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?

    They 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?

    To generalize the report we must assume the sampling method is random and independent and the sample is sufficiently large; in this case it is a reasonable assumption to make.

load("more/atheism.RData")

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

    Each row of table six represents a country. Each row in our dataframe, ‘atheism’, represents an individual participant.

  2. 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")
propAtheist<-sum(us12$response == 'atheist')/nrow(us12)
cat("The proportion of American atheists in our dataset is",round(propAtheist*100,0),"%, which confirms the total of confirmed atheists in Table 6.")
## The proportion of American atheists in our dataset is 5 %, which confirms the total of confirmed atheists in Table 6.

  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?

    The sample is randomly selected.
    Each individual is an independent ‘trial’ (atheist vs non-atheist).
    To satisfy success-failure criterion:

#US, p & n from Table 6
ath <- .05*1002
nath <- (1-.05)*1002
cat("Since",ath,"&",nath,"are sufficiently large, the success-failure criterion is met.")
## Since 50.1 & 951.9 are sufficiently large, the success-failure criterion is met.
inference(us12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Warning: package 'openintro' was built under R version 3.4.4
## Warning: package 'BHH2' was built under R version 3.4.4
## 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 )
  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?
#for confidence interval 95%, z-score is 1.96; SE is .0069 from above output

ME <- 1.96*.0069
cat("The margin of error is",ME)
## The margin of error is 0.013524
  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.
#Argentina and Australia, subsets
Arg12 <- subset(atheism, nationality == "Argentina" & year == "2012")
Aus12 <- subset(atheism, nationality == "Australia" & year == "2012")

Argentina

propArgAtheist<-sum(Arg12$response == 'atheist')/nrow(Arg12)
cat("The proportion of Argentinian atheists in our dataset is",round(propArgAtheist*100,0),"%, which confirms the total of confirmed atheists in Table 6.")
## The proportion of Argentinian atheists in our dataset is 7 %, which confirms the total of confirmed atheists in Table 6.
#Argentina, p & n from Table 6
ath <- .07*1002
nath <- (1-.07)*1002
cat("Argentina: since",ath,"&",nath,"are sufficiently large, the success-failure criterion is met.")
## Argentina: since 70.14 & 931.86 are sufficiently large, the success-failure criterion is met.
#inference, Argentina
inference(Arg12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.0706 ;  n = 991 
## Check conditions: number of successes = 70 ; number of failures = 921 
## Standard error = 0.0081 
## 95 % Confidence interval = ( 0.0547 , 0.0866 )
#Argentina: for confidence interval 95%, z-score is 1.96; SE is .0081 from above output

MEarg <- 1.96*.0081
cat("Argentina: the margin of error is",MEarg)
## Argentina: the margin of error is 0.015876

AUSTRALIA

#Australia
propAusAtheist<-sum(Aus12$response == 'atheist')/nrow(Aus12)
cat("The proportion of Australian atheists in our dataset is",round(propAusAtheist*100,0),"%, which confirms the total of confirmed atheists in Table 6.")
## The proportion of Australian atheists in our dataset is 10 %, which confirms the total of confirmed atheists in Table 6.
#Australia, p from text & n from Table 6
ath <- .10*1040
nath <- (1-.10)*1040
cat("Australia: since",ath,"&",nath,"are sufficiently large, the success-failure criterion is met.")
## Australia: since 104 & 936 are sufficiently large, the success-failure criterion is met.
#inference, Australia
inference(Aus12$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 )
#Australia: for confidence interval 95%, z-score is 1.96; SE is .0093 from above output

MEaus <- 1.96*.0093
cat("Australia: the margin of error is",MEaus)
## Australia: the margin of error is 0.018228
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.

We see that me is at a maximum when p = .5; as p varies to 1 or 0, me decreases.

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))

  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.
summary(p_hats)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## 0.07019 0.09327 0.09904 0.09969 0.10577 0.12981
IQR(p_hats)
## [1] 0.0125

The distribution is normal, unimodal, centered at 0.1, with mean and median approximately equal (.09969 and .09904 respectively), and IQR 0.0125.

  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_hats400.1 <- 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
}


p <- 0.02
n <- 1040
p_hats1040.02 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats1040.02[i] <- sum(samp == "atheist")/n
}


p <- 0.02
n <- 400
p_hats400.02 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats400.02[i] <- sum(samp == "atheist")/n
}

par(mfrow = c(2, 2))

hist(p_hats, main = "p = 0.1, n = 1040", xlim = c(0, 0.18))
hist(p_hats400.1, main = "p = 0.1, n = 400", xlim = c(0, 0.18))
hist(p_hats1040.02, main = "p = 0.02, n = 1040", xlim = c(0, 0.18))
hist(p_hats400.02, main = "p = 0.02, n = 400", xlim = c(0, 0.18))

As expected, an increase in sample size constricts spread, while a decrease in probability simultaneously shifts the center toward 0 and constricts the spread of the distribution.

par(mfrow = c(1, 1))

  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?

    Australia satifies the conditions of inference, as show above, and repeated below.

#Australia, p & n from Table 6
ath <- .10*1039
nath <- (1-.10)*1039
cat("Australia: since",ath,"&",nath,"are sufficiently large, the success-failure criterion is met.")
## Australia: since 103.9 & 935.1 are sufficiently large, the success-failure criterion is met.

Checking Ecuador yields:

#Ecuador, p & n from Table 6
ath <- .02*400
nath <- (1-.02)*400
cat("Since",ath,"&",nath,"are not sufficiently large, the success-failure criterion is not met for Ecuador.")
## Since 8 & 392 are not sufficiently large, the success-failure criterion is not met for Ecuador.

Since the value of 8 calculated for Ecuador is close to 10, we can allow a generalization for the population at large with a responsibly delivered caveat; otherwise inference is not sensible and more data need to be collected.

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.

    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.

\(H_0\): no convincing evidence that difference exists in the atheism indexes
\(H_A\): there is convincing evidence that difference exists in the atheism indexes

#Spain, subsets for 2005 and 2012
Esp05 <- subset(atheism, nationality == "Spain" & year == 2005)
Esp12 <- subset(atheism, nationality == "Spain" & year == 2012)

propEsp05Atheist<-sum(Esp05$response == 'atheist')/nrow(Esp05)
propEsp12Atheist<-sum(Esp12$response == 'atheist')/nrow(Esp12)

cat("The proportion of Spanish atheists in our dataset by year are 2005:",round(propEsp05Atheist*100,0),"% and 2012:",round(propEsp12Atheist*100,0),"%, which approximates the total of confirmed atheists in Table 6.\n\n")
## The proportion of Spanish atheists in our dataset by year are 2005: 10 % and 2012: 9 %, which approximates the total of confirmed atheists in Table 6.
#Spain, p & n from Tables in article
esp05.ath <- .1*1146
esp05.nath <- (1-.1)*1146
cat("Since",esp05.ath,"&",esp05.nath,"are sufficiently large, the success-failure criterion is met for Spain in 2005.\n\n")
## Since 114.6 & 1031.4 are sufficiently large, the success-failure criterion is met for Spain in 2005.
esp12.ath <- .09*1146
esp12.nath <- (1-.09)*1146
cat("Since",esp12.ath,"&",esp12.nath,"are sufficiently large, the success-failure criterion is met for Spain in 2012.\n\n")
## Since 103.14 & 1042.86 are sufficiently large, the success-failure criterion is met for Spain in 2012.
#inference, Spain 2005
inference(Esp05$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, Spain 2012
inference(Esp12$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 )

We see that the confidence intervals overlap; therefore, we accept the null hypothesis, and assert that there is no convincing evidence that a change exists in the atheism indexes of Spain between 2005 and 2012.

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

\(H_0\): no convincing evidence that difference exists in the atheism indexes
\(H_A\): there is convincing evidence that difference exists in the atheism indexes

#US, subsets for 2005 and 2012
US05 <- subset(atheism, nationality == "United States" & year == 2005)
US12 <- subset(atheism, nationality == "United States" & year == 2012)

propUS05Atheist<-sum(US05$response == 'atheist')/nrow(US05)
propUS12Atheist<-sum(US12$response == 'atheist')/nrow(US12)

cat("The proportion of American atheists in our dataset by year are 2005:",round(propUS05Atheist*100,0),"% and 2012:",round(propUS12Atheist*100,0),"%, which approximates the total of confirmed atheists in Table 6.\n\n")
## The proportion of American atheists in our dataset by year are 2005: 1 % and 2012: 5 %, which approximates the total of confirmed atheists in Table 6.
#US, p & n from Tables in article
US05.ath <- .01*1002
US05.nath <- (1-.01)*1002
cat("Since",US05.ath,"&",US05.nath,"are sufficiently large, the success-failure criterion is met for America in 2005.\n\n")
## Since 10.02 & 991.98 are sufficiently large, the success-failure criterion is met for America in 2005.
US12.ath <- .05*1002
US12.nath <- (1-.05)*1002
cat("Since",US12.ath,"&",US12.nath,"are sufficiently large, the success-failure criterion is met for America in 2012.\n\n")
## Since 50.1 & 951.9 are sufficiently large, the success-failure criterion is met for America in 2012.
#inference, US 2005
inference(US05$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 )
#inference, US 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 )

We see that the confidence intervals do not overlap; therefore, we reject the null hypothesis, and assert that there is convincing evidence that a change exists in the atheism indexes of the United States between 2005 and 2012.

  • 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.

    With significance level of .05 we would expect to detect a change in \(.05 \times 39 \text{countries} \approx\) `r round(.05 * 39,0)’

  • 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.

\[ ME = z \times SE\implies ME=z \times\sqrt{\bigg(\frac{p(1-p)}{n}\bigg)}\implies n=\frac{p\times(1-p)\times z^2}{ME^2}\]

#ME is at a maximum with p = .5, from above
#z-score for 95% confidence interval, from text
#margin of error set to .01, from text

n <- (.5*.5*(1.96^2))/(.01^2)
cat("We would have to sample",ceiling(n),"people to remain within the guidelines.")
## We would have to sample 9604 people to remain within the guidelines.