Ex 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?
Answer: The percentages are generated from results of the survey which is a sample of the population, so they are sample statistics.
Ex 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?
Answer: We must assume that the observations are independent. It is a reasonable assumption provided that the survey picked individuals at random and at 51,927 observations we can confidently say that the sample is less than 10% of the population. Since we are comparing countries and religiosity there may be minimum sample size requirements for inference for some smaller countries with small religious minorities.
download.file("http://www.openintro.org/stat/data/atheism.RData", destfile = "atheism.RData")
load("atheism.RData")
Ex 3, What does each row of Table 6 correspond to? What does each row of atheism correspond to?
Answer : Each row in table 6 is aggreate response for each countries and each row in data frame athesim is individual response.
ex 4, 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")
athestprop = sum(us12$response=="atheist") / length(us12$response=="atheist")
round(athestprop*100,2)
## [1] 4.99
Answe, Yes, they are almost the same.
Ex 5, 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?
inference(us12$response, est = "proportion", type = "ci", method = "theoretical", success = "atheist")
## Warning: package 'BHH2' was built under R version 3.5.3
## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.0499 ; n = 1002
## Check conditions: number of successes = 50 ; number of failures = 952
## Standard error = 0.0069
## 95 % Confidence interval = ( 0.0364 , 0.0634 )
Answe: Ans : Observations are independent and each part of proportion are pretty large. The atheist count -50 , so the conditions are satisfied.
Ex 6, Based on the R output, what is the margin of error for the estimate of the proportion of the proportion of atheists in US in 2012?
SE=0.0069
# MOE
MOE=1.96*SE
round(MOE*100,2)
## [1] 1.35
Ex 7, Using the inference function, calculate confidence intervals for the proportion of atheists in 2012 in two other countries of your choice, and report the associated margins of error. Be sure to note whether the conditions for inference are met. It may be helpful to create new data sets for each of the two countries first, and then use these data sets in the inference function to construct the confidence intervals.
Lebanon = subset(atheism, nationality == "Lebanon" & year == "2012")
table(Lebanon$response)
##
## atheist non-atheist
## 10 495
inference(Lebanon$response, est = "proportion", type = "ci", method = "theoretical", success = "atheist")
## Single proportion -- success: atheist
## Summary statistics:
## p_hat = 0.0198 ; n = 505
## Check conditions: number of successes = 10 ; number of failures = 495
## Standard error = 0.0062
## 95 % Confidence interval = ( 0.0077 , 0.032 )
MOE = 1.96*0.0062
round(MOE*100,2)
## [1] 1.22
japan = subset(atheism, nationality == "Japan" & year == "2012")
table(japan$response)
##
## atheist non-atheist
## 372 840
inference(japan$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 )
MOE=1.96*0.0132
round(MOE*100,2)
## [1] 2.59
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=p(1???p)/n??????????????????????????????. This is then used in the formula for the margin of error for a 95% confidence interval: ME=1.96×SE=1.96×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.
Ex 8 Describe the relationship between p and me.
Answer: When large numbers of the proportion move in one direction or the other, the margin of error decreases, as the proportion moves to the middle, the margin of error is at its peak.
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???10 and n(1???p)???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 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 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.
Ex 9, 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.
mean(p_hats)
## [1] 0.09969
sd(p_hats)
## [1] 0.009287382
Ex 10, 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 p^? How does p affect the sampling distribution?
p1 = 0.1
p2 = 0.2
n1 = 400
n2 = 1040
p_hats11 = rep(0, 5000)
p_hats12 = rep(0, 5000)
p_hats21 = rep(0, 5000)
p_hats22 = rep(0, 5000)
for(i in 1:5000){
samp11 = sample(c("atheist", "non_atheist"), n1, replace = TRUE, prob = c(p1, 1-p1))
p_hats11[i] = sum(samp11 == "atheist")/n1
samp12 = sample(c("atheist", "non_atheist"), n2, replace = TRUE, prob = c(p1, 1-p1))
p_hats12[i] = sum(samp12 == "atheist")/n2
samp21 = sample(c("atheist", "non_atheist"), n1, replace = TRUE, prob = c(p2, 1-p2))
p_hats21[i] = sum(samp21 == "atheist")/n1
samp22 = sample(c("atheist", "non_atheist"), n2, replace = TRUE, prob = c(p2, 1-p2))
p_hats22[i] = sum(samp22 == "atheist")/n2
}
par(mfrow = c(2, 2))
hist(p_hats11, main = "p = 0.1, n = 400", xlim = c(0.05, 0.4))
hist(p_hats12, main = "p = 0.1, n = 1040", xlim = c(0.05, 0.4))
hist(p_hats21, main = "p = 0.2, n = 400", xlim = c(0.05, 0.4))
hist(p_hats22, main = "p = 0.2, n = 1040", xlim = c(0.05, 0.4))
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.
par(mfrow = c(1, 1))
All Plots looks normal
Ex 11, 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?
Answer : Australia looks good. But Ecuador has a success =8 which is less than 10, hence it fails success-failure test. So we cannot move ahead with inference.
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.
1, Answer: 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.
Null Hypothesis = H0: There is no difference between the atheism index in Spain from 2005 to 2012.
Alternative Hypothesie = HA: There is a difference between the atheism index in Spain from 2005 to 2012.
spain=subset(atheism, nationality == "Spain")
table(spain$response, spain$year)
##
## 2005 2012
## atheist 115 103
## non-atheist 1031 1042
inference(spain$response, spain$year, est = "proportion", type = "ht", null = 0, alternative = "twosided", method = "theoretical", success = "atheist")
## Warning: Explanatory variable was numerical, it has been converted to
## categorical. In order to avoid this warning, first convert your explanatory
## variable to a categorical variable using the as.factor() function.
## Response variable: categorical, Explanatory variable: categorical
## Two categorical variables
## Difference between two proportions -- success: atheist
## Summary statistics:
## x
## y 2005 2012 Sum
## atheist 115 103 218
## non-atheist 1031 1042 2073
## Sum 1146 1145 2291
## Observed difference between proportions (2005-2012) = 0.0104
##
## H0: p_2005 - p_2012 = 0
## HA: p_2005 - p_2012 != 0
## Pooled proportion = 0.0952
## Check conditions:
## 2005 : number of expected successes = 109 ; number of expected failures = 1037
## 2012 : number of expected successes = 109 ; number of expected failures = 1036
## Standard error = 0.012
## Test statistic: Z = 0.848
## p-value = 0.3966
Since p-value =0.3966 which is greater than 0.05 we accept null hypothesis
b. Is there convincing evidence that the United States has seen a change in its atheism index between 2005 and 2012?
Null Hypothesis = H0: There is no difference between the atheism index in US from 2005 to 2012.
Alternate Hypothesis = HA: There is a difference between the atheism index in US from 2005 to 2012.
us=subset(atheism, nationality == "United States")
table(us$response, us$year)
##
## 2005 2012
## atheist 10 50
## non-atheist 992 952
inference(us$response, us$year, est = "proportion", type = "ht", null = 0, alternative = "twosided", method = "theoretical", success = "atheist")
## Warning: Explanatory variable was numerical, it has been converted to
## categorical. In order to avoid this warning, first convert your explanatory
## variable to a categorical variable using the as.factor() function.
## 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
Since p-value =0 which is less than 0.05 we reject null hypothesis
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.
round(0.05*39,1)
## [1] 2
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.
p=0.5
me=.01
z = 1.96
se = me/z
n = round(p * (1-p) / se^2)
n
## [1] 9604
