LabOnCategoricalDataInference
Mary Glantz
10/28/2021
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.
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?
These appear to be sample statistics. The text says: “according to the latest global poll released by WIN-Gallup International…”, and then introduces the percentages, which are apparently drawn from that poll.
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 sample is random and unbiased. Also, that it represents the populations present in the globe. Given the information about the network of pollsters and where they conducted the interviews, this 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 base our analysis on the original data 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?
head(atheism, 30)## 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
## 11 Afghanistan non-atheist 2012
## 12 Afghanistan non-atheist 2012
## 13 Afghanistan non-atheist 2012
## 14 Afghanistan non-atheist 2012
## 15 Afghanistan non-atheist 2012
## 16 Afghanistan non-atheist 2012
## 17 Afghanistan non-atheist 2012
## 18 Afghanistan non-atheist 2012
## 19 Afghanistan non-atheist 2012
## 20 Afghanistan non-atheist 2012
## 21 Afghanistan non-atheist 2012
## 22 Afghanistan non-atheist 2012
## 23 Afghanistan non-atheist 2012
## 24 Afghanistan non-atheist 2012
## 25 Afghanistan non-atheist 2012
## 26 Afghanistan non-atheist 2012
## 27 Afghanistan non-atheist 2012
## 28 Afghanistan non-atheist 2012
## 29 Afghanistan non-atheist 2012
## 30 Afghanistan non-atheist 2012Each row of Table 6 corresponds to a country. Each row of atheism corresponds to a response from an individual.
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.
Exercise 4
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")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 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 must be random, independent, and no more than 10% of the population. It appears that all those conditions 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$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.0499 ; n = 1002
## Check conditions: number of successes = 50 ; number of failures = 952
## Standard error = 0.0069
## 95 % Confidence interval = ( 0.0364 , 0.0634 )Note that since the goal is to construct an interval estimate for a proportion, it’s necessary to specify what constitutes a “success”, which here is a response of “atheist”.
Although formal confidence intervals and hypothesis tests don’t show up in the report, suggestions of inference appear at the bottom of page 7: “In general, the error margin for surveys of this kind is ± 3-5% at 95% confidence”.
Exercise 6
Based on the R output, what is the margin of error for the estimate of the proportion of the proportion of atheists in US in 2012?
+ or - 1.4%
Exercise 7
Using the inference function, calculate confidence intervals for the proportion of atheists in 2012 in two other countries of your choice, and report the associated margins of error. Be sure to note whether the conditions for inference are met. It may be helpful to create new data sets for each of the two countries first, and then use these data sets in the inference function to construct the confidence intervals.
canada12 <- subset(atheism, nationality == "Canada" & year == "2012")
rf12 <- subset(atheism, nationality == "Russian Federation" & year == "2012")
germany12 <- subset(atheism, nationality == "Germany" & year == "2012")inference(canada12$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.0898 ; n = 1002
## Check conditions: number of successes = 90 ; number of failures = 912
## Standard error = 0.009
## 95 % Confidence interval = ( 0.0721 , 0.1075 )I have no reason to believe the samples in Canada are not random or independent, and we know they are not greater than 10% of the population. The margin of error is + or - 1.8%.
inference(rf12$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.06 ; n = 1000
## Check conditions: number of successes = 60 ; number of failures = 940
## Standard error = 0.0075
## 95 % Confidence interval = ( 0.0453 , 0.0747 )I have no reason to believe the samples from the Russian Federation are not random or independent, and we know they are not greater than 10% of the population. The margin of error is + or - 1.5%.
inference(germany12$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.1494 ; n = 502
## Check conditions: number of successes = 75 ; number of failures = 427
## Standard error = 0.0159
## 95 % Confidence interval = ( 0.1182 , 0.1806 )I have no reason to believe the samples in Canada are not random or independent, and we know they are not greater than 10% of the population. The margin of error is + or - 3.1%.
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=p(1−p)/n‾‾‾‾‾‾‾‾‾‾√. This is then used in the formula for the margin of error for a 95% confidence interval: ME=1.96×SE=1.96×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×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.
The margin of error increases as the population proportion increases from 0 to .5 and then it decreases as the proportion increases to 1.
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≥10 and n(1−p)≥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 p̂ 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 p̂ 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.
## 0.07019 0.09327 0.09904 0.09969 0.10577 0.12981sd(p_hats)## [1] 0.009287382The sampling distribution of sample proportions at n=1040 and p=0.1 has a center at mean = 0.09969 and median at 0.09904. The standard distribution is .009287. The shape appears to be normal.
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 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 p̂ ? How does p affect the sampling distribution?
p <- 0.1
n <- 400
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
}
p <- 0.02
n <- 1040
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
}
p <- 0.02
n <- 400
p_hats4 <- rep(0, 5000)
for(i in 1:5000){
samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
p_hats4[i] <- sum(samp == "atheist")/n
}
par(mfrow = c(2,2))
hist(p_hats2, main = "p = 0.1, n = 400", xlim = c(0, 0.18))
hist(p_hats3, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))
hist(p_hats4, main = "p = 0.02, n = 400", xlim = c(0, 0.18))
With a larger sample size, the distribution seems to be less spread out. P also appears to affect the spread, with a lower p causing a narrower spread.
summary(p_hats2)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.05250 0.09000 0.10000 0.09976 0.11000 0.15500summary(p_hats3)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.005769 0.017308 0.020192 0.019954 0.023077 0.039423summary(p_hats4)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.00000 0.01500 0.02000 0.01988 0.02500 0.04750The summary also appears to show the same thing regarding the spread.
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 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?
It is somewhat misleading, but the report does give a margin of error with a range from 3 to 5%. The plots, though, give a better picture of how the spread varies.
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.
- 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.
# create new datasets
spain12 <- subset(atheism, nationality == "Spain" & year == "2012")
spain05 <- subset(atheism, nationality == "Spain" & year == "2005")inference(spain12$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.09 ; n = 1145
## Check conditions: number of successes = 103 ; number of failures = 1042
## Standard error = 0.0085
## 95 % Confidence interval = ( 0.0734 , 0.1065 )inference(spain05$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 )Null Hypothesis: Spain has not seen a change in its atheism index between 2005 and 2012 Alternative Hypothesis: Spain has seen a change in its atheism index between 2005 and 2012
In 2005, we are 95% confident that the mean proportion of atheists was 9% plus or minus 1.7% In 2012, we are 95% confident that the mean proportion of atheists was 10% plus or minus 1.7%
Those mean proportions are basically the same, so there is insufficient evidence to reject the hypothesis that Spain has not seen a change in its atheism index between 2005 and 2012.
- Is there convincing evidence that the United States has seen a change in its atheism index between 2005 and 2012?
# create new datasets
us12 <- subset(atheism, nationality == "United States" & year == "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 )Null Hypothesis: The United States has not seen a change in its atheism index between 2005 and 2012 Alternative Hypothesis: The United States has seen a change in its atheism index between 2005 and 2012
In 2005, we are 95% confident that the mean proportion of atheists was 1% plus or minus .7% In 2012, we are 95% confident that the mean proportion of atheists was 5% plus or minus 1.4%
Those mean proportions are different, so there is sufficient evidence to reject the hypothesis that the United States has not seen a change in its atheism index between 2005 and 2012.
- 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.
With a 95% confidence interval, this type of error happens about 5% of the time. 39 countries are in table 4, and 1.95 is 5% of 39. Thus, we’d expect to detect a change simply by chance in about 2 countries.
- 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.
As p increases to 0.5 percent, the margin of error increases. As it increases from 0.5 percent to 1, the margin of error decreases. In this case, a guessed p tilde of 0.5 will be conservative. So solve for n with the following equation: *desired SE (.01) = sqrt((.5**.5)/n + 4). n will be the desired sample size.*