Inference for categorical data

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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.

Getting Started

Load packages

In this lab we will explore the data using the dplyr package and visualize it using the ggplot2 package for data visualization. The data can be found in the companion package for this course, statsr.

Let’s load the packages.

library(statsr)
library(dplyr)

## Warning: package 'dplyr' was built under R version 4.0.2

library(ggplot2)

## Warning: package 'ggplot2' was built under R version 4.0.2

The survey

The press release for the poll, conducted by WIN-Gallup International, can be accessed here.

Take a moment to review the report then address the following questions.

1. How many people were interviewed for this survey?

A poll conducted by WIN-Gallup International surveyed 51,000 people from 57 countries.
A poll conducted by WIN-Gallup International surveyed 52,000 people from 57 countries.
A poll conducted by WIN-Gallup International surveyed 51,917 people from 57 countries.
A poll conducted by WIN-Gallup International surveyed 51,927 people from 57 countries. THIS

##2. Which of the following methods were used to gather information?**

Face to face
Telephone
Internet
All of the above THIS

##3. True / False: In the first paragraph, several key findings are reported. These percentages appear to be -> sample statistics**.

True THIS
False

4. True / False:The title of the report is “Global Index of Religiosity and Atheism”. To generalize the report’s findings to the global human population, We must assume that the sample was a random sample from the entire population in order to be able to generalize the results to the global human population. This seems to be a reasonable assumption.

True THIS
False THIS IS the correct, meu unico errro

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.

data(atheism)

atheism

## # A tibble: 88,032 x 3
##    nationality response     year
##    <fct>       <fct>       <int>
##  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
##  7 Afghanistan non-atheist  2012
##  8 Afghanistan non-atheist  2012
##  9 Afghanistan non-atheist  2012
## 10 Afghanistan non-atheist  2012
## # ... with 88,022 more rows

5. What does each row of Table 6 correspond to?

Countries THIS
Individual Persons
Religions

6. What does each row of atheism correspond to?

Countries
Individual Persons THIS
Religions

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.

Create a new dataframe called us12 that contains only the rows in atheism associated with respondents to the 2012 survey from the United States:

us12 <- atheism %>%
  filter(nationality == "United States" , atheism$year == "2012")

7. Next, calculate the proportion of atheist responses in the United States in 2012, i.e. in us12. True / False: This percentage agrees with the percentage in Table~6.

True THIS
False

# type your code for Question 7 here, and Knit
us12

## # A tibble: 1,002 x 3
##    nationality   response     year
##    <fct>         <fct>       <int>
##  1 United States non-atheist  2012
##  2 United States non-atheist  2012
##  3 United States non-atheist  2012
##  4 United States non-atheist  2012
##  5 United States non-atheist  2012
##  6 United States non-atheist  2012
##  7 United States non-atheist  2012
##  8 United States non-atheist  2012
##  9 United States non-atheist  2012
## 10 United States non-atheist  2012
## # ... with 992 more rows

summary(us12$response)

##     atheist non-atheist 
##          50         952

n = 1002 casos igual - lá na tabela 6 diz que tem 90% nao ateista(60+30) talvez 95%, se contar 5% de missing e 5% ateísta.

50 e 592 apareceu agora

n <- 1002
50/n

## [1] 0.0499002

igual resposta questão 7 - TRUE

Inference on proportions

As was hinted earlier, Table 6 provides sample statistics, that is, calculations made from the sample of 51,927 people. What we’d like, though, is insight into the population 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 lab: the confidence interval and the hypothesis test.

Exercise: 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?

INDEPEDENCE E SIMPLE SIZE/SKEW (i2)

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(y = response, data = us12, statistic = "proportion", type = "ci", method = "theoretical", success = "atheist")

## Single categorical variable, success: atheist
## n = 1002, p-hat = 0.0499
## 95% CI: (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 ofatheist`.

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.”

Exercise: Imagine that, after reading a front page story about the latest public opinion poll, a family member asks you, “What is a margin of error?” In one sentence, and ignoring the mechanics behind the calculation, how would you respond in a way that conveys the general concept?

A) É aquilo que permite dar confiança a esse intervalo. Essa margem dá 95% (nesse caso) de que a % de americanos ateístas deve ser entre 3.6% e 6.3% (It is what gives confidence to that interval. This margin gives 95% (in this case) that the% of American atheists must be between 3.6% and 6.3%)

QUESTION 8. 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?

The margin of error for the estimate of the proportion of atheists in the US in 2012 is 0.05.
The margin of error for the estimate of the proportion of atheists in the US in 2012 is 0.025.
The margin of error for the estimate of the proportion of atheists in the US in 2012 is 0.0135. THIS

# type your code for Question 8 here, and Knit
n <- 1002
p <- 0.05
z <- 1.96
me <- z* sqrt((p*(1-p))/n)
me

## [1] 0.01349488

Exercise: 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.

str(atheism)

## tibble [88,032 x 3] (S3: tbl_df/tbl/data.frame)
##  $ nationality: Factor w/ 57 levels "Afghanistan",..: 1 1 1 1 1 1 1 1 1 1 ...
##  $ response   : Factor w/ 2 levels "atheist","non-atheist": 2 2 2 2 2 2 2 2 2 2 ...
##  $ year       : int [1:88032] 2012 2012 2012 2012 2012 2012 2012 2012 2012 2012 ...

#Brasil

# type your code for the part 1 Exercise here, and Knit
bra12 <- atheism %>%
  filter(nationality == "Brazil" , atheism$year == "2012")
summary(bra12$response)

##     atheist non-atheist 
##          20        1982

inference(y = response, data = bra12, statistic = "proportion", type = "ci", method = "theoretical", success = "atheist")

## Single categorical variable, success: atheist
## n = 2002, p-hat = 0.01
## 95% CI: (0.0056 , 0.0143)

#brazil margin of error

n_bra <- 2002
p_bra <- 20/n_bra
me_bra <- z* sqrt((p_bra*(1-p_bra))/n_bra)
me_bra

## [1] 0.00435639

#CAMEROON

cmr12 <- atheism %>%
  filter(nationality == "Cameroon" , atheism$year == "2012")
summary(cmr12$response)

##     atheist non-atheist 
##          15         489

inference(y = response, data = cmr12, statistic = "proportion", type = "ci", method = "theoretical", success = "atheist")

## Single categorical variable, success: atheist
## n = 504, p-hat = 0.0298
## 95% CI: (0.0149 , 0.0446)

##Cameroon Margin of error

n_cmr <- 504
p_cmr <- 15/n_bra
me_cmr <- z* sqrt((p_cmr*(1-p_cmr))/n_cmr)
me_cmr

## [1] 0.007528723

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 (\(ME = 1.96 \times SE\)). Lastly, we plot the two vectors against each other to reveal their relationship.

d <- data.frame(p <- seq(0, 1, 0.01))
n <- 1000
d <- d %>%
  mutate(me = 1.96*sqrt(p*(1 - p)/n))
ggplot(d, aes(x = p, y = me)) +
  geom_line()

9. Which of the following is false about the relationship between \(p\) and \(ME\).

The \(ME\) reaches a minimum at \(p = 0\).
The \(ME\) reaches a minimum at \(p = 1\).
The \(ME\) is maximized when \(p = 0.5\).
The most conservative estimate when calculating a confidence interval occurs when \(p\) is set to 1. THIS

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.

10. True / False: There is 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. Then use their responses as the first input on the inference, and use year as the grouping variable.

True
False THIS

# type your code for Question 10 here, and Knit
spain <- atheism %>%
  filter(nationality == "Spain")

summary(spain$response)

##     atheist non-atheist 
##         218        2073

prop.table(table(spain$response,spain$year),2)

##              
##                     2005       2012
##   atheist     0.10034904 0.08995633
##   non-atheist 0.89965096 0.91004367

#H0 diz que p05 = p12, ou p05 - p12 = 0 e HA diz que é diferente

spa12 <- atheism %>%
  filter(nationality == "Spain" , atheism$year == "2012")
spa12

## # A tibble: 1,145 x 3
##    nationality response     year
##    <fct>       <fct>       <int>
##  1 Spain       non-atheist  2012
##  2 Spain       non-atheist  2012
##  3 Spain       non-atheist  2012
##  4 Spain       non-atheist  2012
##  5 Spain       non-atheist  2012
##  6 Spain       non-atheist  2012
##  7 Spain       non-atheist  2012
##  8 Spain       non-atheist  2012
##  9 Spain       non-atheist  2012
## 10 Spain       non-atheist  2012
## # ... with 1,135 more rows

n2012 <- 1145
n2012

## [1] 1145

spa05 <- atheism %>%
  filter(nationality == "Spain" , atheism$year == "2005")
spa05

## # A tibble: 1,146 x 3
##    nationality response     year
##    <fct>       <fct>       <int>
##  1 Spain       non-atheist  2005
##  2 Spain       non-atheist  2005
##  3 Spain       non-atheist  2005
##  4 Spain       non-atheist  2005
##  5 Spain       non-atheist  2005
##  6 Spain       non-atheist  2005
##  7 Spain       non-atheist  2005
##  8 Spain       non-atheist  2005
##  9 Spain       non-atheist  2005
## 10 Spain       non-atheist  2005
## # ... with 1,136 more rows

n2005 <- 1146
n2005

## [1] 1146

p05 <- 0.10
p12 <- 0.09
cr1 <- p05 - p12
cr2 <- z * sqrt((p05*(1 - p05))/(n2005) + (p12*(1 - p12))/(n2012))
interv1 <- cr1 + cr2
interv2 <- cr1 - cr2
interv1

## [1] 0.03400999

interv2

## [1] -0.01400999

#o valor null = 0 consta no intervalo, por isso coloquei **false**

11. True / False: There is convincing evidence that the United States has seen a change in its atheism index between 2005 and 2012.

True THIS
False

# type your code for Question 11 here, and Knit
#H0 diz que pus05 = pus12 ou pus05-pus12 = 0 e HA diz que é diferente
us <- atheism %>%
  filter(nationality == "United States")
summary(us$response)

##     atheist non-atheist 
##          60        1944

prop.table(table(us$response,us$year),2)

##              
##                     2005       2012
##   atheist     0.00998004 0.04990020
##   non-atheist 0.99001996 0.95009980

pus05 <- 0.001
pus12 <- 0.05
nus2012 <- 1002

summary(us12$response)

##     atheist non-atheist 
##          50         952

us05 <- atheism %>%
  filter(nationality == "United States" , atheism$year == "2005")
us05

## # A tibble: 1,002 x 3
##    nationality   response     year
##    <fct>         <fct>       <int>
##  1 United States non-atheist  2005
##  2 United States non-atheist  2005
##  3 United States non-atheist  2005
##  4 United States non-atheist  2005
##  5 United States non-atheist  2005
##  6 United States non-atheist  2005
##  7 United States non-atheist  2005
##  8 United States non-atheist  2005
##  9 United States non-atheist  2005
## 10 United States non-atheist  2005
## # ... with 992 more rows

nus2005 <- 1002

summary(us05$response)

##     atheist non-atheist 
##          10         992

crus1 <- pus05 - pus12
crus2 <- z * sqrt((pus05*(1 - pus05))/(nus2005) + (pus12*(1 - pus12))/(nus2012))
intervus1 <- crus1 + crus2
intervus2 <- crus1 - crus2
intervus1

## [1] -0.03536395

intervus2

## [1] -0.06263605

#o valor null = 0 não consta no intervalo, por isso coloquei **true**

ANTES DA QUESTÃO DOZE, ABAIXO OUTRAS FORMAS DE SOLUCIONAR AS QUESTÕES 10 ESPANHA E 11 EUA (QUE CHEGARAM NO MESMO RESULTADO QUE EU, MAS DE OUTRO JEITO)

#OUTRO JEITO DE RESPONDER A 10(SPAIN) RESPOSTA-> FALSE
inference(y = response, x = as.factor(year), data = spain, statistic = 'proportion', type = 'ht', null = 0, alternative = 'twosided', success = 'atheist', method = 'theoretical')

## Response variable: categorical (2 levels, success: atheist)
## Explanatory variable: categorical (2 levels) 
## n_2005 = 1146, p_hat_2005 = 0.1003
## n_2012 = 1145, p_hat_2012 = 0.09
## H0: p_2005 =  p_2012
## HA: p_2005 != p_2012
## z = 0.8476
## p_value = 0.3966

#OUTRO JEITO DE RESPONDER A 10(UNITED STATES) - resposta TRUE
inference(y = response, x = as.factor(year), data = us, statistic = 'proportion', type = 'ht', null = 0, alternative = 'twosided', success = 'atheist', method = 'theoretical')

## Response variable: categorical (2 levels, success: atheist)
## Explanatory variable: categorical (2 levels) 
## n_2005 = 1002, p_hat_2005 = 0.01
## n_2012 = 1002, p_hat_2012 = 0.0499
## H0: p_2005 =  p_2012
## HA: p_2005 != p_2012
## z = -5.2431
## p_value = < 0.0001

12. 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: Type 1 error.

0
1
1.95 t
5

# type your code for Question 12 here, and Knit
# são 39 países na table 4 do relatório
39*0.05

## [1] 1.95

13. 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.

2401 people
At least 2401 people
9604 people
At least 9604 people this

# type your code for Question 13 here, and Knit
p_maxME <- 0.5
me <- 0.01
n_min = (p_maxME*(1-p_maxME))/(me^2/1.96^2)
n_min

## [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.