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?

Answer: The percentages are due to the results of the survey which is a sample of the population (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: You would have to assume that the findings were randomly selected and that the groups are independent of each other. Also, an assumption that the sample size is large.

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.

load("more/atheism.RData")
  1. What does each row of Table 6 correspond to? What does each row of atheism correspond to?

Answer: Within Table 6, each row corresponds to results of the survey for each country. Each row of atheism corresponds to a single observation- one individual thats surveyed.

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")

# Let's remove unused levels
us12$nationality <- as.factor(as.character(us12$nationality))

# Let's get the proportions
( us12prop <- prop.table(table(us12$nationality, us12$response)) )
##                
##                   atheist non-atheist
##   United States 0.0499002   0.9500998

The proportion of atheists in the US is 0.0499002 which is less than the report which had a value of 0.05

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: The observations must be independent. If we are having the assumption that individuals were chosen using a random sample and if the sample is less than 10% of the population, the condition is satisfied. Also the observations must come from a nearly normal distribution. If we are considering the percentage of atheists at 0.05 and number of observations at 1,002 the observered number of atheists is 50 which is more than 10.

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?

Answer: Confidence Interval is (0.0364, 0.0634), so margin of error is (0..0634 - 0.0364) / 2 = 0.0135. The margin of error is 1.96 * Standard Error = 1.96 * 0.0069 = 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.
# Let's generate proportions for Colombia
col12 <- subset(atheism, nationality == "Colombia" & year == "2012")
col12$nationality <- as.factor(as.character(col12$nationality))
( col12prop <- prop.table(table(col12$nationality, col12$response)) )
##           
##               atheist non-atheist
##   Colombia 0.02970297  0.97029703
inference(col12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.0297 ;  n = 606 
## Check conditions: number of successes = 18 ; number of failures = 588 
## Standard error = 0.0069 
## 95 % Confidence interval = ( 0.0162 , 0.0432 )

Margin of error for Colombia is (0.0432 - 0.0162) / 2 = 0.0135.

# Generate proportions for Lithuania
lit12 <- subset(atheism, nationality == "Lithuania" & year == "2012")
lit12$nationality <- as.factor(as.character(lit12$nationality))
( lit12prop <- prop.table(table(lit12$nationality, lit12$response)) )
##            
##                 atheist non-atheist
##   Lithuania 0.009852217 0.990147783
inference(lit12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.0099 ;  n = 1015 
## Check conditions: number of successes = 10 ; number of failures = 1005 
## Standard error = 0.0031 
## 95 % Confidence interval = ( 0.0038 , 0.0159 )

The margin of error for Lithuania equals: (0.0159 - 0.0038) / 2 = 0.00605.

Similar to conditions for US sample, we can assume that samples for Colombia and Lithuania are independent and follow nearly normal distribution. Although the percentage of atheists in Lithuania is small, the number of atheists in the sample is 10, so it is borderline acceptable to assume nearly normal distribution.

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: Looking at the graph, the proportion of 0.50 is the proportion that will provide the largest margin of error possible. So, if we have p = 0.5 and we calculate p(1 - p) = 0.5 (1 - 0.5) = 0.5 * 0.5 is the maximum value for the numerator, making it the biggest possible value for the ME. Also, for p = 0 and p = 1 will be returning the minimums since 1 * 0 = 0 In other words, the nearer p comes to 0.5 the bigger ME will be and the nearer p becomes to 0 or 1 the lower ME will be.

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.

Answer:

library(psych)
describe(p_hats)
##    vars    n mean   sd median trimmed  mad  min  max range skew kurtosis
## X1    1 5000  0.1 0.01    0.1     0.1 0.01 0.07 0.13  0.06 0.06    -0.09
##    se
## X1  0
summary(p_hats)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## 0.07019 0.09327 0.09904 0.09969 0.10577 0.12981

The simulation to produce 5000 samples of size 1040 sample proportions, each follow normal distribution since they follow the conditions on Inference for proportions, some samples might have some outliers due to chance. The median and mean of the distribution are near identical at 0.1 with an standard deviation of 0.01. The sampling distribution of sample of proportions has a bell shape and follow a normal distribution.

  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?
  1. 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
}
  1. 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
}
  1. 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_hats, main = "p = 0.1, n = 1040", xlim = c(0, 0.18))
hist(p_hats1, main = "p = 0.1, n = 400", xlim = c(0, 0.18))
hist(p_hats2, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))
hist(p_hats3, main = "p = 0.02, n = 400", xlim = c(0, 0.18))

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: By analyzing the conditions on inference proportions, we have as follows:

# Australia

n_au <- 1040
p_au <- 0.1

cond_au <- c(n_au * p_au >= 10, n_au * (1 - p_au) >= 10)

# Ecuador

n_ecu <- 400
p_ecu <- 0.02

cond_ecu <- c(n_ecu * p_ecu >= 10, n_ecu * (1 - p_ecu) >= 10)

Based on Australia’s conditions are TRUE for n * p and TRUE n * ( 1 - p). Based on Ecuador’s conditions are FALSE for n * p and TRUE n * ( 1 - p).

Since one of the Ecuador’s conditions are not met, we might be inclined to reject the results since one of the conditions is not met. However, in my opinion this could be counted as valid result since n_ecu * p_ecu = 8 and the difference with the normal distribution should not be significant different.

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 == "Spain" & year == "2005")
spain05$nationality <- as.factor(as.character(spain05$nationality))
table(spain05$nationality, spain05$response)
##        
##         atheist non-atheist
##   Spain     115        1031
spain12 <- subset(atheism, nationality == "Spain" & year == "2012")
spain12$nationality <- as.factor(as.character(spain12$nationality))
table(spain12$nationality, spain12$response)
##        
##         atheist non-atheist
##   Spain     103        1042

We can assume observations to be independent. The number of atheists in 2005 is 115 and in 2012 it is 103. Both are greater than 10, so we can assume near normal distribution.

H0: The number of atheists in Spain did not change between 2005 and 2012, or p12=p05=0.1 HA: The number of atheists in Spain changed between 2005 and 2012, or p12≠0.1.

inference(spain05$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(spain12$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 )

There is significant overlap between confidence interval for 2005 sample and 2012 sample. Additionally, p12=0.09 and it is within the confidence interval for 2005 - (0.081,0.1177) - so we fail to reject the null hypothesis. The change in atheism is likely due to chance.

**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 == "United States" & year == "2005")
us05$nationality <- as.factor(as.character(us05$nationality))
table(us05$nationality, us05$response)
##                
##                 atheist non-atheist
##   United States      10         992
table(us12$nationality, us12$response)
##                
##                 atheist non-atheist
##   United States      50         952

We can assume observations to be independent. The number of atheists in 2005 is 10 and in 2012 it is 50. The nubmer of atheists in 2005 is borderline enough to assume near normal distribution.

H0: The number of atheists in the United States did not change between 2005 and 2012, or p12=p05=0.01 HA: The number of atheists in the United States changed between 2005 and 2012, or p12≠0.01

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 )

There is no overlap between confidence interval for 2005 sample and 2012 sample. Additionally, p12=0.05 and it is outside of the confidence interval for 2005 - (0.0038,0.0161)- so we reject the null hypothesis. The change in atheism is not likely due to chance.

If there has been no change in the atheism index, but we detect a change due to chance and reject a null hypothesis even though it is true, that means we have made a Type I error. At a significant level of 0.05 and considering that we have 39 countries, we would expect to make a Type I error with 39 * 0.05 = 1.95 or, rounding up, with 2 countries.

If the margin of error is 0.01, then at 95% confidence SE=0.011.96=0.0051. Since we do not know the value of p, we assume the worst case scenario with p=0.5. Considering, SE=p(1−p)n‾‾‾‾‾‾√, then n=p(1−p)SE2=0.5∗0.50.00512=9604. Sample must include at least 9,604 people. This is a worst case scenario. It is possible that there is some good estimate of what proportion of residents attend services, so p can be lowered in the above calculation.

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.