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:

https://github.com/jbryer/DATA606/blob/master/inst/labs/Lab6/more/Global_INDEX_of_Religiosity_and_Atheism_PR__6.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?

Answer

These findings appear to be population parameters as reported in the paragraph. There’s no metion of any sample.
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

We must assume that: (1) The sample size represented no more than 10% of the world’s population. (2) Respondents were chosen randomly. (3) The sample size followed a normal distribution.

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")
head(atheism,10)

##    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
## 7  Afghanistan non-atheist 2012
## 8  Afghanistan non-atheist 2012
## 9  Afghanistan non-atheist 2012
## 10 Afghanistan non-atheist 2012

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

Answer

Each row in Table 6 of the report corresponds to sample statistics for each country. Each row of of the atheism table corresponds to each respondent from each country and whether they are atheist or non-atheist.

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?

Answer

Yes. It does. Count of Aetheists is five percent of the total US respondents in the sample.

us12 <- subset(atheism, nationality == "United States" & year == "2012")

countofUSAtheist <- length(us12$response[us12$response=="atheist"])
countofUSRespondents <- length(us12$response)
countofUSAtheist / countofUSRespondents

## [1] 0.0499002

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 conditions are:

The sample observations are independent which requires that the respondents are chosen at random and represent less than 10% of the population. Both conditions for the US respondents have been met.
The “success-failure” condition has also been met. The sample size is sufficiently large. The count of “successes” is 50 and the count of failures is 952. Both counts are greater than 10.

The conditions for the US respondents has been met.

countofUSAtheist

## [1] 50

countofUSRespondents - countofUSAtheist

## [1] 952

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 3.5.1

## 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”.

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

The MOE is 0.0135

MOE <- abs(.0364 - 0.0499)
Lower_Bound <- 0.0499 - MOE
Upper_Bound <- 0.0499 + MOE

Lower_Bound

## [1] 0.0364

Upper_Bound

## [1] 0.0634

MOE

## [1] 0.0135

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. ## How does the proportion affect the margin of error?

ANSWER

Looking at Ireland and Ukraine. Taking Ireland first. The conditions for inference are met:

The sample observations are independent which requires that the respondents are chosen at random and represent less than 10% of Ireland’s population. Both conditions for the Ireland respondents have been met.
The “success-failure” condition has also been met. The sample size is sufficiently large. The count of “successes” is 100 and the count of failures is 910. Both counts are greater than 10.

The MOE for Ireland is 0.0184 or 1.84%, and the MOE for the Ukraine is 0.0104 or 1.04%. The porportion of successes in Ireland is 0.09 or roughly 10%. The porportion of successes in Ukraine is 0.0296 or 3%. When the porportion of success is higher, so to is the MOE (Ireland success porportion is 10% and MOE is 1.84%) (US success porportion is 5% and MOE is 1.35%) (Ukraine success porportion is 3% and the MOE is 1.04%)

Ireland12 <- subset(atheism, nationality == "Ireland" & year == "2012")

countofIrelandAtheist <- length(Ireland12$response[Ireland12$response=="atheist"])
countofIrelandRespondents <- length(Ireland12$response)
countofIrelandAtheist / countofIrelandRespondents

## [1] 0.0990099

countofIrelandAtheist

## [1] 100

countofIrelandRespondents - countofIrelandAtheist

## [1] 910

inference(Ireland12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")

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

## p_hat = 0.099 ;  n = 1010 
## Check conditions: number of successes = 100 ; number of failures = 910 
## Standard error = 0.0094 
## 95 % Confidence interval = ( 0.0806 , 0.1174 )

MOE <- abs(.0806 - 0.099)
Lower_Bound <- 0.099 - MOE
Upper_Bound <- 0.099 + MOE

Lower_Bound

## [1] 0.0806

Upper_Bound

## [1] 0.1174

MOE

## [1] 0.0184

The sample observations are independent which requires that the respondents are chosen at random and represent less than 10% of Ukraine’s population. Both conditions for the Ireland respondents have been met.
The “success-failure” condition has also been met. The sample size is sufficiently large. The count of “successes” is 30 and the count of failures is 983. Both counts are greater than 10.

Ukraine12 <- subset(atheism, nationality == "Ukraine" & year == "2012")
head(Ukraine12,10)

##       nationality    response year
## 48913     Ukraine non-atheist 2012
## 48914     Ukraine non-atheist 2012
## 48915     Ukraine non-atheist 2012
## 48916     Ukraine non-atheist 2012
## 48917     Ukraine non-atheist 2012
## 48918     Ukraine non-atheist 2012
## 48919     Ukraine non-atheist 2012
## 48920     Ukraine non-atheist 2012
## 48921     Ukraine non-atheist 2012
## 48922     Ukraine non-atheist 2012

countofUkraineAtheist <- length(Ukraine12$response[Ukraine12$response=="atheist"])
countofUkraineRespondents <- length(Ukraine12$response)
countofUkraineAtheist / countofUkraineRespondents

## [1] 0.029615

countofUkraineAtheist

## [1] 30

countofUkraineRespondents - countofUkraineAtheist

## [1] 983

inference(Ukraine12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")

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

## p_hat = 0.0296 ;  n = 1013 
## Check conditions: number of successes = 30 ; number of failures = 983 
## Standard error = 0.0053 
## 95 % Confidence interval = ( 0.0192 , 0.0401 )

MOE <- abs(.0192 - 0.0296)
Lower_Bound <- 0.0296 - MOE
Upper_Bound <- 0.0296 + MOE

Lower_Bound

## [1] 0.0192

Upper_Bound

## [1] 0.04

MOE

## [1] 0.0104

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.

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.

ANSWER

The spread of the data is unimodal, symmetrical, normally distributed. The highest frequency is centered in the middle around p=0.1.

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?

Answer

For the first two sampling distributions with n=1040, p_hats=0.1 and n=400 p_hats01=0.1, the histograms both show a unimodal, symmetrical normal distribution with the highest frequency centered in the middle over 0.1. “n” seems to be affecting the spread of the distribution. When “n” is 1040, the spread is more tight around the center, conversely when n is large at 1040, the spread is wider.

As for the two sampling distributions with n=1040, p_hats02=0.02 and n=400 p_hats03=0.02, the histograms both show a unimodal, symmetrical normal distribution with the highest frequency centered in the middle over 0.02. Again, “n” seems to be affecting the spread of the distribution. When “n” is 1040, the spread is more tight around the center, conversely when n is large at 1040, the spread is wider.

p <- 0.1
n <- 400
p_hats01 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats01[i] <- sum(samp == "atheist")/n
}

p <- 0.02
n <- 1040
p_hats02 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats02[i] <- sum(samp == "atheist")/n
}

p <- 0.02
n <- 400
p_hats03 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats03[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_hats01, main = "p = 0.1, n = 400", xlim = c(0, 0.18))
hist(p_hats02, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))
hist(p_hats03, 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.

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

Yes because the plots show a normal distribution of the samples, and the Central Limit Theorem which holds that the distribution of sample means–if they meet the established conditions–will follow a normal distribution and from there we can make inferences about the population.

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.

Answer

The sample observations are independent which requires that the respondents are chosen at random and represent less than 10% of Spain’s population. Both conditions for the Spain respondents have been met.
The “success-failure” condition has also been met. The sample size is sufficiently large. The count of “successes” is 115 for 2005 and 103 for 2012. The count of failures is 1031 and 1042 for each year. All counts are greater than 10.

The hypothesis test is as follows:

Null Hypothesis: Spain has not seen a change in its atheism index between 2005 and 2012 Alt Hypothesis: Spain has seen a change in its atheism index between 2005 and 2012

The Confidence Intervals over lap which means that we cannot conclude that the difference in the two years is statistically significant. This was further confirmed by finding the p-value for the difference in populations which is 0.8017235, and because the p-value is greater than 0.05, we cannot reject the null hypothesis.

Spain05 <- subset(atheism, nationality == "Spain" & year == "2005" )

Spain12 <- subset(atheism, nationality == "Spain" & year == "2012" )

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 )

p_hats04 =  (115/(115+1031)) - (103/(103+1042)) 
p_Spain05 =  ((.1003)*(1-.1003)) / 1146
p_Spain12 =  ((0.09)*(1-0.09)) / 1145

SE = sqrt(p_Spain05 + p_Spain12)
SE

## [1] 0.01225854

Zscore = (p_hats04 - 0) / SE
pnorm(Zscore)

## [1] 0.8017235

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

Answer

CONDITIONS

The sample observations are independent which requires that the respondents are chosen at random and represent less than 10% of Spain’s population. Both conditions for the Spain respondents have been met.
The “success-failure” condition has also been met. The sample size is sufficiently large. The count of “successes” is 115 for 2005 and 103 for 2012. The count of failures is 1031 and 1042 for each year. All counts are greater than 10.

The hypothesis test is as follows:

Null Hypothesis: The US has not seen a change in its atheism index between 2005 and 2012 Alt Hypothesis: The US has seen a change in its atheism index between 2005 and 2012

The Confidence Intervals over lap which means that we cannot conclude that the difference in the two years is statistically significant. This was further confirmed by finding the p-value for the difference in populations which is 0.9999, and because the p-value is greater than 0.05, we cannot reject the null hypothesis.

No, there is no convincing evidence that the US has seen a change in its atheism index between 2005 and 2012.

us05 <- subset(atheism, nationality == "United States" & year == "2005")
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 )

p_hats05 =  (50/(50+952)) - (10/(10+992)) 
p_US05 =  ((.01)*(1-.01)) / 1002
p_US12 =  ((0.0499)*(1-0.0499)) / 1002

SE = sqrt(p_US05 + p_US12)
Zscore = (p_hats05 - 0) / SE
Zscore

## [1] 5.278505

pnorm(Zscore)

## [1] 0.9999999

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.

Answer

Using the 0.05 significance level, we would expect to detect a change in any country that saw a 5% or more change of which there are 4. However, by doing so, we would be comitting a Type 1 error without a hypothesis test.

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.

Answer

You would need to sample more than 1000 people to get a margin of error of 1% with 95% confidence. At a sample size of 1000, the MOE is at 1% when the proportion is 0% or 100%. As the proportion moves between these two values, the MOE grows.

We can confirm this by substituting p = .5 which gives a sample size of 9,604 needed for a MOE of 1% at a 95% Confidence Interval.

p = .5
MOE = .01
CI = 1.96

n = .25 / (.01/1.96)^2
n

## [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.