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.
To access the press release for the poll, conducted by WIN-Gallup International, click on the following link:
Take a moment to review the report then address the following questions.
These appear to be sample statistics since the numbers are calculated from the sample of 50,000 people, and not directly from the human population of 7.55 billion people.
The sample would have to be randomly selected from people across the Earth. They claim that they surveyed 50,000 people from 57 Countries on 5 continents. There are 233 countries in the world according to https://en.wikipedia.org/wiki/List_of_countries_by_population_(United_Nations), but the majority of the currently 7.55 billion people are concentrated in the 13 most populous countries. Cases also need to be independent, so you need to sample less than 10% of the population, which is also true. We also need at least 10 ‘successes’ and ‘failures’, in this case ‘atheists’ and ‘non-atheists’, this condition is also true. It seems reasonable to generalize these data to the global human population.
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")
atheism
correspond to?Each row in table 6 is a country where the survey was conducted. Each row in the ‘atheism’ table is a person who was interviewed for the survey.
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
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.
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")
# true = 1, false = 0, the numerator gives total number of trues
percent_ath <- sum(us12[,2] == 'atheist')/length(us12[,2]) #gives percent by dividing by length of the vector.
percent_ath
## [1] 0.0499002
This rounds to 5%, which is consistent with table 6.
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.
Cases are independent is true, 51,927 is less than 10% the world population. We also have at least 10 ‘successes’ and 10 ‘failures’, that is atheists and non-atheists or visa-versa. The conditions for inference have been 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(us12$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")
## Warning: package 'openintro' was built under R version 3.4.1
## Warning: package 'BHH2' was built under R version 3.4.1
## 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”.
The margin of error is half the width of the confidence interval.
ME = (0.0634-0.0364)/2
ME
## [1] 0.0135
1.35% margin of error.
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.Note the inference function does a condition check. First I selected Sweden since Scandinavian countries tend to be less religious.
sw12 <- subset(atheism, nationality == "Sweden" & year == "2012")
inference(sw12$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")
## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.0808 ; n = 495
## Check conditions: number of successes = 40 ; number of failures = 455
## Standard error = 0.0122
## 95 % Confidence interval = ( 0.0568 , 0.1048 )
The sample size is small enough for the cases to be independent, and the success-failure criteria is met.
ME2 = (0.1048-0.0568)/2
ME2
## [1] 0.024
Sweden has a 2.4% Margin of Error.
I then selected Fiji since it is the most religious country I’ve visited.
fj12 <- subset(atheism, nationality == "Fiji" & year == "2012")
inference(fj12$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")
## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.0098 ; n = 1018
## Check conditions: number of successes = 10 ; number of failures = 1008
## Standard error = 0.0031
## 95 % Confidence interval = ( 0.0038 , 0.0159 )
The sample size is small enough for the cases to be independent, and the success-failure criteria is met, barely.
ME3 = (0.0159-0.0038)/2
ME3
## [1] 0.00605
Fiji has a 0.6% 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")
p
and me
.It appears to be a parabola with a maximum at 50%, however it terminates at ME = 0 and p = 0 or 1.
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.
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
IQR(p_hats)
## [1] 0.0125
sd(p_hats)
## [1] 0.009287382
From this code we can determine two different measures of center and spread, median: 0.09904, IQR: 0.0125 mean: 0.09969, sd: 0.009287382
The shape of the distribution is bell-shaped.
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_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 = 400", xlim = c(0, 0.18))
p <- 0.02
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 = 0.02, n = 1040", xlim = c(0, 0.18))
p <- 0.02
n <- 400
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 = 0.02, n = 400", xlim = c(0, 0.18))
Decreasing p or increasing n makes the distribution more narrow.
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.
par(mfrow = c(1, 1))
Both the above histograms seem to follow a Normal Distribution. This is what the conditions of inference are checking; that the data are Normally distributed. It is sensible to proceed with inference calculations.
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.
Answer the following two questions using the inference
function. As always, write out the hypotheses for any tests you conduct and outline the status of the conditions for inference.
a. Is there convincing evidence that Spain has seen a change in its atheism index between 2005 and 2012?
Hint: Create a new data set for respondents from Spain. Form confidence intervals for the true proportion of athiests in both years, and determine whether they overlap.
sp12 <- subset(atheism, nationality == "Spain" & year == "2012")
sp05 <- subset(atheism, nationality == "Spain" & year == "2005")
inference(sp12$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")
## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.09 ; n = 1145
## Check conditions: number of successes = 103 ; number of failures = 1042
## Standard error = 0.0085
## 95 % Confidence interval = ( 0.0734 , 0.1065 )
inference(sp05$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 )
The 95% confidence intervals for both years do overlap for Spain. Therefore there is not convincing evidence that the proportion of Atheists have changed 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 == "United States" & year == "2005")
inference(us12$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")
## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.0499 ; n = 1002
## Check conditions: number of successes = 50 ; number of failures = 952
## Standard error = 0.0069
## 95 % Confidence interval = ( 0.0364 , 0.0634 )
inference(us05$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 )
The 95% confidence intervals for both years do not overlap for the US. Therefore there is convincing evidence that the proportion of Atheists have changed between 2005 and 2012.
The significance level, \(\alpha\), gives the probability of incorrectly rejecting the null hypothesis, which is a Type 1 error. In this case we would expect 5% to change by chance.
Margin of error is at a maximum at \(p = 0.5\). Therefore we will assume this since a different value of p will reduce overall Margin of Error.
\[ ME = 1.96*\sqrt{\frac{p(1-p)}{n}} \\ n = \frac{p(1-p)}{(\frac{ME}{1.96})^2} \\ n = \frac{0.5^2}{(\frac{0.01}{1.96})^2} \\ n = (\frac{0.5*1.96}{0.01})^2 \]
n = (0.5*1.96/0.01)^2
n
## [1] 9604
9,604 people.
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.