9_MarcelaRendon_Lab8
Marcela Rendon
10/28/2021
Inference for categorical data
In August of 2012, news outlets ranging from the Washingt 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 and then address the following questions.
Exercise 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 percentages reported in the first paragraph are sample statistics.
Exercise 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?
We must assume that the people polled were selected randomly, independently, and are representative of the overall population of each region or country. We also assume that the countries selected were stratified or selected in some way that doesn’t exclude a major region of the world. I think it seems like a reasonable assumption.
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 lose our analysis on the original set of individual responses to the survey. Load this data set into R with the following command.
download.file("http://www.openintro.org/stat/data/atheism.RData", destfile = "atheism.RData")
load("atheism.RData")Exercise 3
What does each row of Table 6 correspond to? What does each row of atheism correspond to?
Each row of Table 6 shows a country and the percentage responses categorized as a religious person, not a religious person, a convinced atheist, or don’t know/no response as well as the sample size for the country. Each row of atheism lists the answer categorized either as atheist or not atheist and lists the country and year the answer was obtained.
Exercise 4
Using the command below, create a new dataframe calle us12 that contains only the rows in atheism associated with respondents to the 2012 survey from the United States. Next, calcualte 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")
nrow(us12)## [1] 1002atheistUS12 <- subset(us12, response == "atheist")
nrow(atheistUS12)## [1] 50propAtheistUS12 <- nrow(atheistUS12)/nrow(us12)
propAtheistUS12## [1] 0.0499002The propportion of of atheists in the US in 2012 is 0.0499002. This matches the 5% shown in Table 6 if we round.
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.
Exercise 5
Write out the conditions for interference 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 that the observations are independent of each other, the sample should be normal, and that they are randomly selected. I think that all the conditions are met.
If the conditions for the inference are reasonable, we can either calculate the standard error and construct the interval by hand, or allow the inderence 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 +/- 3-5% at 95% confidence”.
Exercise 6
Based on the R output, what is the margin of error for the estimate of the proportion of atheists in the US in 2012?
zscore <- 1.96 #for 95% confidence interval
se <- 0.0069 # from problem above
marginError <- zscore*se
marginError## [1] 0.013524The margin of error for the estimate of the proportion of atheists in the US in 2012 is 0.013524.
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 used in the formula for the margin of error for a 95% confidence interval: ME = 1.96xSE = 1.96xsqrt(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 associated with each of these values of p using the familiar approximate formula (ME = 2 x 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")
Exercise 8
Describe the relationship between p and me.
It looks like the largest margin of error occurs at 0.5 population proportion. As the population proportion increases or decreases from 0.5, the margin of error is lower.
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 noth $np $ and \(n(1-p) \ge 10\). 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 should really 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 p-hat and then plot a histogram to visualize their distribution.
set.seed(01)
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 p-hat 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.
Exercise 9
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.
## 0.07212 0.09327 0.10000 0.09981 0.10577 0.13269The sampling looks liek a normal distribution with a median and mean of about 0.1, which is the same as the p-value. Its shape looks like a normal bell shape. There are a few outliers but overall it seems like a normal distribution.
Exercise 10
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 accomodate 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 p-hat? How does p affect the sampling distribution?
- n = 400 and p = 0.01:
set.seed(02)
p <- 0.1
n <- 400
p_hats1 <- rep(0, 5000)
for(i in 1: 5000) {
samp <- sample(c("atheist", "non-atheist"), n, replace = TRUE, prob = c(p, 1-p))
p_hats1[i] <- sum(samp == "atheist")/n
}- n = 1040 and p = 0.02:
set.seed(03)
p <- 0.02
n <- 1040
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
}- n = 400 and p = 0.02:
set.seed(04)
p <- 0.02
n <- 400
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
}Histograms:
par(mfrow = c(2,2))
hist(p_hats, main = "p = 0.1, n = 1040", xlim = c(0, 0.18))
hist(p_hats1, main = "p = 0.1, n = 400", xlim = c(0, 0.18))
hist(p_hats, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))
hist(p_hats, main = "p = 0.02, n = 400", xlim = c(0, 0.18))
The three new distributions are similar in the sense that the center of are the same. However, the spread or variability decreases when the sample size increases. They all follow normal distributions.
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.
Exercise 11
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 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 report does?
I think it would be reasonable to proceed with inference reporting the margin of error for Australia but not Ecuador. The Australia sample has met the conditions for sampling distributions whereas the Ecuador sample does not meet the conditions. The np for Ecuador is only 8 but it is 10 for Australia.
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.
1. 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 atheists in both years, and determine whether they overlap.
Null hypothesis: There is no convincing evidence that Spain has seen a change in its atheism index between 2005 and 2012. Alternative hypothesis: There is convincing evidence that Spain has seen a change in its atheism index between 2005 and 2012.
spain2005 <- subset(atheism, nationality == "Spain" & year == "2005")
spain2005atheist <- subset(spain2005, response == "atheist")
spain2012 <- subset(atheism, nationality == "Spain" & year == "2012")
spain2012atheist <- subset(spain2012, response == "atheist")
inference(spain2005$response, est = "proportion", type = "ci", method = "theoretical", success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.1003 ; n = 1146
## Check conditions: number of successes = 115 ; number of failures = 1031
## Standard error = 0.0089
## 95 % Confidence interval = ( 0.083 , 0.1177 )inference(spain2012$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 confidence intervals for 2005 and 2012 overlap, meaning that there is no convincing evidence that there was a change in levels of atheism in Spain between 2005 and 2012.
b. Is there convincing evidence that the United States has seen a change in atheism between 2005 and 2012?
usatheist2005 <- subset(atheism, nationality == "United States" & year == "2005")
inference(usatheist2005$response, est = "proportion", type = "ci", method = "theoretical", success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.01 ; n = 1002
## Check conditions: number of successes = 10 ; number of failures = 992
## Standard error = 0.0031
## 95 % Confidence interval = ( 0.0038 , 0.0161 )# from Exercise 5
inference(us12$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 intervals for 2005 and 2012 do not overlap. This gives us evidence that the population proportion of atheists in the United States is different between the two years. It looks like atheism levels increased in the United States from 2005 to 2012.
2. 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 significant level of 0.05) simply by chance? Hint: Look in the textbook index under Type 1 error.
A type I error is when we incorrectly reject the null hypothesis (meaning that the null hypothesis is actually true). In this case, we assume that about 5% of the countries in table 4 would detect a change by chance (about 2 countries).
3. 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.
p <- 0.5
Z <- 1.96
ME <- 0.01
num <- (p * (1-p) * Z ^2) / ME^2
num## [1] 9604We would need at a sample of at least 9604 residents to ensure that the sample proportion is within a 1% margin of error of the true proportion with 95% confidence.