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Student Name : Sachid Deshmukh

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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?

Answer : The key statistics reported in paragraph-1 are sample statistics. Numbers presented are from the sample collected in global poll by Win-Gallup International by interviewing more than 50,000 mean and women selected from 57 countries across the globe in five continents.

  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?

We should assume following conditions for sampling methods

Based on the paragraph, we can say that these are the resonable assumptions. Win-Gallup international collected 50,000 samples which is large enough to derive population parameters correctly. Sample population is representative of actual population since they involved men and women from 57 countries in five continents

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?

Answer - Each row of table 6 corresponds to sample proportions grouped by country for year 2012. Each row of atheism dataset correspond to individual response of the survey conducted in year 2005 and 2012

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")
us12.atheist = subset(us12, response == "atheist")
us12.atheist.prop = nrow(us12.atheist)*100/nrow(us12)
paste("Atheist proportions calculated from atheism data set = ", us12.atheist.prop, "%")
## [1] "Atheist proportions calculated from atheism data set =  4.99001996007984 %"

Answer : - The calculated proportion is 4.99% which matches with percentage (5%) given in table 6. It is given that table has a rounding error of 1%. Keeping that in mind we can conclude that calculated proportion matches with table proportion

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?

Answer : Following are the conditions for inferencing 95% confidence interval for sample proportions

It looks like all the conditions mentioned above satisfies in our case. We can infer 95% confidence interval of the proportions of atheists in United State in 2012

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?

Answer : - Based on R output, margin of error for the estimate of the proportions of the atheists in US in 2012 is 1.96 * SE = 1.96 * 0.0069 = 1.35%

  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.
aus12 <- subset(atheism, nationality == "Australia" & year == "2012")
inference(aus12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.1001 ;  n = 1039 
## Check conditions: number of successes = 104 ; number of failures = 935 
## Standard error = 0.0093 
## 95 % Confidence interval = ( 0.0818 , 0.1183 )

Answer : 95% confidence interval of the proportions of atheists in Australia is 8.1% to 11.8% with margin of error = 1.82%

can12 <- subset(atheism, nationality == "Canada" & year == "2012")
inference(can12$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 )

Answer : 95% confidence interval of the proportions of atheists in Canada is 7.2% to 10.7% with margin of error = 1.76%

In both the cases, all the conditions for samples needed for confidence interval inference seems to be satisfied.

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.

Answer - There is a non linear relationship between proportion and margin of error. It increases quadratically as proportion increases from 0 to 0.5. There after it decreaes quadratically

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.

**Answer : - Sampling distribution of the sample proportions for n = 1040 and p = 0.1 is normally distributed. It has center at 10% which is (1040 * 10 ) / 100 = value of 104. The standard deviation of the distribution (spread) = sqrt(np(1-p)) = sqrt(1040 * 0.1 * 0.9) = 9.67 which is 9.67*100/1040 = 0.9%**

  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?
p <- 0.1
n <- 400
p_hats400 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats400[i] <- sum(samp == "atheist")/n
}

p <- 0.02
n <- 1040
p_hats1040 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats1040[i] <- sum(samp == "atheist")/n
}

p <- 0.02
n <- 400
p_hats400.1 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats400.1[i] <- sum(samp == "atheist")/n
}

par(mfrow = c(2, 2))
hist(p_hats400, main = "p = 0.1, n = 400", xlim = c(0, 0.18))
hist(p_hats1040, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))
hist(p_hats400.1, main = "p = 0.02, n = 400", xlim = c(0, 0.18))
par(mfrow = c(1, 1))

Answer : As sample size increases spread of the sampling distribution decreases. As p increases center of the sampling distribution increases

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?

**Answer : One of the important criteria for inferencing confidence interval is to assume samples are approximately normally distributed. As per text book the rule of thumb is to check for np and n(1-p) and make sure those are greater than 10 to ensure this. For Austrelia ). np = 0.1 * 1040 = 104 and n(1-p) = 936 and for Ecuador np = 0.02 400 = 8 and n*(1-p) = 392 Even though rule doen’t satisfy for Ecudor, we can still go ahead and and proceed to report margin of errors. In this case standard error for Ecudor will be higher compared to Austrelia since sample size is smaller.**

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.

spain5 <- subset(atheism, nationality == "Spain" & year == "2005")
inference(spain5$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 )
spain12 <- subset(atheism, nationality == "Spain" & year == "2012")
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 )

Answer - Based on the inference test for Spain atheism proportion for year 2005 and 2012 we can say that. * Atheism in spain in 2005 was between 8.3% to 11.7% * Atheism in spain in 2012 was between 7.3% to 10.65%

We can see that confidence interval for Spain atheism proportion for year 2005 and 2012 overlap with each other and are not distinctly separated. We can conclude that there is no change in the atheism level for Spain between year 2005 and 2012

**b.** Is there convincing evidence that the United States has seen a
change in its atheism index between 2005 and 2012?
us5 <- subset(atheism, nationality == "United States" & year == "2005")
inference(us5$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 )
us12 <- subset(atheism, nationality == "United States" & year == "2012")
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 )

Answer - Based on the inference test for United States atheism proportion for year 2005 and 2012 we can say that. * Atheism in United States in 2005 was between 0.3% tp 1.6% * Atheism in United States in 2012 was between 3.6% to 6.3%

We can see that confidence interval for United States atheism proportion for year 2005 and 2012 are distinctly separated. We can conclude that there is a change in the atheism level for United States between year 2005 and 2012. We can conclude that atheism proportion is increaed in United States between 2005 and 2012

Answer - Even though there is no change in the atheism index for a given country, we can detect a change 5% of the time at 0.05 significance level. This is Type-1 error also known as false positive. We can expect one in 20 times to observe a change in the atheism proportion for a given country even though there is no change in reality. This is resonable given a significance level of 0.05. At this significance level we expect to see 5% false positive or type-1 error.

Answer - If the margin of error is 0.01, then at 95% confidence SE= 0.01/1.96 =0.0051. Since we do not know the value of p, we assume the worst case scenario with p=0.5. Considering, SE= sqrt(p(1 - p)/n), then n= (p(1 - p))/(SE^2) = (0.5 X 0.5)/(0.0051^2)= 9604. Sample must include at least 9,604 people.

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.