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:

http://www.wingia.com/web/files/richeditor/filemanager/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?

A: Percentages reported appear to be calculated from actual population parameters.

  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?

A: Since the poll is based on a simple random sample and consists of less than 10% of the world’s population, it satisfies condition of independence.

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?

A: Each row in the table corresponds to data about individual’s beliefs if they consider themselves as a religious person or not. In this case, rows with value of atheist in response column can be considered as success(1) and rows with other values as failure(0).

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?

A: Atheist proportion from the us12 dataframe = 50/1002 = 0.05. Atheist proportion from Table 6 = 0.05. Calculation from dataframe and table match.

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.

  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?

A: Atheist proportion from the us12 dataframe = 50/1002 = 0.0499. Sample size n = 1002 and point estimate \(\hat{p}\) = 0.0499. Following are conditions

1. Independence condition: The poll is based on a simple random sample and consists of less than 10% of the United States population, which verifies independence.

2. Success-failure condition: Sample size of the population that considers themselves to be atheists, successes = 50 and sample size of the population that considers themselves to be non-atheists failures = 952. Since successes and failures are higher than 10, it meets the condition.

With the conditions met, I am confident that sampling distribution of \(\hat{p}\) is nearly normal.

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”.

  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?

A: From above calculation, Standard error SE = 0.0069, Margin of error for 95% confidence level = \(\pm {1.96*SE}\) = \(\pm\) 0.0135

  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.
# Create dataset for Germany
g12 <- subset(atheism, nationality == "Germany" & year == "2012")

# Create dataset for Hong Kong
hk12 <- subset(atheism, nationality == "Hong Kong" & year == "2012")

A: For Germany, sample size n = 502, which less than 10% of Germany’s population. Satisfies condition of independence. Sample size of the population that considers themselves to be atheists, successes = 75 and sample size of the population that considers themselves to be non-atheists failures = 852. Since successes and failures are higher than 10, it meets success-failure the condition.

With the conditions met, sampling distribution for Germany \(\hat{p}\) is nearly normal.

From below calculation, Standard error SE = 0.0159, Margin of error for 95% confidence level = \(\pm {1.96*SE}\) = \(\pm\) 0.0312 and Confidence interval = ( 0.1182 , 0.1806 )

# Germany
inference(g12$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 )

For Hong Kong, sample size n = 500, which is less than 10% of Hong Kong’s population. Satisfies condition of independence. Sample size of the population that considers themselves to be atheists, successes = 45 and sample size of the population that considers themselves to be non-atheists failures = 912. Since sucesses and failures are higher than 10, it meets success-failure the condition.

With the conditions met, sampling distribution for Hong Kong \(\hat{p}\) is nearly normal.

From below calculation, Standard error SE = 0.0128, Margin of error for 95% confidence level = \(\pm {1.96*SE}\) = \(\pm\) 0.0251 and Confidence interval = ( 0.0649 , 0.1151 )

inference(hk12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.09 ;  n = 500 
## Check conditions: number of successes = 45 ; number of failures = 455 
## Standard error = 0.0128 
## 95 % Confidence interval = ( 0.0649 , 0.1151 )

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.

A: Without knowing data and no information from the prior survey, 50% chance of success is a good guess for events with only two outcomes (success or failure). In our case survey questions being are you female? and are you left-handed?. There are only two possible outcomes for each question, yes as “success” and no as “failure.” Proportion \(\hat{p} = 0.50\), provides with the largest margin of error possible. As the proportion, \(\hat{p}\) moves closer towards 0 or 1, the margin of error decreases. Hence relationship between p and me follows bell curve.

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))

summary(p_hats)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## 0.07019 0.09327 0.09904 0.09969 0.10580 0.12980

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.

A: Histogram of sample proportions, fits under the normal curve. It is unimodal and skew is similar on both sides. The median and mean of the distribution are near identical at 0.1 (round to 1), which is also the population proportion.

  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?

A: Since n is the denominator of p, a greater value of n results in a smaller value of p. Looking at all four histogram, they all fit under the normal curve, this tells n does not have a big impact on the center but affects the spread and shape of the distribution of sampling proportions. As sample size n increases, spread decreases. Sample size and spread are inversely proportional. This explains larger samples generate estimates closer to population proportion p.

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))
summary(p_hats)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## 0.06923 0.09327 0.10000 0.09991 0.10580 0.13560
p <- 0.1
n <- 400
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 = 400", xlim = c(0, 0.18))
summary(p_hats)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.0550  0.0900  0.1000  0.1002  0.1100  0.1600
p <- 0.02
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 = 0.02, n = 1040", xlim = c(0.0001, 0.05))
summary(p_hats)
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
## 0.007692 0.017310 0.020190 0.019940 0.023080 0.038460
p <- 0.02
n <- 400
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 = 0.02, n = 400", xlim = c(0.0001, 0.05))

summary(p_hats)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## 0.00250 0.01500 0.02000 0.02005 0.02500 0.05000

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?

A: Australia matches both conditions. Independence, sample size n of 1039 represents less than 10% of total population of Australia. Proportion value \(\hat{p}\) = 0.1. Success = 104 and Failure = 935, both values are greater than 10. So, Australia meets both conditions, and we can proceed with inference and report margin of errors.

Ecuador matches only one condition. Independence, sample size n of 404 represents less than 10% of total population of Ecuador. Proportion value \(\hat{p}\) = 0.02. Success = 8 and Failure = 396, as success is less than 10 it fails success-failure condition. Ecuador fails to meet conditions, and we cannot proceed with inference and report margin of errors.

par(mfrow = c(1, 1))
# Create dataset for Australia
a12 <- subset(atheism, nationality == "Australia" & year == "2012")
summary(a12)
##       nationality          response        year     
##  Australia  :1039   atheist    :104   Min.   :2012  
##  Afghanistan:   0   non-atheist:935   1st Qu.:2012  
##  Argentina  :   0                     Median :2012  
##  Armenia    :   0                     Mean   :2012  
##  Austria    :   0                     3rd Qu.:2012  
##  Azerbaijan :   0                     Max.   :2012  
##  (Other)    :   0
# Create dataset for Ecuador
e12 <- subset(atheism, nationality == "Ecuador" & year == "2012")
summary(e12)
##       nationality         response        year     
##  Ecuador    :404   atheist    :  8   Min.   :2012  
##  Afghanistan:  0   non-atheist:396   1st Qu.:2012  
##  Argentina  :  0                     Median :2012  
##  Armenia    :  0                     Mean   :2012  
##  Australia  :  0                     3rd Qu.:2012  
##  Austria    :  0                     Max.   :2012  
##  (Other)    :  0

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.

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.

A: We can use sample distribution of the difference of two proportions.

Null Hypothesis: There is no difference in atheism index between years 2005 and 2012.

\(H_{0}: \hat{p}_{2005} = \hat{p}_{2012}\)

Alternative Hypothesis: There is difference in atheism index between years 2005 and 2012.

\(H_{A}: \hat{p}_{2005} = \hat{p}_{2012}\)

Using inference function, \(p-value = 0.3966\) is greater than 0.05, we accept null hypothesis, there is no convincing evidence that atheism index is different between years 2005 and 2012.

# Create dataset for Spain
spain.data <- subset(atheism, atheism$nationality == "Spain")

#As year is stored as int, inference function was returning a message
#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

#Convert year to character values
spain.data$chr_year = as.character(spain.data$year)

inference(spain.data$response, spain.data$chr_year, est = "proportion",type = "ht",null = 0, alternative = "twosided", method = "theoretical", success = "atheist")
## 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

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

A: We can use sample distribution of the difference of two proportions.

Null Hypothesis: There is no difference in atheism index between years 2005 and 2012.

\(H_{0}: \hat{p}_{2005} = \hat{p}_{2012}\)

Alternative Hypothesis: There is difference in atheism index between years 2005 and 2012.

\(H_{A}: \hat{p}_{2005} = \hat{p}_{2012}\)

Using inference function, \(p-value = 0\) is less than 0.05, we reject null hypothesis, there is convincing evidence that atheism index has changed between years 2005 and 2012.

# Create dataset for US
us.data <- subset(atheism, atheism$nationality == "United States")

#As year is stored as int, inference function was returning a message
#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

#Convert year to character values
us.data$chr_year = as.character(us.data$year)

inference(us.data$response, us.data$chr_year, est = "proportion",type = "ht",null = 0, alternative = "twosided", method = "theoretical", success = "atheist")
## 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

A: A type 1 error is rejecting the null hypothesis when \(H_{0}\) is actually true. We typically do not want to incorrectly reject \(H_{0}\) more than 5% of the time. This corresponds to a significance level of 0.05. In other words, the decision is incorrectly made 5% of the time even though \(H_{0}\) seems correct. Since there are 39 countries in Table 4, if summary of the survey results from 2005 to 2012 is incorrect by 5% there will at least 0.05 * 39 = 2 countries.

A: Assuming we do not any prior information or any earlier results to compare, \(\hat{p} = 0.5\). At 95% confidence level z-value = 1.96. Margin of error(ME) = \(z*\sqrt{p(p-1)/n}\)

Sample population size would be at least 9604.

p <- 0.5
z <- 1.96
me <- 0.01

n = ((z^2)*p*(1-p))/(me^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.