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.
JR Answer: These are sample statistics. They are not population parameters as the number of people are very samll relative to the population size.
JR Answer: Well, the title infers a global view, therefore the study would need to be need to be preferable to cover multiple locations to represent a good population sample
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?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.
JR Answer: Each row is the response from the country. The atheism represents the sample from that country.
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
## United States:1002 atheist : 50 Min. :2012
## Afghanistan : 0 non-atheist:952 1st Qu.:2012
## Argentina : 0 Median :2012
## Armenia : 0 Mean :2012
## Australia : 0 3rd Qu.:2012
## Austria : 0 Max. :2012
## (Other) : 0
## [1] 50
## [1] 1002
## [1] 0.0499002
JR Answer: The answer is just a rounding percentage of the actual calc.
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.
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”.
JR Answer:
Random, normal and independent.
## [1] 0.0135
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.Japan12 <- subset(atheism, nationality == "Japan" & year == "2012")
JI <-inference(Japan12$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.3069 ; n = 1212
## Check conditions: number of successes = 372 ; number of failures = 840
## Standard error = 0.0132
## 95 % Confidence interval = ( 0.281 , 0.3329 )
J1_lbound <- 0.0281
J1_ubound <- 0.3329
J1_d <- J1_ubound - J1_lbound
J1_emargin<- J1_d/2
J1_emargin## [1] 0.1524
Russia12 <- subset(atheism, nationality == "Russian Federation" & year == "2012")
rI<-inference(Russia12$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.06 ; n = 1000
## Check conditions: number of successes = 60 ; number of failures = 940
## Standard error = 0.0075
## 95 % Confidence interval = ( 0.0453 , 0.0747 )
r1_lbound <- 0.0453
r1_ubound <- 0.0747
r1_d <- r1_ubound - r1_lbound
r1_emargin<- r1_d/2
r1_emargin## [1] 0.0147
JR Answer:
Japan lower/upper confidence level = ( 0.281 , 0.3329 )
Russin lower/upper confidence level = ( 0.0453 , 0.0747 )
JR Answer: The Margin of Error goes higher with a higher proportion.
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.JR Answer: When p increases from 0 to 0.5, margin of error also increases to a maximum ~ 0.030. With p goes from 0.5 - 1, margin of error decreases back to zero. ## 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.
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
JR Answer:
IT has a normal distribution. the median is .1 and mean of .10 (center). The spread is .025. It is bimodal.
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?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.
p0 <- 0.10
n0 <- 1040
p1 <- 0.10
n1 <- 400
p2 <-.20
n2 <-1040
p3 <-.02
n3 <-400
p0_hats <- rep(0, 5000)
p1_hats <- rep(0, 5000)
p2_hats <- rep(0, 5000)
p3_hats <- rep(0, 5000)
for(i in 1:5000){
samp0 <- sample(c("atheist", "non_atheist"), n0, replace = TRUE, prob = c(p0, 1-p0))
p0_hats[i] <- sum(samp0 == "atheist")/n
}
for(i in 1:5000){
samp1 <- sample(c("atheist", "non_atheist"), n1, replace = TRUE, prob = c(p1, 1-p1))
p1_hats[i] <- sum(samp1 == "atheist")/n
}
for(i in 1:5000){
samp2 <- sample(c("atheist", "non_atheist"), n2, replace = TRUE, prob = c(p2, 1-p2))
p2_hats[i] <- sum(samp2 == "atheist")/n
}
for(i in 1:5000){
samp3 <- sample(c("atheist", "non_atheist"), n3, replace = TRUE, prob = c(p3, 1-p3))
p3_hats[i] <- sum(samp3 == "atheist")/n
}
par(mfrow = c(2, 2))
hist(p0_hats, main = "p = 0.1, n = 1040", xlim = c(0, 0.25))
hist(p1_hats, main = "p = 0.1, n = 400", xlim = c(0, 0.25))
hist(p2_hats, main = "p = 0.2, n = 1040", xlim = c(0, 0.25))
hist(p3_hats, main = "p = 0.2, n = 400", xlim = c(0, 0.25))JR Answer: They all have a normal distribution. p changes cause a shifts the location of the curce center to the p value. n affects will move the curve spread whereby the greater the n, the larger the spread.
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?
No. Ecuador sample size it too small under 10 percent, so we would want to increase the sample size of Ecuador.
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.
Spain5 <- subset(atheism, nationality == "Spain" & year == "2005")
SI2 <-inference(Spain5$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.1003 ; n = 1146
## Check conditions: number of successes = 115 ; number of failures = 1031
## Standard error = 0.0089
## 95 % Confidence interval = ( 0.083 , 0.1177 )
## [1] 0.01735
Spain12 <- subset(atheism, nationality == "Spain" & year == "2012")
SI <-inference(Spain12$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 )
S12_lbound <- 0.0734
S12_ubound <- 0.1065
S12_d <- S12_ubound - S12_lbound
S12_emargin<- S12_d/2
S12_emargin## [1] 0.01655
JR Answer: Spain 2005 conf. interval is 0.083 Lbound and 0.1177 Ubound Spain 2012 conf. interval is 0.0734 Lbound and 0.1065 Ubound I belive there is overlap between 0.083 and 0.1065 from conf. interval no.
**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")
USI5 <-inference(US5$response, est = "proportion", type = "ci", method = "theoretical",
success = "atheist")## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.01 ; n = 1002
## Check conditions: number of successes = 10 ; number of failures = 992
## Standard error = 0.0031
## 95 % Confidence interval = ( 0.0038 , 0.0161 )
US5_lbound <- 0.0038
US5_ubound <- 0.0161
US5_d <- US5_ubound - US5_lbound
US5_emargin<- US5_d/2
US5_emargin## [1] 0.00615
US12 <- subset(atheism, nationality == "United States" & year == "2012")
USI12 <-inference(US12$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 )
US12_lbound <- 0.0364
US12_ubound <- 0.0634
US12_d <- US12_ubound - US12_lbound
US12_emargin<- US12_d/2
US12_emargin## [1] 0.0135
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.
JR Answer:
Type 1 error is rejecting the null hypothosis when H0 is true. H0 being that countries with no change are not likely to have significant level of changes
JR answer: The p would have to be .10