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:

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?

    These are sample statistics.

  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?

    The sample should be stratified and random, though the basis of stratification could certainly be arguable.

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.

## [1] 88032     3

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

    Each row of Table 6 corresponds to a country.

    One would assume that each row in atheism refers to a respondant, though it is not labelled. More confusing is that there are 88,032 rows and according to the statement of sample size on page 7 and the summary on page 15 there were 51,927 respondants. “88,032” does not appear anywhere in the document. I checked the tail, too, to make sure that there weren’t footnotes included in the data.

    Further examination shows that the data set includes both the 2012 and 2005 results, leading to the larger sample size (though technically this is two samples). The 2005 study appears to have had 36,105 respondants.

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?
##         nationality    response year
## 49926 United States non-atheist 2012
## 49927 United States non-atheist 2012
## 49928 United States non-atheist 2012
## 49929 United States non-atheist 2012
## 49930 United States non-atheist 2012
## 49931 United States non-atheist 2012
## # A tibble: 3 x 3
##   response    no_rows    pct
##   <fct>         <int>  <dbl>
## 1 atheist          50 0.0499
## 2 non-atheist     952 0.950 
## 3 Total          1002 1
It looks like the US 2012 sample was 5% atheist and 95% theist. Yes, it agrees with the table (though I always get nervous when there's an "if not, why?" when I don't get that response).

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?

    All responses are independent. There are more than 30 respondants. It is unlikely that we would not get a normal curve with tens of thousands of respondants.

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.

## 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?

    0.0135

  2. 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.

## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.3 ;  n = 1000 
## Check conditions: number of successes = 300 ; number of failures = 700 
## Standard error = 0.0145 
## 95 % Confidence interval = ( 0.2716 , 0.3284 )
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.1395 ;  n = 509 
## Check conditions: number of successes = 71 ; number of failures = 438 
## Standard error = 0.0154 
## 95 % Confidence interval = ( 0.1094 , 0.1696 )
The same assumptions that make the general survey meet the conditions for inference make these countries meet them as well.
Country n ME
Czech Republic 1,000 .0284
Netherlands 509 .0301
US 1,002 .0135

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.

  1. Describe the relationship between p and me.

    At first I thought you were asking for a mathematical relationship, which would be ME2 x n = p - p2, but that seems like a weird thing (and I’m curious as to whether that’s correct).

    It looks like the margin of error grows symetrically to a peak as p approaches 50%, then it declines when we have 100% of the population.

    This doesn’t make sense, though–why would the ME increase as you get more data? How can it be that if you have data for 20% of the sample your ME is the same as if you have 80% of the sample?

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.

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.
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## 0.07019 0.09327 0.09904 0.09969 0.10577 0.12981
## [1] "standard deviation is 0.00928738233415376"
Mean and Median are both around 0.1, as is clear from the graph, ranging from a low of 0.07 to a high of 0.13. It looks to be of a fairly normal shape.
  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?

n appears to narrow the distribution as it grows. As p increases so do the central measurements (mean/median)

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?

    Unfortunately 2% of 400 respondants is 8–not enough to satisfy our (arbitrary but accepted) minimums.

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.
## 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 )
**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.
## 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 )

Hypotheses:

H0 = Atheism levels in Spain did not change between 2005 and 2012.
HA = Atheism levels have changed between those years.

Hypotheses (mathematical):

H0 = paeth05 = paeth12
HA = paeth05 \(\neq\) paeth12

Conditions for inference

Conditions are met. Responses are independent and there are more than 10 success and failures in each sample.

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

CI Esp2005 is (0.0830, 0.1177)
CI Esp2012 is (0.0734, 0.1065)

There is overlap between the CIs from 0.0830 - 0.1065 so no, there is not convincing evidence as the proportions of both years could fall within the same range.

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

Hypotheses:

H0 = Atheism levels in the US did not change between 2005 and 2012.
HA = Atheism levels have changed between those years.

Hypotheses (mathematical):

H0 = paeth05 = paeth12
HA = paeth05 \(\neq\) paeth12

Conditions for inference

Conditions are barely met. Responses are independent but there were only 10 aethists reported in the US in 2005.

## 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 )
CI US2005 is (0.0038, 0.0161)
CI US2012 is (0.0364, 0.0634)

There is no overlap between confidence intervals so I'll move forward with the hypothesis testing.

Pooled proportion is 60 / 2004 = 0.03 (3%)

test statistic = \(\frac{point\ estimate - null\ value}{SE}\)

SE = \(\sqrt{\frac{0.03\ \cdot\ 0.97}{1002} + \frac{0.03\ \cdot\ 0.97}{1002}}\)

Z = \(\frac{(0.0499 - 0.01) - 0}{0.0076}\)

## [1] 5.235344

As was suggested by the CI not overlapping, the z score for the comparison is over 5–easily enough to reject the null hypothesis.

  • 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.

    1

  • 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.

0.01 > 1.96 \(\cdot \sqrt{\frac{p(1-p)}{n}}\)

0.01 > 1.96 \(\cdot \sqrt{\frac{0.5^2}{n}}\)

0.01^2 > 1.96^2 \(\cdot \frac{0.25}{n}\)

n > \(\frac{1.96^2 \cdot 0.25}{0.0001}\)

n > 9,604

## [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.