Now, we have to check the third condition.

p_australia <- 0.1  
n_australia <- 1040

#check np≥10 condition
p_australia*n_australia
## [1] 104
#check n(1−p)≥10 condition
n_australia*(1-p_australia)
## [1] 936

The success-failure condition is satisfied for Australia.

p_ecuador <- 0.02  
n_ecuador <- 400

#check np≥10 condition
p_ecuador*n_ecuador
## [1] 8
#check n(1−p)≥10 condition
n_ecuador*(1-p_ecuador)
## [1] 392

The success-failure condition is NOT satisfied for Ecuador because n(1−p) is less than 10.

So, we can apply inference function only to Australia.

au12 <- subset(atheism, nationality == "Australia" & year == "2012")
inference(au12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.1001 ;  n = 1039 
## Check conditions: number of successes = 104 ; number of failures = 935 
## Standard error = 0.0093 
## 95 % Confidence interval = ( 0.0818 , 0.1183 )

The average proportion of atheists in Australia in 2012 falls in the range of (0.4263 , 0.5137) 95% of the time.

lower_bound <- 0.4263
upper_bound <- 0.5137
MAE <- (upper_bound-lower_bound)/2
MAE
## [1] 0.0437

Margin of error for Australia is about 4.37%.

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.

sp05 <- subset(atheism, nationality == "Spain" & year == "2005")
sp12 <- subset(atheism, nationality == "Spain" & year == "2012")

Let’s check the success-failure condition for data set that represents atheism index in Spain in 2005.

p_sp05 <- nrow(subset(sp05,response=="atheist"))/nrow(sp05)
n_sp05 <- nrow(sp05)

#check np≥10 condition
p_sp05*n_sp05
## [1] 115
#check n(1−p)≥10 condition
n_sp05*(1-p_sp05)
## [1] 1031

The success-failure condition is satisfied.

Next let’s check the success-failure condition for data set that represents atheism index in Spain in 2012.

p_sp12 <- nrow(subset(sp12,response=="atheist"))/nrow(sp12)
n_sp12 <- nrow(sp12)


#check np≥10 condition
p_sp12*n_sp12
## [1] 103
#check n(1−p)≥10 condition
n_sp12*(1-p_sp12)
## [1] 1042

The success-failure condition is satisfied.

Now we can apply inference function to both data sets to find confidence levels.

inference(sp05$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.1003 ;  n = 1146 
## Check conditions: number of successes = 115 ; number of failures = 1031 
## Standard error = 0.0089 
## 95 % Confidence interval = ( 0.083 , 0.1177 )
inference(sp12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.09 ;  n = 1145 
## Check conditions: number of successes = 103 ; number of failures = 1042 
## Standard error = 0.0085 
## 95 % Confidence interval = ( 0.0734 , 0.1065 )

We can see that there is a huge overlap in confidence intervals.

If the confidence intervals do not overlap then one can conclude that there is a statistical difference between means. However, the opposite can not be concluded. When the confidence intervals do overlap then there still may be a difference. Since the confidence intervals overlap there might be a change in Spain’s atheism index between 2005 and 2012. However, there is NO convincing evidence that Spain has seen a 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?

Let's create two new data sets.
us05 <- subset(atheism, nationality == "United States" & year == "2005")
us12 <- subset(atheism, nationality == "United States" & year == "2012")

Let’s check the success-failure condition for data set that represents atheism index in the US in 2005.

p_us05 <- nrow(subset(us05,response=="atheist"))/nrow(us05)
n_us05 <- nrow(us05)

#check np≥10 condition
p_us05*n_us05
## [1] 10
#check n(1−p)≥10 condition
n_us05*(1-p_us05)
## [1] 992

The success-failure condition is satisfied.

Next let’s check the success-failure condition for data set that represents atheism index in the US in 2012.

p_us12 <- nrow(subset(us12,response=="atheist"))/nrow(us12)
n_us12 <- nrow(us12)


#check np≥10 condition
p_us12*n_us12
## [1] 50
#check n(1−p)≥10 condition
n_us12*(1-p_us12)
## [1] 952

The success-failure condition is satisfied.

Now we can apply inference function to both data sets to find confidence levels.

inference(us05$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(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 )

We can see that the confidence intervals do not overlap. If the confidence intervals do not overlap then one can conclude that there is a statistical difference between means. So, we can conclude that there is convincing evidence that the United States has seen a change in its atheism index between 2005 and 2012.

# 95% acceptance area
round(0.95*39,0)
## [1] 37
# 5% rejection area
round(0.05*39,0)
## [1] 2

There are two countries that reject null hypotheses when it’s true. In other words, two countries would detect a change in the atheism index when there is no change.

Z <- 1.96
p <- 0.04

#find n for interval when p is less or equal 0.04
n <- p*(1-p)/(0.01/Z)^2
round(n,0)
## [1] 1475
#find n for interval when p is greater or equal 0.76
p <- 0.76
n <- p*(1-p)/(0.01/Z)^2
round(n,0)
## [1] 7007

The number of people that can generate margin of error of 1% <=1475(approximatly) or >=7007(approximatly).

This is a product of OpenIntro that is released under a Creative Commons Attribution-ShareAlike 3.0 Unported. This lab was written for OpenIntro by Andrew Bray and Mine Çetinkaya-Rundel.