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.

library(plyr)
library(knitr)

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.

  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?

Ans: These appear to be sample statistics since the numbers are calculated from the sample of 50,000 people, and not directly from the human population of 7.55 billion people.

  1. 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?

Ans: The sample would have to be randomly selected from people across the Earth. They claim that they surveyed 50,000 people from 57 Countries on 5 continents. There are 233 countries in the world according to https://en.wikipedia.org/wiki/List_of_countries_by_population_(United_Nations), but the majority of the currently 7.55 billion people are concentrated in the 13 most populous countries. Cases also need to be independent, so you need to sample less than 10% of the population, which is also true. We also need at least 10 ‘successes’ and ‘failures’, in this case ‘atheists’ and ‘non-atheists’, this condition is also true. It seems reasonable to generalize these data to the global human population.

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")
  1. What does each row of Table 6 correspond to? What does each row of atheism correspond to?

Ans: Each row in table 6 is a country where the survey was conducted. Each row in the ‘atheism’ table is a person who was interviewed for the survey.

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.

  1. 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")

# true = 1, false = 0, the numerator gives total number of trues
percent_ath <- sum(us12[,2] == 'atheist')/length(us12[,2]) #gives percent by dividing by length of the vector.
percent_ath
## [1] 0.0499002

This rounds to 5%, which is consistent with table 6. ## 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.

  1. 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?

Ans: Cases are independent is true, 51,927 is less than 10% the world population. We also have at least 10 ‘successes’ and 10 ‘failures’, that is atheists and non-atheists or visa-versa. The conditions for inference have been met. We assume that the samples are randomly selected.

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

That is: np = 1002 * 0.05 = 50.1 and n(1-p) = 1002 * 0.95 = 951.9 so this condition is 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 \(\pm\) 3-5% at 95% confidence”.

  1. 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?

Ans: Our z value for 95% confidence will be z=1.96

From above we got Standard error: SE = 0.0069

z <- 1.96
SE <- 0.0069
ME <- z * SE
ME
## [1] 0.013524

The margin of error for US is 0.013524.

  1. 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.

Ans:

The two countries are India and USA.

The conditions for the inference for India and USA are met.

The sample size are much less than 10% of the total populations for each country.

The sample sizes are big enough to satisfy the Success_failure condition.

Country 1: India

ind12 <- subset(atheism, nationality == "India" & year == "2012")
pind12athe <- count(ind12$response == 'atheist')
names(pind12athe) <- c("atheist", "total")
pind12athe$percent <- pind12athe$total / sum(pind12athe$total) * 100
kable(pind12athe)
atheist total percent
FALSE 1059 96.978022
TRUE 33 3.021978 
inference(ind12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.0302 ;  n = 1092 
## Check conditions: number of successes = 33 ; number of failures = 1059 
## Standard error = 0.0052 
## 95 % Confidence interval = ( 0.0201 , 0.0404 )

Our z value for 95% confidence will be z=1.96

From above we got Standard error: SE = 0.0052

z <- 1.96
SE <- 0.0052
ME <- z * SE
ME
## [1] 0.010192

The margin of error for India is 0.010192.

Country 2: United States

us12 <- subset(atheism, nationality == "United States" & year == "2012")
pus12athe <- count(us12$response == 'atheist')
names(pus12athe) <- c("atheist", "total")
pus12athe$percent <- pus12athe$total / sum(pus12athe$total) * 100
kable(pus12athe)
atheist total percent
FALSE 952 95.00998
TRUE 50 4.99002 
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 )

Our z value for 95% confidence will be z=1.96

From above we got Standard error: SE = 0.0069

z <- 1.96
SE <- 0.0069 
ME <- z * SE
ME
## [1] 0.013524

The margin of error for United States 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 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")

  1. Describe the relationship between p and me.

Ans: Based on the graph, the proportion of 0.50 is the proportion that will provide the largest margin of error possible.

That is, if we have p = 0.5 and we do p(1 - p) = 0.5 (1 - 0.5) = 0.5 * 0.5 is the maximum value for the numerator, making it the biggest possible value for the ME.

Similar cases for p = 0 and p = 1 will be returning the minimums since 1 * 0 = 0

In other words, the nearer p comes to 0.5 the bigger ME will be and the nearer p becomes to 0 or 1 the lower ME will be.

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.

  1. 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.

Ans:

summary(p_hats)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## 0.07019 0.09327 0.09904 0.09969 0.10577 0.12981
IQR(p_hats)
## [1] 0.0125
sd(p_hats)
## [1] 0.009287382

From this code we can determine two different measures of center and spread, median: 0.09904, IQR: 0.0125 mean: 0.09969, sd: 0.009287382

The shape of the distribution is bell-shaped.

  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?

Ans:

a)n=400 and p=0.1

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
}

b)n=1040 and p=0.02

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
}

c)n=400 and p=0.02

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
}

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_hats2, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))
hist(p_hats3, main = "p = 0.02, n = 400", xlim = c(0, 0.18))

par(mfrow = c(1, 1))

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.

  1. 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?

Ans:

By analyzing the conditions on inference proportions, we have a s follows:

# Australia

n_au <- 1040
p_au <- 0.1

cond_au <- c(n_au * p_au >= 10, n_au * (1 - p_au) >= 10)

cond_au
## [1] TRUE TRUE
# Ecuador

n_ecu <- 400
p_ecu <- 0.02

cond_ecu <- c(n_ecu * p_ecu >= 10, n_ecu * (1 - p_ecu) >= 10)

cond_ecu
## [1] FALSE  TRUE

Based on Australia’s conditions are TRUE for n * p and TRUE n * ( 1 - p).

Based on Ecuador’s conditions are FALSE for n * p and TRUE n * ( 1 - p).

Since one of the Ecuador’s conditions are not met, we might be inclined to reject the results since one of the conditions is not met. However, in my opinion this could be counted as valid result since n_ecu * p_ecu = 8 and the difference with the normal distribution should not be significant different.

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.

Ans:

\[{ H }_{ 0 }:\quad There\quad is\quad no\quad convincing\quad evidence\quad that\quad Spain\quad has\quad seen\quad a\quad change\quad in\quad its\quad atheism\quad index\quad between\quad 2005\quad and\quad 2012\quad (p_{ 2005 }=p_{ 2012 }).\]

\[{ H }_{ A }:\quad There\quad is\quad convincing\quad evidence\quad that\quad Spain\quad has\quad seen\quad a\quad change\quad in\quad its\quad atheism\quad index\quad between\quad 2005\quad and\quad 2012\quad (p_{ 2005 }\neq p_{ 2012 }).\]

2005:

esp05 <- subset(atheism, nationality == "Spain" & year == "2005")
pesp05athe <- count(esp05$response == 'atheist')
names(pesp05athe) <- c("atheist", "total")
pesp05athe$percent <- pesp05athe$total / sum(pesp05athe$total) * 100
kable(pesp05athe)
atheist total percent
FALSE 1031 89.9651
TRUE 115 10.0349

2012:

esp12 <- subset(atheism, nationality == "Spain" & year == "2012")
pesp12athe <- count(esp12$response == 'atheist')
names(pesp12athe) <- c("atheist", "total")
pesp12athe$percent <- pesp12athe$total / sum(pesp12athe$total) * 100
kable(pesp12athe)
atheist total percent
FALSE 1042 91.004367
TRUE 103 8.995633

Inference:

spain <- subset(atheism, nationality == "Spain" & year == "2005"  | nationality == "Spain" & year == "2012")

inference(y = spain$response, x = spain$year, est = "proportion",type = "ht", null = 0, alternative = "twosided", method = "theoretical", success = "atheist")
## Warning: Explanatory variable was numerical, it has been converted to
## categorical. In order to avoid this warning, first convert your explanatory
## variable to a categorical variable using the as.factor() function.
## Response variable: categorical, Explanatory variable: categorical
## Two categorical variables
## Difference between two proportions -- success: atheist
## Summary statistics:
##              x
## y             2005 2012  Sum
##   atheist      115  103  218
##   non-atheist 1031 1042 2073
##   Sum         1146 1145 2291
## Observed difference between proportions (2005-2012) = 0.0104
## 
## H0: p_2005 - p_2012 = 0 
## HA: p_2005 - p_2012 != 0 
## Pooled proportion = 0.0952 
## Check conditions:
##    2005 : number of expected successes = 109 ; number of expected failures = 1037 
##    2012 : number of expected successes = 109 ; number of expected failures = 1036 
## Standard error = 0.012 
## Test statistic: Z =  0.848 
## p-value =  0.3966

Since our returned p-value is more that 0.05; we fail to reject our NULL hypothesis in favor of the alternative HA hypothesis; that is: There is No convincing evidence that Spain has seen a change in its atheism index between 2005 and 2012.

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

Ans:

\[{ H }_{ 0 }:\quad There\quad is\quad no\quad convincing\quad evidence\quad that\quad United \quad States\quad has\quad seen\quad a\quad change\quad in\quad its\quad atheism\quad index\quad between\quad 2005\quad and\quad 2012\quad (p_{ 2005 }=p_{ 2012 }).\]

\[{ H }_{ A }:\quad There\quad is\quad convincing\quad evidence\quad that\quad United\quad States\quad has\quad seen\quad a\quad change\quad in\quad its\quad atheism\quad index\quad between\quad 2005\quad and\quad 2012\quad (p_{ 2005 }\neq p_{ 2012 }).\]

2005:

us05 <- subset(atheism, nationality == "United States" & year == "2005")
pus05athe <- count(us05$response == 'atheist')
names(pus05athe) <- c("atheist", "total")
pus05athe$percent <- pus05athe$total / sum(pus05athe$total) * 100
kable(pus05athe)
atheist total percent
FALSE 992 99.001996
TRUE 10 0.998004

2012:

us12 <- subset(atheism, nationality == "United States" & year == "2012")
pus12athe <- count(us12$response == 'atheist')
names(pus12athe) <- c("atheist", "total")
pus12athe$percent <- pus12athe$total / sum(pus12athe$total) * 100
kable(pus12athe)
atheist total percent
FALSE 952 95.00998
TRUE 50 4.99002

Inference:

us <- subset(atheism, nationality == "United States" & year == "2005"  | nationality == "United States" & year == "2012")

inference(y = us$response, x = us$year, est = "proportion",type = "ht", null = 0, alternative = "twosided", method = "theoretical", success = "atheist")
## Warning: Explanatory variable was numerical, it has been converted to
## categorical. In order to avoid this warning, first convert your explanatory
## variable to a categorical variable using the as.factor() function.
## Response variable: categorical, Explanatory variable: categorical
## Two categorical variables
## Difference between two proportions -- success: atheist
## Summary statistics:
##              x
## y             2005 2012  Sum
##   atheist       10   50   60
##   non-atheist  992  952 1944
##   Sum         1002 1002 2004
## Observed difference between proportions (2005-2012) = -0.0399
## 
## H0: p_2005 - p_2012 = 0 
## HA: p_2005 - p_2012 != 0 
## Pooled proportion = 0.0299 
## Check conditions:
##    2005 : number of expected successes = 30 ; number of expected failures = 972 
##    2012 : number of expected successes = 30 ; number of expected failures = 972 
## Standard error = 0.008 
## Test statistic: Z =  -5.243 
## p-value =  0

Since our returned p-value equals 0; we reject our NULL hypothesis in favor of the alternative HA hypothesis; that is: There is convincing evidence that United States has seen a change in its atheism index between 2005 and 2012.

Ans:

A type 1 error is rejecting the null hypothesis when H0 is actually true.

Typically we do not want to incorrectly reject H0 more than 5% of the time; this resumes to a 0.05 significance level.

Since there are 39 countries listed in Table 4; all we need to do is to multiply 0.05 by 39 to estimate how many countries we would expect to detect a change in the atheism index simply by chance.

The result is 1.95, or about 2 countries would be expected to detect a change in atheism just by chance.

Ans:

There are two unknown variables in this question: p and n.

When we do not have and estimate for p we follow the guideline that the margin of error is largest when p is 0.5. As reference we typically use this as the worst case estimate if no other estimate is available. The estimate must have a margin of error no greater than 1%, this means that if use the formula \(ME=zcotSE\) with a 95% confidence interval we obtain \(z=1.96\); hence \(ME=1\cdot 96\ast sqrt(p(1-p)/n));\) since our \(ME\le 0.01\).

\[SE=\sqrt { \frac { p*(1-p) }{ n } }\] \[=>\quad ME=z???SE\] \[=>\quad M=z*\sqrt { \frac { p*(1-p) }{ n } }\]

\[=>\quad\frac { ME }{ z } =\sqrt { \frac { p*(1-p) }{ n } }\] \[=>\quad { \left( \frac { ME }{ z } \right) }^{ 2 }=\frac { p*(1-p) }{ n }\] \[=>\quad n=\frac { p*(1-p) }{ { \left( \frac { ME }{ z } \right) }^{ 2 } }\] \[=>\quad n=\frac { p*(1-p)*{ z }^{ 2 } }{ { \left( ME \right) }^{ 2 } }\]

p <-0.5 # Will generate the maximum ME
z <-1.96 # 95% Confidence interval
ME <- 0.01 # Marging of error no greater than 0.01
n <- p * (1 - p) * z ^ 2  / ME ^ 2
n
## [1] 9604

And based on the results we will need at least 9604 participants to ensure the sample proportion is within 0.01 of the true proportion with 95% confidence.

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.