Inference for categorical data

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 * https://sidmennt.is/wp-content/uploads/Gallup-International-um-tr%C3%BA-og-tr%C3%BAleysi-2012.pdf

Take a moment to review the report then address the following questions.

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?

Sample statistics (derived from the data sample) taken from a poll.
A poll conducted by WIN-Gallup International surveyed 51,927 people from 57 countries

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 basic requirements for generalization is that the sample needs to be a. independent and b. random. We assume that the samples were randomly selected and the groups are independent of each other.

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")

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

Table 6 contains each row that represent a country. And in each country, a survey was performed that asked whether you are religious, not-religious, or aetheist. Each row corresponds to the percentage of religious, not-religious, atheist, no response per each countries.

Each row of atheism corresponds to the response whether you are atheist or non-atheist.

colnames(atheism)

## [1] "nationality" "response"    "year"

table(atheism$response)

## 
##     atheist non-atheist 
##        5498       82534

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.

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")

We can state that it reasonably agrees the percentage in Table 6 (5%) with rounding.

library(data.table)
library(knitr)
us12 <- as.data.table(us12)
total <- nrow(us12)
q4 <-us12[, .(count = .N), by = response]

kable(q4[, prop := round((count / total) * 100,2), by = .(response)])

response	count	prop
non-atheist	952	95.01
atheist	50	4.99

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.

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?

The sample observations are independent – 1002 for the US is less than 10% of the total population in the US.
We assume that the samples are random.
To have at least 10 pos/neg responses:

np = 1002 * 0.05 
n.1p <- 1002 * 0.95

The success - failure conditions require np >= 10, 50.1 and n(1-p) >= 10, 951.9

np >= 10

## [1] TRUE

n.1p >= 10

## [1] TRUE

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”.

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?

Margin of error = z * standard of error for 95%, z = 1.96

se <- 0.0069
me <- 1.96 * se
me

## [1] 0.013524

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.

atheism <- as.data.table(atheism)
k12 <- atheism[nationality == 'Korea, Rep (South)' & year == '2012']
a12 <- atheism[nationality == 'Argentina' & year == '2012']

inference(k12$response, est = "proportion", type = "ci", method = "theoretical", success = "atheist")

## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.1504 ;  n = 1523 
## Check conditions: number of successes = 229 ; number of failures = 1294 
## Standard error = 0.0092 
## 95 % Confidence interval = ( 0.1324 , 0.1683 )

k.se <- 0.0092
k.me <- 1.96 * k.se
k.me

## [1] 0.018032

table(k12$response)

## 
##     atheist non-atheist 
##         229        1294

inference(a12$response, est = "proportion", type = "ci", method = "theoretical", success = "atheist")

## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.0706 ;  n = 991 
## Check conditions: number of successes = 70 ; number of failures = 921 
## Standard error = 0.0081 
## 95 % Confidence interval = ( 0.0547 , 0.0866 )

a.se <- 0.0081
a.me <- 1.96 * k.se
a.me

## [1] 0.018032

table(a12$response)

## 
##     atheist non-atheist 
##          70         921

Both South Korea and Argentina satisfy the conditions for inference and have np, n(1-p) >= 10.

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")

Describe the relationship between p and me.

Margin of error increases with increasing population until it reaches a maximum at 0.5 (50%) then Margin of error decreases as population increases to 1 (100%)

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.

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.

The sampling distribution is near normal. (Mean of 0.09, sd of 0.009.)

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

boxplot(p_hats)

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?

par(mfrow = c(2,2))

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))

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))

All the distributions appear to be normally distributed. The smaller the n, the larger the standard of error, and the larger the n, the smaller the standard of error, the smaller the spread, the smaller the margin error. The value of p affects the center of sampling distribution. For p is the opposite. The larger the p, the smaller the spread, the larger the margin of error (until it reaches 0.5)

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.

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?

au.n <- 1040
au.p <- 0.1

au.n * au.p >= 10

## [1] TRUE

au.n * (1-au.p) >= 10

## [1] TRUE

ed.n <- 400
ed.p <- 0.02

ed.n * ed.p >= 10

## [1] FALSE

ed.n * (1-ed.p) >= 10

## [1] TRUE

Both have a normal distribution with similar spreads, thus the margin of errors presumably near. Ecuador conditions are not met for success-failure(8) though, thus we shall not proceed with inference.

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.

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.

H0: Spain atheism index 05 == spain atheism index 12
Ha: spain atheism index 05 != spain atheism index 12

s05 <- atheism[nationality == "Spain" & year == "2005"]
s12 <- atheism[nationality == "Spain" & year == "2012"]
inference(s05$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(s12$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 CI for each year overlap, and there is no evidence that the indicies have changed significantly.

**b.** Is there convincing evidence that the United States has seen a
change in its atheism index between 2005 and 2012?

H0: United States atheism index 05 == United States atheism index 12
Ha: United States atheism index 05 != United States atheism index 12

us05 <- atheism[nationality == "United States" & year == "2005"]
us12 <- atheism[nationality == "United States" & year == "2012"]
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 )

The confidence interval for ’05 and ’12 in the states don’t poverlap, Thus H0 is rejected. There appears to be some chnages in the atheism index.

If in fact there has been no change in the atheism index in the countries listed in Table 4, in how many of those countries would you expect to detect a change (at a significance level of 0.05) simply by chance?
Hint: Look in the textbook index under Type 1 error.

A Type 1 Error is rejecting the null hypothesis when H0 is actually true. As a general rule of thumb, for those cases where the null hypothesis is actually true, we do not want to incorrectly reject H0 more than 5% of the time. With an alpha of 0.05, we expect about 5% of the countries would expect a change simply by chance. There are 39 countries in Table 4, therefore ~2 countries are expected to change simply by chance.

Suppose you’re hired by the local government to estimate the proportion of residents that attend a religious service on a weekly basis. According to the guidelines, the estimate must have a margin of error no greater than 1% with 95% confidence. You have no idea what to expect for \(p\). How many people would you have to sample to ensure that you are within the guidelines?
Hint: Refer to your plot of the relationship between \(p\) and margin of error. Do not use the data set to answer this question.

Both p, n variables are missing.

When we need to estimate for p as we do not have the information, we assume that the margin of error is largest when p is 0.5. This is the worst case of estimate. The estimate must have a margin of error < 0.01.

p <- 0.5
me <- 0.01
z <- 1.96 # 95% confidence or alpha - 0.05

#margin of error  = (p*(1-p)/n)^2
#n = (p*(1-p)/standard of error) ^2
se <- me/z

n <- (p*(1-p)) / se^2

We would need to ask 9604 people in order to be able to create 95% confidence interval with less than 1% of margin error.

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.