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.

0.1 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.

0.1.1 Exercise 1

  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?
These data were taken from a poll  conducted by WIN-Gallup International surveyed 51,927 people from 57 countries so they are based on sample statistics. It would not be feasible to know the exact population parameters in this case however percentages appear to apply to the population.

0.1.2 Exercise 2

  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?
For reasonable assumption requires following:
1. Independence : The sample observations are independent. 
2. Randomization : The sample size must have been picked through random sampling 
3. Normally Distributed : The sample size must be < 10% of the population 

- It is reasonable to assume independence as the poll is conducted by reputed WiN (WorldWide Independent Network of Market Research) Gallup International.
- We can assume randomization as it includes 57 countries
- In this poll, the survey included within and across countries. There are over 7 billion people and 195 countries in the world today and 10% of the total population is 700,000,000 (700 million) and 10% of countries is 20 countries. The sample consists of 50,000 people (0.07%) across 57 countries (30%) which is not ideal but the sample size is large enough to assume normality

0.2 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")

0.2.1 Exercise 3

  1. What does each row of Table 6 correspond to? What does each row of atheism correspond to?
names(atheism)
## [1] "nationality" "response"    "year"
colnames(atheism)
## [1] "nationality" "response"    "year"
#unique(atheism$nationality)
unique(atheism$response)
## [1] non-atheist atheist    
## Levels: atheist non-atheist
Each row of Table 6 data represents Countries in alphabetical order, Sample Size Unweighted, A religious person, Not a religious person, A convinced atheist and Don't know / no response.
Each row of atheism data represents nationality, response to whether the person is atheist or non-atheist and the year the poll was taken.

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.

0.2.2 Exercise 4

  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")
us12atheist <- subset(atheism, nationality == "United States" & year == "2012" & response == "atheist")
#us12nonatheist <- subset(atheism, nationality == "United States" & year == "2012" & response == "non-atheist")
nrow(us12atheist)/nrow(us12)
## [1] 0.0499002
5% of the responses in the dataset are of the category "atheist".

In Table 6, 5% of the US sample is "a convinced atheist".

Both of these percentages agree.

0.3 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.

0.3.1 Exercise 5

  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?
3 conditions needs to be satisfied
1. Independence
2. Randomization
3. Normally Distributed

- We can safely assume observations are independent. 
- The poll is based on a simple random sample as it consists of fewer than 10% of the U.S. population
- For normal distribution, at least 10 expected successes and failures condition has to be met. The sample size of 1002 is large enough, as a success rate of 5% (atheism rate) is greater than 10. 5% of 1,002 is ~= 50.

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”.

0.3.2 Exercise 6

  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?
# margin of error = z-score (assuming a normal distribution) for 95%
# confidence interval * Standard Error

SE = 0.0069 # Standard Error
Z = qnorm(0.975) # Z score for 95% confidence level, right side = 1.96
ME = SE * Z # Margin of Error
ME
## [1] 0.01352375

0.3.3 Exercise 7

  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.
chn12 <- subset(atheism, nationality == "China" & year == "2012")
inference(chn12$response, est = "proportion", type = "ci", method = "theoretical", 
          success = "atheist")
## Single proportion -- success: atheist 
## Summary statistics:

## p_hat = 0.47 ;  n = 500 
## Check conditions: number of successes = 235 ; number of failures = 265 
## Standard error = 0.0223 
## 95 % Confidence interval = ( 0.4263 , 0.5137 )
pkt12 <- subset(atheism, nationality == "Pakistan" & year == "2012")
inference(pkt12$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 )
Z = qnorm(0.975) # Z score for 95% confidence level, right side = 1.96

#China
SE_china = 0.0223 # Standard Error
ME_china = SE_china * Z # Margin of Error
ME_china
## [1] 0.0437072
#Pakistan
SE_pakistan = 0.0027 # Standard Error
ME_pakistan = SE_pakistan * Z # Margin of Error
ME_pakistan
## [1] 0.005291903
Inference For China :
1. Independence : Observations are independent. 
2. Randomization : The poll is based on a simple random sample and sample size for this study was 500 which is fewer than 10% of the Chinese population confirms random.
3. Normally Distributed : Success-failure condition. The sample size of 500 is large enough, as a success rate of 47% (atheism rate) is greater than 10. 47% of 500 is ~= 235.

Margin of Error = 0.0437072

Inference For Pakistan :
1. Independence : Observations are independent. The poll is based on a simple random sample and consists of fewer than 10% of the Pakistani population, which verifies independence. 
2. Randomization : The sample size for this study was 2705.
3. Normally Distributed : Success-failure condition. The sample size of 2705 is large enough, as a success rate of 2% (atheism rate) is greater than 10. 2% of 2705 is ~= 54.

Margin of Error = 0.005291903

0.4 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")

0.4.1 Exercise 8

  1. Describe the relationship between p and me.
It is parabolic. 
Margin of Error (ME) increases with increasing p from 0 until it reaches a peak maximum when the population proportion p is around 0.5 (50%). Then ME decreases as p increases to 1 (100%).

0.5 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.

0.5.1 Exercise 9

  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
sd(p_hats)
## [1] 0.009287382
boxplot(p_hats,y_lab="p_hats",x_lab="proportions")

The sampling distribution has a near normal distribution with the mean close to the population mean of 0.1. There are a few outliers on both tails but these are small compared to the total sample size.

0.5.2 Exercise 10

  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?
par(mfrow = c(2, 2))

#first histogram
hist(p_hats, main = "p = 0.1, n = 1040", xlim = c(0, 0.25))

p <- 0.1
n <- 400
p_hats2 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats2[i] <- sum(samp == "atheist")/n
}

#second histogram
hist(p_hats2, main = "p = 0.1, n = 400", xlim = c(0, 0.25))

p <- 0.02
n <- 1040
p_hats3 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats3[i] <- sum(samp == "atheist")/n
}

#third histogram
hist(p_hats3, main = "p = 0.02, n = 1040", xlim = c(0, 0.25))

p <- 0.02
n <- 400
p_hats4 <- rep(0, 5000)

for(i in 1:5000){
  samp <- sample(c("atheist", "non_atheist"), n, replace = TRUE, prob = c(p, 1-p))
  p_hats4[i] <- sum(samp == "atheist")/n
}

#fourth histogram
hist(p_hats4, main = "p = 0.02, n = 400", xlim = c(0, 0.25))

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.

0.5.3 Exercise 11

  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
n_au <- 1040
p_au <- 0.1
au <- c(n_au * p_au >= 10, n_au * (1 - p_au) >= 10)
au
## [1] TRUE TRUE
# Ecuador
n_ec <- 400
p_ec <- 0.02
ec <- c(n_ec * p_ec >= 10, n_ec * (1 - p_ec) >= 10)
ec
## [1] FALSE  TRUE
Australia has 0.1×1040=104 successes, which passes.
Ecuador has 0.02×400=8 successes, which does not meet the 10 success threshold.

They both have a normal distribution with almost similar spreads, so their margin of errors should be close. Ecaudor though violates the success-failure condition of 10 successful smaples since 0.02 of 400 is only 8. Hence, Ecuador's sampling distribution appears skewed to the right so we should not proceed with inference for Ecuador.

0.6 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.

0.6.1 Question 1

  • 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?

    #H0: p12 - p05 = 0 #There was no change in the atheism index in Spain from 2005 to 2012 
#HA: p12 - p05 != 0 #There was a change in the atheism index in Spain from 2005 to 2012

spn05 <- subset(atheism, nationality == "Spain" & year == "2005")
atheist_spn05 <- count(spn05, response == "atheist")
non_atheists_spn05 <- count(spn05, response != "atheist")
total_spn05 <- atheist_spn05 + non_atheists_spn05 
atheist_spn05/total_spn05 *100
    ##   response == "atheist"       n
## 1                   NaN 89.9651
## 2                    50 10.0349
    spn12 <- subset(atheism, nationality == "Spain" & year == "2012")
atheist_spn12 <- count(spn12, response == "atheist")
non_atheists_spn12 <- count(spn12, response != "atheist")
total_spn12 <- atheist_spn12 + non_atheists_spn12
atheist_spn12/total_spn12 *100
    ##   response == "atheist"         n
## 1                   NaN 91.004367
## 2                    50  8.995633
    par(mfrow = c(1, 2))
inference(spn05$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(spn12$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 )
    spn_05_12 <- subset(atheism, nationality == "Spain" & year == "2005"  | nationality == "Spain" & year == "2012")
inference(y = spn_05_12$response, x = spn_05_12$year, est = "proportion",type = "ht", null = 0, alternative = "twosided", method = "theoretical", success = "atheist")
    ## Response variable: categorical, Explanatory variable: categorical
## Two categorical variables
## Difference between two proportions -- success: atheist
## Summary statistics:
##              x
## y             2005 2012  Sum
##   atheist      115  103  218
##   non-atheist 1031 1042 2073
##   Sum         1146 1145 2291
    ## Observed difference between proportions (2005-2012) = 0.0104
## 
## H0: p_2005 - p_2012 = 0 
## HA: p_2005 - p_2012 != 0 
## Pooled proportion = 0.0952 
## Check conditions:
##    2005 : number of expected successes = 109 ; number of expected failures = 1037 
##    2012 : number of expected successes = 109 ; number of expected failures = 1036 
## Standard error = 0.012 
## Test statistic: Z =  0.848 
## p-value =  0.3966

    p_spn05 = 0.1003
n_spn05 = 1146 
p_spn12 = 0.09
n_spn12 = 1145 

PE_spn = p_spn12 - p_spn05

SE_spn = sqrt((p_spn05*(1-p_spn05)/n_spn05)+(p_spn12*(1-p_spn12)/n_spn12))
SE_spn
    ## [1] 0.01225854
    #Control interval for difference between proportion in 2005 and 2012
PE_spn + (1.96*SE_spn)
    ## [1] 0.01372674
    PE_spn - (1.96*SE_spn)
    ## [1] -0.03432674
    For inference in Spain :    
1. Independence : Observations are independent. 
2. Randomization : The poll is based on a simple random sample and consists of fewer than 10% of the Spanish population in 2005 and 2012, which verifies independence. The sample size for this study was 1046 in 2005 and 1045 in 2012.
3. Normal Distribution  : Success-failure condition. The sample size of 1046 in 2005 and 1045 in 2012 is large enough, as a success rate of 10% and 9% (atheist rate) in 2005 and 2012 respectively are greater than 10. 

Since the p-value is greater than .05, we can fail to reject the null hypothesis and we conclude there is no convincing evidence that there is a change in the atheism index in Spain from 2005 to 2012 since the confidence interval (95%) overlap. In addition, the control interval showing the difference in their proportion include zero. This means that there is evidence that the true atheism index between this two years are the same.

    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.

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

    usa05 <- subset(atheism, nationality == "United States" & year == "2005")
atheist_usa05 <- count(usa05, response == "atheist")
non_atheists_usa05 <- count(usa05, response != "atheist")
total_usa05 <- atheist_usa05 + non_atheists_usa05 
atheist_usa05/total_usa05 *100
    ##   response == "atheist"         n
## 1                   NaN 99.001996
## 2                    50  0.998004
    usa12 <- subset(atheism, nationality == "United States" & year == "2012")
atheist_usa12 <- count(usa12, response == "atheist")
non_atheists_usa12 <- count(usa12, response != "atheist")
total_usa12 <- atheist_usa12 + non_atheists_usa12 
atheist_usa12/total_usa12 *100
    ##   response == "atheist"        n
## 1                   NaN 95.00998
## 2                    50  4.99002
    par(mfrow = c(1, 2))
inference(usa05$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(usa12$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 )
    usa_05_12 <- subset(atheism, nationality == "United States" & year == "2005"  | nationality == "United States" & year == "2012")
inference(y = usa_05_12$response, x = usa_05_12$year, est = "proportion",type = "ht", null = 0, alternative = "twosided", method = "theoretical", success = "atheist")
    ## Response variable: categorical, Explanatory variable: categorical
## Two categorical variables
## Difference between two proportions -- success: atheist
## Summary statistics:
##              x
## y             2005 2012  Sum
##   atheist       10   50   60
##   non-atheist  992  952 1944
##   Sum         1002 1002 2004
    ## Observed difference between proportions (2005-2012) = -0.0399
## 
## H0: p_2005 - p_2012 = 0 
## HA: p_2005 - p_2012 != 0 
## Pooled proportion = 0.0299 
## Check conditions:
##    2005 : number of expected successes = 30 ; number of expected failures = 972 
##    2012 : number of expected successes = 30 ; number of expected failures = 972 
## Standard error = 0.008 
## Test statistic: Z =  -5.243 
## p-value =  0

    p_usa05 = 0.01
n_usa05 = 1002 
p_usa12 = 0.05
n_usa12 = 1002 

PE_usa = p_usa12 - p_usa05

SE_usa = sqrt(((p_usa05*(1-p_usa05))/n_usa05)+((p_usa12*(1-p_usa12))/n_usa12))
SE_usa
    ## [1] 0.007568714
    #Control interval for difference between proportion in 2005 and 2012
PE_usa + (1.96*SE_usa)
    ## [1] 0.05483468
    PE_usa - (1.96*SE_usa)
    ## [1] 0.02516532
    Since p value is effectively 0, we can reject the Null Hypothesis and there is evidence that there is a change in the atheism index in the USA from 2005 to 2012. The control interval does NOT include 0 so we reject the null hypohtesis that p_2012 - p_2005 is zero.

0.6.2 Question 2

  • 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.
Those with point difference (difference between 2005 and 2012 proportions) that are relatively large (about +/-3% difference) and a small margin of error (large n sample size) will probably detect a change at the 0.05 significance level. If indeed the population atheism shows that there is no change, then these will be examples of Type 1 errors - rejecting the null hypothesis p_2012 - p_2005 = 0, when in fact it is true.

A Type 1 error occures when we reject the null hypothesis when it is true. Since we are using a .05 significance level, we would expect 5% of countries to have a Type 1 error. 
paste("This would amount to ",.05*39,"countries.")
## [1] "This would amount to  1.95 countries."

0.6.3 Question 3

  • 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.
#SE = sqrt((p*(1-p))/n)
#ME = z * SE = z * sqrt((p*(1-p))/n)
# n = (z/ME)^2 / (p*(1-p))


# Using a p-value to generate the maximum ME, .5 and a .01 ME.
p <- 0.5
z <- qnorm(0.975) # 95% confidence (0.95 + ((1-0.95)/2))
ME = 0.01 # 1% 
n = ( (z/ME)^2 )* ( p*(1-p) )
paste("The sample size to meet the guideline : ", n)
## [1] "The sample size to meet the guideline :  9603.64705173531"