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/EPSY530Summer2015/blob/master/Labs/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?

    The statistics reported in the first paragraph are sample statistics (derived from the data sample) since it says, “according to the latest 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?

    To generalize the results of the poll to the global human population we must assume that the survey respondents were chosen at random from the population and that there was no bias in the sample. It seems unlikely to me that there is no non-response bias, or convenience bias in a poll that is global in scale. Since there are 195 countries in the world and the survey only sampled from 57 countries, and there are 6 continents (if you exclude Antartica) and the survey only sampled 5, I don’t think you could generalize the results to the global population as a whole, only possibly to each individual country that was sampled. Even that is a stretch since the summary at the end says that 11 out of the 57 countries were only polled in urban areas.

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?

    Each row in table 6 corresponds to the aggragated results from one country that was polled.

    Each row in the aetheism dataset corresponds to one respondent in the poll, and includes the country they are from, whether they are an aetheist or not and the year they were polled.

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")
us12$nationality <- factor(us12$nationality)
prop.table(table(us12$response))
    ## 
##     atheist non-atheist 
##   0.0499002   0.9500998

    Yes, it does agree with the percentage in tabe 6 if you round to two digits.

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?

    Each proportion separately must follow a normal model, in order to check conditions for normality, the sample observations must be independent and have at least 10 successes and at least 10 failures. Additionally, the two samples must be independent of each other. It seems reasonable to assume these conditions are met since:

    • there are 1002 observations in our sample of US residents surveyed in 2012, so if the proportion of aetheists is 5% and non-aetheists are 95% each proportion would meet the success-failure condition
    • the two proportions are independent of each other
    • the observations can be assumed to be independent since they were randomly selected.

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")
## Warning: package 'BHH2' was built under R version 3.4.4
## 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 atheists in US in 2012?

    ME <- 0.0634 - 0.0499
ME
    ## [1] 0.0135

    The margin of error is 1.35%.

  2. 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.

    Germany

    It seems reasonable to assume the conditions for inference are met since:

    • there are 502 observations in our sample and the proportion of aetheists is 15% and non-aetheists is 85% so each proportion would meet the success-failure condition
    • the two proportions are independent of each other
    • the observations can be assumed to be independent since they were randomly selected
    germany12 <- subset(atheism, nationality == "Germany" & year == "2012")
inference(germany12$response, est = "proportion", type = "ci", method = "theoretical", 
      success = "atheist")
    ## Single proportion -- success: atheist 
## Summary statistics:

    ## p_hat = 0.1494 ;  n = 502 
## Check conditions: number of successes = 75 ; number of failures = 427 
## Standard error = 0.0159 
## 95 % Confidence interval = ( 0.1182 , 0.1806 )
    ME <- 0.1806 - 0.1494

    The margin of error is 3.12%.

    Italy

    It seems reasonable to assume the conditions for inference are met since:

    • there are 987 observations in our sample and the proportion of aetheists is 8% and non-aetheists is 92% so each proportion would meet the success-failure condition
    • the two proportions are independent of each other
    • the observations can be assumed to be independent since they were randomly selected
    italia12 <- subset(atheism, nationality == "Italy" & year == "2012")
inference(italia12$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 )
    ME <- 0.097 - 0.08

    The margin of error is 1.7%.

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 margin of error me increases as the proportion p approaches 0.5 and then decreases from 0.5 to 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
}

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
    sd(p_hats)
    ## [1] 0.009287382

    The sampling distribution of sample proportions at \(n = 1040\) and \(p = 0.1\) appears normal in shape with it’s center at approximately \(p\) or .10, a spread from .07 to .13, and a standard deviation of .009.

  2. 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?

    p <- 0.1
n <- 400
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
}
    p <- 0.02
n <- 1040
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
}
    p <- 0.02
n <- 400
p_hats4 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats4[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_hats2, main = "p = 0.1, n = 400", xlim = c(0, 0.18))
hist(p_hats3, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))
hist(p_hats4, main = "p = 0.02, n = 400", xlim = c(0, 0.18))

    par(mfrow = c(1, 1))

    Each distribution has it’s center at p. The two with sample size of 400 have wider spreads while the two with samples of size 1040 have narrower spreads. The plot with the smaller proportion and larger sample size has the narrowest spread and the distribution with the larger proportion and smaller sample size has the widest spread. A change in p moves the center of the sampling distribution to the new p. An increase in n narrows the spread while a decrease in n widens the spread of the sampling 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?

    For Australia this makes sense. The proportion and sample size correspond to the first of the four plots we did above, so the sample size is sufficiently large enough to have a relatively small margin of error for that proportion. For Ecuador we don’t even meet the success-failure condition. We would only have 8 successes at .02 p in 400 observations. So it would not normally make sense to preceed with inference, however, since it is just a small part of a much larger study, it may make sense to report it as long as you highlight the large margin of error, rather than excluding it from your results.

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.

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.