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:
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?
They are sample statistics.
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?
We must assume that they used a random sampling method. I think that is reasonable since it was conducted by Gallup.
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
atheismcorrespond to?
Each row in Table 6 corresponds to a country from the 1012 survey. Each row in atheism corresponds to a survey respondent from the years 2005 and 2012
str(atheism)## 'data.frame': 88032 obs. of 3 variables:
## $ nationality: Factor w/ 57 levels "Afghanistan",..: 1 1 1 1 1 1 1 1 1 1 ...
## $ response : Factor w/ 2 levels "atheist","non-atheist": 2 2 2 2 2 2 2 2 2 2 ...
## $ year : int 2012 2012 2012 2012 2012 2012 2012 2012 2012 2012 ...table(atheism$year)##
## 2005 2012
## 36105 51927t(t(table(atheism$nationality)))##
## [,1]
## Afghanistan 1031
## Argentina 1993
## Armenia 495
## Australia 1039
## Austria 2005
## Azerbaijan 509
## Belgium 527
## Bosnia and Herzegovina 2000
## Brazil 2002
## Bulgaria 1997
## Cameroon 1008
## Canada 2005
## China 500
## Colombia 1212
## Czech Republic 2000
## Ecuador 804
## Fiji 1018
## Finland 1969
## France 3359
## Georgia 1000
## Germany 1004
## Ghana 2995
## Hong Kong 500
## Iceland 1704
## India 2183
## Iraq 1000
## Ireland 1010
## Italy 1974
## Japan 2412
## Kenya 2000
## Korea, Rep (South) 3047
## Lebanon 505
## Lithuania 2040
## Macedonia 2418
## Malaysia 1040
## Moldova 2170
## Netherlands 1014
## Nigeria 2098
## Pakistan 5409
## Palestinian territories (West Bank and Gaza) 627
## Peru 2414
## Poland 1045
## Romania 2089
## Russian Federation 2000
## Saudi Arabia 500
## Serbia 2073
## South Africa 402
## South Sudan 1020
## Spain 2291
## Sweden 495
## Switzerland 1020
## Tunisia 498
## Turkey 1032
## Ukraine 2026
## United States 2004
## Uzbekistan 500
## Vietnam 1000table(atheism$response)##
## atheist non-atheist
## 5498 82534To investigate the link between these two ways of organizing this data, take a look at the estimated proportion of atheists in the United States. Towards the bottom of Table 6, we see that this is 5%. We should be able to come to the same number using the atheism data.
- Using the command below, create a new dataframe called
us12that contains only the rows inatheismassociated with respondents to the 2012 survey from the United States. Next, calculate the proportion of atheist responses. Does it agree with the percentage in Table 6? If not, why?
us12 <- subset(atheism, nationality == "United States" & year == "2012")
prop.table(table(us12$response))##
## atheist non-atheist
## 0.0499002 0.9500998print("It agrees with the PDF table.")## [1] "It agrees with the PDF table."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 we expect to see at least 10 successes & 10 failures in the sample. Yes. Since Gallup conducted the survey, I think we can assume that the observations are independent. Also, we do have at least 10 successes and failures.
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 3.3.3## 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?
ath_US <- 0.0499002
ath_US - 0.0364 ## [1] 0.0135002- Using the
inferencefunction, 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 theinferencefunction to construct the confidence intervals.
Finland
Yr12 <- subset(atheism, year == "2012")
Fin12 <- subset(Yr12, nationality == "Finland")
table(Fin12$response)##
## atheist non-atheist
## 59 926print("Yes, the conditions for inference are met.")## [1] "Yes, the conditions for inference are met."inference(Fin12$response, est = "proportion", type = "ci", method = "theoretical", success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.0599 ; n = 985
## Check conditions: number of successes = 59 ; number of failures = 926
## Standard error = 0.0076
## 95 % Confidence interval = ( 0.0451 , 0.0747 )Georgia
Geor12 <- subset(Yr12, nationality == "Georgia")
table(Geor12$response)##
## atheist non-atheist
## 10 990print("Yes, the conditions for inference are met.")## [1] "Yes, the conditions for inference are met."inference(Geor12$response, est = "proportion", type = "ci", method = "theoretical", success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.01 ; n = 1000
## Check conditions: number of successes = 10 ; number of failures = 990
## Standard error = 0.0031
## 95 % Confidence interval = ( 0.0038 , 0.0162 )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
pandme.
The ME approaches zero at proportions near zero and one. The ME is increases as the proportion approaches .5 from each direction.
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
meanto calculate summary statistics.
It’s a symetrical, normal distribution with its mean and median being very close to .1. It’s SD is 0.009.
options(scipen = 999, digits = 4)
t(psych::describe(p_hats))## X1
## vars 1.0000000
## n 5000.0000000
## mean 0.0996900
## sd 0.0092874
## median 0.0990385
## trimmed 0.0996224
## mad 0.0085535
## min 0.0701923
## max 0.1298077
## range 0.0596154
## skew 0.0567685
## kurtosis -0.0911565
## se 0.0001313- 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))
n = 400
p = 0.1
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 = paste("p =", p, "n =", n), xlim = c(0, 0.18))
n = 1040
p = 0.02
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 = paste("p =", p, "n =", n), xlim = c(0, 0.18))
n = 400
p = 0.02
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 = paste("p =", p, "n =", n), xlim = c(0, 0.18))
AS \(p\) approaches .5, the spread increases. As \(n\) increases, the spread decreases.
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?
Yes, for Australia. However, Ecuador does NOT have at least 10 successes and failures.
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
inferencefunction. 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.
\(H_{0}: \mu_{Spain2005} - \mu_{Spain2012} = 0\) \(H_{A}: \mu_{Spain2005} - \mu_{Spain2012} \neq0\)
spain <- subset(atheism, nationality == "Spain")
spain05 <- subset(spain, year == 2005)
spain12 <- subset(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 )Since confidence intervals overlap, there is not convincing evidence that there was a change in Spain’s atheism index.
**b.** Is there convincing evidence that the United States has seen a change in its atheism index between 2005 and 2012?\(H_{0}: \mu_{US2005} - \mu_{US2012} = 0\) \(H_{A}: \mu_{US2005} - \mu_{US2012} \neq0\)
US <- subset(atheism, nationality == "United States")
US05 <- subset(US, year == 2005)
US12 <- subset(US, 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 )Since the confidence intervals do not overlap, it’s unlikely that this difference is due to chance. Thus, we have convincing evidence that there was change 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.
39 * .05## [1] 1.95- 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.
max_me <- .01
cL <- .95
cv <- abs(qnorm((1 - cL)/2))
max_p_hat <- .5
se <- max_me / cv
n <- max_p_hat * (1 - max_p_hat) / se^2
ceiling(n)## [1] 9604This 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.