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.
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.
###Answer
The finding appears to be derived from Survey results of the Population and are Sample statistics.
###Answer
We must assume that the random sample from the entire population and that the observations were independant and that the sample was large enough for Inference.
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.
atheism correspond to?###Answer
Each row in Table 6 corresponds to a country and the sample size taken from that country , the percentage of the sample that reported as a religious person , not a religious person , A convinced Atheist or ones that Don't know or didnt respond to the question.
## 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
As we can see above from the first couple of rows of the `atheism` data set that each row corresponds to a country the year the survey was conducted and whether the person fromt hat country reported to be an atheist or non-atheist.
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.
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")
total_us12_atheist<- nrow(us12_atheist)
totalus12<-nrow(us12)
proportion_us12_atheiest<- total_us12_atheist/totalus12
proportion_us12_atheiest## [1] 0.0499002
As we can see the proportion is 0.0499002 which is around 5% and it is the same as the table 6 because we are using the same data set.
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.
The Conditions for inference are the following :
1. The Sample observations are indepependant.
2. The sample must be large enough that n*p>10 and n*(1-p)>10
We know that the observations are indpenedant. Now let us calculate the second condition below.
## [1] 50
## [1] 952
Looks like the conditions for inference are satisfied. Since both n*p and n*(1-p) are both greater than 10. I am pretty confident that the conditions for inference are met.
Now we will calculate the Standard Error and upper and lower intervals for the Confidence Interval.
## [1] 0.006878629
## [1] 0.03641809 0.06338231
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 3.6.3
## 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”.
Calculating the margin of Error now
margin_of_error_upper<-upper-p
margin_of_error_lower<-p-lower
c(margin_of_error_lower,margin_of_error_upper)## [1] 0.01348211 0.01348211
## [1] 0.02696423
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.The countries I will choose are Brazil and Ghana
Brazil_2012<-subset(atheism, nationality == "Brazil" & year == "2012")
Pakistan_2012<-subset(atheism, nationality == "Pakistan" & year == "2012")
Total_Brazil_2012<-nrow(Brazil_2012)
Total_Pakistan_2012<-nrow(Pakistan_2012)
Total_Atheist_Brazil_2012<-nrow(subset(Brazil_2012,response=="atheist"))
Total_Atheist_Pakistan_2012<-nrow(subset(Pakistan_2012,response=="atheist"))
Total_Brazil_2012## [1] 2002
## [1] 20
## [1] 2704
## [1] 54
Proportion_Brazil<-Total_Atheist_Brazil_2012/Total_Brazil_2012
Proportion_Pakistan<-Total_Atheist_Pakistan_2012/Total_Pakistan_2012
Proportion_Brazil## [1] 0.00999001
## [1] 0.01997041
Now calculating the Confidence Interval for Brazil
inference(Brazil_2012$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.01 ; n = 2002
## Check conditions: number of successes = 20 ; number of failures = 1982
## Standard error = 0.0022
## 95 % Confidence interval = ( 0.0056 , 0.0143 )
Now calculating the Confidence Interval for Pakistan
inference(Pakistan_2012$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.02 ; n = 2704
## Check conditions: number of successes = 54 ; number of failures = 2650
## Standard error = 0.0027
## 95 % Confidence interval = ( 0.0147 , 0.0252 )
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")p and me. The margin of error is dependant on proportion and it is a parabolic relationship we can see as the proportion incrases the margin increases.
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.
mean to calculate summary statistics.## [1] 5000
## [1] 0.09969
## [1] 0.009307152
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.07115 0.09327 0.10000 0.10011 0.10673 0.12404
The mean is 0.10011 the standard deviation is 0.0093072 the shape is nearly normal.
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?get.getSample <- function(p,n){
p_sample <- rep(0, 5000)
for(i in 1:5000){
samp <- sample(c("male", "female"), n, replace = TRUE, prob = c(p, 1-p))
p_sample[i] <- sum(samp == "male")/n
}
return(p_sample)
}
p_sample_1 <- get.getSample(0.1, 1040)
p_sample_2 <- get.getSample(0.1, 400)
p_sample_3 <- get.getSample(0.02, 1040)
p_sample_4 <- get.getSample(0.02, 400)
par(mfrow = c(2, 2))
hist(p_sample_1, main = "p = 0.1, n = 1040", xlim = c(0, 0.18))
hist(p_sample_2, main = "p = 0.1, n = 400", xlim = c(0, 0.18))
hist(p_sample_3, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))
hist(p_sample_4, main = "p = 0.02, n = 400", xlim = c(0, 0.18))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.
###Answer
Yes for Australia and No for Ecuador because for Australia the distribution looks like it is normal and the conditions for inference are satisfied however for ecuador the Conditions for inference does not satisfy because n*p<10
## [1] 104
## [1] 936
## [1] 8
## [1] 392
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.
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.
spain2005 <- subset(atheism, nationality == "Spain" & year == "2005")
spain2012 <- subset(atheism, nationality == "Spain" & year == "2012")
inference(spain2005$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.1003 ; n = 1146
## Check conditions: number of successes = 115 ; number of failures = 1031
## Standard error = 0.0089
## 95 % Confidence interval = ( 0.083 , 0.1177 )
inference(spain2012$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 )
As we can see above for spain that the confidence intervals do overlap so there is not any convincing evidence that the proportion of atheism has changed.
**b.** Is there convincing evidence that the United States has seen a
change in its atheism index between 2005 and 2012?
us2005 <- subset(atheism, nationality == "United States" & year == "2005")
inference(us2005$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.01 ; n = 1002
## Check conditions: number of successes = 10 ; number of failures = 992
## Standard error = 0.0031
## 95 % Confidence interval = ( 0.0038 , 0.0161 )
us2012 <- subset(atheism, nationality == "United States" & year == "2012")
inference(us2012$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 )
Yes based on the above results we can easily say that the proportions have changed as there is no overlap in the confidence intervals.
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.
If there is no change we will be rejecting the null hypotheses and we will detect a change simply by chance at 0.05 significant level.
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.
In order to answer this question we can easily say that the margin of error for a 95% confidence interval is 1.96 x sqrt(p(1-p)/n).
So for the margin of error to be no greater than 1% our calculation for the confidence interval should give us a value <0.01 however here we do not have a value for p and I would atleast sample 5000 people to remain in guidlelines and get the margin of error of 0.01 see below the results if we round we get close to 0.01
Lets assume p=0.5
## [1] 0.01385929