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

Take a moment to review the press release for the poll conducted by WIN-Gallup International 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?

Because these percentages are referring to number gathered from pollsters, they would be considered 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?

In order to generalize our findings, our data set must be representative of the population to which we wish to generalize. At this point, it’s unclear if that’s a good assumption. At the end of the report, it summarizes the methods used in each country, including a field for whether the sample was national or urban. One should be skeptical about generalizing from an urban population to an entire country with regards to religious outlook.

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 the necessary packages and the data set into R with the following command.

library(oilabs)
library(ggplot2)
library(dplyr)  
data(atheism)
  1. What does each row of Table 6 correspond to? What does each row of atheism correspond to?

Each row corresponds to a country in Table 6. In the atheism data set, each row corresponds to a respondent.

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 data frame 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 <- 
  atheism %>% 
  filter(nationality == "United States" & year == 2012)
us12 %>%
  select(response) %>%
  table()/1002
## .
##     atheist non-atheist 
##   0.0499002   0.9500998
# OR
table(us12$response)/1002
## 
##     atheist non-atheist 
##   0.0499002   0.9500998

Yes, it agrees very closely. The calculated proportion of atheists is 0.0499.

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?

1) Each observation is independent of one another. This seems reasonable as long as the pollsters didn’t, for example, survey multiple people in the same family.
2) There are at least 10 expected successes and failures. \(.05 \times 1002 = 50.1\), which is greater than 10, so by that rule of thumb, this is condition is met.

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(y = 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?

We can find the margin of error either by multiplying the reported SE by 1.96 (for a 95% interval) or dividing the length of the confidence interval in two. \(1.96 \times0.0069 = 0.0135\).

  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.
ita12 <- 
  atheism %>% 
  filter(nationality == "Italy" & year == 2012)
inference(y = ita12$response, est = "proportion", type = "ci", 
          method = "theoretical", success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.08 ;  n = 987 
## Check conditions: number of successes = 79 ; number of failures = 908 
## Standard error = 0.0086 
## 95 % Confidence interval = ( 0.0631 , 0.097 )
ind12 <- 
  atheism %>% 
  filter(nationality == "India" & year == 2012)
inference(y = ind12$response, est = "proportion", type = "ci", 
          method = "theoretical", success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.0302 ;  n = 1092 
## Check conditions: number of successes = 33 ; number of failures = 1059 
## Standard error = 0.0052 
## 95 % Confidence interval = ( 0.0201 , 0.0404 )

We can assume independent observations (these samples are much smaller than 10% of the national populations) and have met the success/fail conditions, so it is reasonable to use the normal approximation to the sampling 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. \(p\) for all values of \(p\) between 0 and 1.

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 (\(ME = 1.96 \times SE\)). Lastly, we plot the two vectors against each other to reveal their relationship.

n <- 1000
p <- seq(0, 1, 0.01)
ME <- 1.96 * sqrt(p * (1 - p)/n)
qplot(x = p, y = ME, geom = "line",
      ylab = "Margin of Error", xlab = "Population Proportion")

  1. Describe the relationship between p and ME.

Holding the sample size constant, the ME reaches its maximum value when \(p = .50\). It decreases symmetrically as the proportion approaches both 0 and 1, taking a value of 0 when the proportion is equal to both 0 and 1.

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
}

qplot(x = p_hats, geom = "histogram", 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. 
## 0.06635 0.09423 0.10000 0.10020 0.10580 0.13560
mean(p_hats)
## [1] 0.1002267
sd(p_hats)
## [1] 0.009207751

*The sampling distribution is unimodal and symmetric, centered at a mean of about .1, with a standard deviation of .00934

  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 \(p = 0.02\). 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?
p <- 0.1
n <- 400
p_hatsA <- rep(0, 5000)

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

p <- 0.02
n <- 1040
p_hatsB <- rep(0, 5000)

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

p <- 0.02
n <- 400
p_hatsC <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hatsC[i] <- sum(samp == "atheist")/n
}
qplot(x = p_hats, geom = "histogram", main = "p = 0.1, n = 1040", xlim = c(0, 0.18))
## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.

qplot(x = p_hatsA, geom = "histogram", main = "p = 0.1, n = 400", xlim = c(0, 0.18))
## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.

qplot(x = p_hatsB, geom = "histogram", main = "p = 0.02, n = 1040", xlim = c(0, 0.18))
## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.

qplot(x = p_hatsC, geom = "histogram", main = "p = 0.02, n = 400", xlim = c(0, 0.18))
## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.

  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?

Using the normal approximation should be reasonable for Australia, but for Ecuador, the expected number of successes is only \(0.02 \times 400 = 8\), so it doesn’t meet that condition. The effect is apparent in the right skew of the sampling distribution.

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 <- atheism %>%
  filter(nationality == "Spain" & year == 2005)
inference(y = 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 )
spain12 <- atheism %>%
  filter(nationality == "Spain" & year == 2012)
inference(y = 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 )

The 95% confidence intervals for the true proportion of atheists are (.083, .118) for 2005 and (.073 and .107) for 2012. Because there is so much overlap, this doesn’t provide strong evidence that there has been a chance in Spain’s atheism index over this time period.

**b.** Is there convincing evidence that the United States has seen a
change in its atheism index between 2005 and 2012?
us05 <- atheism %>%
  filter(nationality == "United States" & year == 2005)
inference(y = 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 )
us12 <- atheism %>%
  filter(nationality == "United States" & year == 2012)
inference(y = 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 95% CI for the proportion of atheists in 2005 was (.004,.016) and in 2012 it was (.036, .063). These two intervals have no overlap, which suggests that, unlike Spain, there has been a change in the atheism index in the US between 2005 to 2012.

There are 39 countries listed in Table 4. If in fact there had been no change, we still would have detected them with probability 0.05 in each case. Since we’re concerned with the total number of “successes” in a series of n = 39 independent trials, we’re dealing with a binomially distributed random variable. The expected value of the binomial is \(np = 39 \times 0.05 = 1.95\) countries.

The key is to realize that the margin of error is maximized when \(p = 0.5\). So to ensure that we end up with a margine of error less than or equal to 0.01, we need to assume that the true \(p\) is .05, which is the worse case scenario. The formula for the margin of error is \(z^*\times SE\). We set that equal to .01, find the appropriate \(z^*\) for a 95% CI, plug in the worst case choice of \(p\), and solve for \(n\).

\[ 0.01 = 1.96 \times \sqrt{\frac{0.5 \times 0.5}{n}} \\ n = (0.5 \times 0.5) / \left(\frac{0.01}{1.96}\right)^2 \\ \]

(.5*.5)/(.01/1.96)^2
## [1] 9604

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.