** Lab 6 - Exercises, Data Analysis Questions, and Short Answers - Xialing Walla**

Exercise 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 percentages are sample statistics derived from all the interviews conducted accross the world. It is not practical to ask every single person in this world their religious stand.

Exercise 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?

We must assume that the interviewees were simple randomly selected. It is very important to the accuracy of sample statistics that one group is not overly selected vs. the others. Only simple random sampling method can truly represent the different segments of human beliefs close to the appropriate proportions of the populatioin. Sample size almost is of importance.

download.file("http://www.tpmltd.com/ban/atheism.RData", destfile = "atheism.RData")
load("atheism.RData")

Exercise 3 What does each row of Table 6 correspond to? What does each row of atheism correspond to?

Each row of Table 6 correspond to a country with response percentages for all groups, whereas each row of atheism corresponds a country with responses only related to atheist and non-atheist.

Exercise 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, atheism$nationality == "United States" & atheism$year == "2012")
by(us12$nationality, us12$response, length)
us12$response: atheist
[1] 50
-------------------------------------------------------- 
us12$response: non-atheist
[1] 952
50/(952+50)
[1] 0.0499

The proportion of atheist responses for the US is 0.0499 which rounds to 0.05 in table 6.

Exercise 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?

The conditions are:

inference(data = us12$response, est = "proportion", type = "ci", method = "theoretical", success = "atheist")
Warning: package 'openintro' was built under R version 3.1.1
Please visit openintro.org for free statistics materials

Attaching package: 'openintro'

The following object is masked from 'package:datasets':

    cars
Warning: package 'BHH2' was built under R version 3.1.1

Attaching package: 'BHH2'

The following object is masked from 'package:openintro':

    dotPlot
One categorical variable 
Single proportion 
Observed proportion = 0.0499 
Number of successes = 50 ; Number of failures = 952 
Standard error = 0.0069 
95 % Confidence interval = ( 0.04 , 0.06 )

Exercise 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?

ME <- 1.96* 0.0069
ME
ME for est. P [1] 0.01352

Exercise 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.

The two countries I picked are Spain and Vietnam. The conditions for the inference are met for Spain as again the sample size are much less than 10% of the total populations for these two countries. The sample sizes are big enough to also satisfy the Success_failure conditon. However, the condition for Vietnam failed as the proportion for atheist is 0% so np >= 10 condition is not met.

next country to test is Uzbekistan. All conditions are met.

Standard error for Spain = 0.0085

Spain12 = subset(atheism, atheism$nationality == "Spain" & atheism$year == "2012")
inference(data = Spain12$response, est = "proportion", type = "ci", method = "theoretical", success = "atheist")
One categorical variable 
Single proportion 
Observed proportion = 0.09 
Number of successes = 103 ; Number of failures = 1042 
Standard error = 0.0085 
95 % Confidence interval = ( 0.07 , 0.11 )

Standard error for Vietnam = ??

Viet12 = subset(atheism, atheism$nationality == "Vietnam" & atheism$year == "2012")
inference(data = Viet12$response, est = "proportion", type = "ci", method = "theoretical", success = "atheist")
One categorical variable 
Single proportion 
Observed proportion = 0 
Number of successes = 0 ; Number of failures = 500 
Error: There aren't at least 10 successes and 10 failures, use simulation
methods instead.

Standard error for Uzbekistan = 0.0063

Uz12 = subset(atheism, atheism$nationality == "Uzbekistan" & atheism$year == "2012")
inference(data = Uz12$response, est = "proportion", type = "ci", method = "theoretical", success = "atheist")
One categorical variable 
Single proportion 
Observed proportion = 0.02 
Number of successes = 10 ; Number of failures = 490 
Standard error = 0.0063 
95 % Confidence interval = ( 0.01 , 0.03 )

Exercise 8 Describe the relationship between p and me.

Based on the graph below the proportion of 0.50 is the proportion with the largest margin of error possible. the plot follows a bell curve. When the proportions move away from 0.50 and get closer to the extremes of 0.00 or 1.00, the margin of error decreases. There is an inverse Correlation between p and me as they move in opposite directions.

n <- 1000
p <- seq(0, 1, 0.01)
me <- 2 * sqrt(p * (1 - p)/n)
plot(me ~ p)

plot of chunk unnamed-chunk-9

Exercise 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.

The simulation to produce 5000 samples of size 1040 sample proportions follow normal distribution with some outliers. The median and mean of the distribution are near identical at 0.1, which is also the population proportion. The range of the distribution is 0.0668 and the IQR is .0118.

set.seed(724)
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
}
summary(p_hats)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.0692  0.0942  0.1000  0.1000  0.1060  0.1360 
hist(p_hats, main = "p = 0.1, n = 1040", xlim = c(0, 0.18))

plot of chunk unnamed-chunk-11

boxplot(p_hats)

plot of chunk unnamed-chunk-11

Exercise 10 Replicate 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 Once you're done, you can reset the layout of the plotting window by using the command par(mfrow = c(1,1)) or clicking on “Clear All” above the plotting window (if using RStudio). Note that the latter will get rid of all your previous plots.

Since n is the denominator of p the bigger the n the smaller the p. n does not affect the center but affects the spread and shape of the distribution of sampling proportions. The spread (variability) decreases as the sample size increases. So larger samples usually give closer estimates of the population proportion p. As the sample sizes get larger the shape of the distribution also follows more the shape of normal distribtuion.

p <- 0.1
n <- 400
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
}
p <- 0.02
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
}
p <- 0.02
n <- 400
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
}
par(mfrow = c(2,2))
hist(p_hats, main = "p = 0.02, n = 1040", xlim = c(0.0001, 0.05))
hist(p_hats, main = "p = 0.1, n = 400", xlim = c(0, 0.18))
hist(p_hats, main = "p = 0.02, n = 400", xlim = c(0.0001, 0.06))

plot of chunk unnamed-chunk-15

Exercise 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?

I would feel comfortable proceeding with the inference and me for Australia as the data met the conditions for the sampling distribution of p. However, Ecuador data did not meet the success-failure condition as np is only 8, which is smaller than the 10 needed.

** Data Analysis Questions**

The question of atheism was asked by WIN-Gallup International in a similar survey that was conducted in 2005. Table 4 on page 13 of the report summarizes survey results from 2005 and 2012 for 39 countries.

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? 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.

The observed p-value is 0.3966, which is greater than the significance level 0.05. Therefore, we fail to reject the null hypothesis that there is no convincing evidence that Spain has seen a change in its atheism index between 2005 and 2012.

atheism$year = gsub(2005, "2005", atheism$year)
atheism$year = gsub(2012, "2012", atheism$year)
spain <- subset(atheism, atheism$nationality == "Spain")
inference(data = spain$response, group = spain$year, est = "proportion",type = "ht",null = 0, alternative = "twosided", method = "theoretical", success = "atheist")
Two categorical variables
Difference between two proportions
n_2005 = 1146 ; n_2012 = 1145 
Observed difference between proportions = 0.0104
H0: p_2005 - p_2012 = 0 
HA: p_2005 - p_2012 != 0 
Pooled proportion = 0.0952 
Group 1: Number of expected successes = 109 ; Number of expected failures = 1037 
Group 2: Number of expected successes = 109 ; Number of expected failures = 1036 
Standard error = 0.012 
Test statistic: Z =  0.848 
p-value:  0.3966 
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5.463e-07 5.178e-07 4.907e-07

plot of chunk unnamed-chunk-16

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

The observed p-value is 0, which is less than the significance level 0.05. Therefore, we reject the null hypothesis that there is no convincing evidence that US has seen a change in its atheism index between 2005 and 2012.

US<- subset(atheism, atheism$nationality == "United States")
inference(data = US$response, group = US$year, est = "proportion",type = "ht",null = 0, alternative = "twosided", method = "theoretical", success = "atheist")
Two categorical variables
Difference between two proportions
n_2005 = 1002 ; n_2012 = 1002 
Observed difference between proportions = -0.0399
H0: p_2005 - p_2012 = 0 
HA: p_2005 - p_2012 != 0 
Pooled proportion = 0.0299 
Group 1: Number of expected successes = 30 ; Number of expected failures = 972 
Group 2: Number of expected successes = 30 ; Number of expected failures = 972 
Standard error = 0.008 
Test statistic: Z =  -5.243 
p-value:  0 
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7.246e-05 6.94e-05 6.646e-05 6.363e-05 6.093e-05 5.833e-05 5.583e-05 5.344e-05 5.114e-05 4.894e-05 4.683e-05 4.48e-05 4.286e-05 4.1e-05 3.921e-05 3.75e-05 3.586e-05 3.428e-05 3.278e-05 3.133e-05 2.995e-05 2.862e-05 2.735e-05 2.614e-05 2.497e-05 2.385e-05 2.279e-05 2.176e-05 2.079e-05 1.985e-05 1.895e-05 1.809e-05 1.727e-05 1.649e-05 1.574e-05 1.502e-05 1.433e-05 1.367e-05 1.305e-05 1.244e-05 1.187e-05 1.132e-05 1.08e-05 1.029e-05 9.815e-06 9.357e-06 8.919e-06 8.501e-06 8.102e-06 7.721e-06 7.357e-06 7.009e-06 6.677e-06 6.361e-06 6.058e-06 5.769e-06 5.494e-06 5.231e-06 4.98e-06 4.741e-06 4.513e-06 4.295e-06 4.088e-06 3.89e-06 3.701e-06 3.521e-06 3.349e-06 3.186e-06 3.03e-06 2.881e-06 2.74e-06 2.605e-06 2.476e-06 2.354e-06 2.237e-06 2.126e-06 2.021e-06 1.92e-06 1.824e-06 1.733e-06 1.646e-06 1.564e-06 1.485e-06 1.41e-06 1.339e-06 1.271e-06 1.207e-06 1.145e-06 1.087e-06 1.032e-06 9.791e-07 9.29e-07 8.813e-07 8.361e-07 7.93e-07 7.521e-07 7.133e-07 6.764e-07 6.413e-07 6.08e-07 5.763e-07 5.463e-07 5.178e-07 4.907e-07

plot of chunk unnamed-chunk-17

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?

A type 1 error is rejecting the null hypothesis when H0 is actually true. We typically do not want to incorrectly reject H0 more than 5% of the time. This corresponds to a significance level of 0.05. Since there are 39 countries in Table 4 that summarizes survey results from 2005 to 2012 we will need to multiply 0.05 by 39 to estimate how many countries we would expect to detect a change in the atheism index simply by chance. the result is 1.95, or 2 countries.

0.05*39
expected change [1] 1.95

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.

There are two unknown variablies in this question: p and n. When we do not have and estimate for p we follow the guideline that the margin of error is largest when p is 0.5. So we typically use this worst case estimate if no other estimate is available. The estimate must have a margin of error no greater than 1%. We use the formula ME = z*SE = 1,96 * sqrt(p(1-p)/n)) <=0.01. Based on the calculation we would need at least 9604 participants to ensure the sample proportion is within 0.01 of the true porportion with 95% confidence.

P <-0.5
Z.alpha <-1.96
ME <-0.01
N<- Z.alpha^2*P*(1-P)/ME^2
N
[1] 9604

4. What concepts from the textbook are covered in this lab? What concepts, if any, are not covered in the textbook? Have you seen these concepts elsewhere, e.g. lecture, discussion section, previous labs, or homework problems? Be specific in your answer.

IN this lab we covered the concepts of making proportion inferences for categorical data with confidence intervals, hypothesis, and Chi-Square distribution. Most of the theories have been covered in previous chapters such as how to computer the confidence intervals, perform hypothesis tests, and beware of type 1 error. We used Anova again to calculate critical values and learned that When only two factors are involved, ANOVA and the t-test result in the same conclusion.

** Short Answer Questions**

1. When only two factors are involved, ANOVA and the t-test result in the same conclusion. Suppose that 14 selected students were divided into 2 groups. One group was taught using a combination of lecture and programmed instruction and the other group was taught with a combination of lecture and television. At the end of the course, each student was given a 50 item test. The following is a list of the number correct for each student.

Based on the critical values calculated by ANOVa and two sample t test we reject the null hypothesis that there is no difference in the means.

(a) Test the equality of means using ANOVA.

score <-c(19,17,23,22,17,16,27,25,32,28,31,26,23,24)
method=c(rep("LP",8),rep("LT",6))
compare=data.frame(score,method)
compare
   score method
1     19     LP
2     17     LP
3     23     LP
4     22     LP
5     17     LP
6     16     LP
7     27     LP
8     25     LP
9     32     LT
10    28     LT
11    31     LT
12    26     LT
13    23     LT
14    24     LT
boxplot (score ~ method, data = compare)

plot of chunk unnamed-chunk-20

summary(aov(score ~ method, data = compare))
            Df Sum Sq Mean Sq F value Pr(>F)   
method       1    149   148.6    9.65 0.0091 **
Residuals   12    185    15.4                  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(b) Test equality of means using a t test.

landpscore <-c(19,17,23,22,17,16,27,25)
landtscore <-c(32,28,31,26,23,24)
t.test(landpscore,landtscore,mu = 0,)

    Welch Two Sample t-test

data:  landpscore and landtscore
t = -3.159, df = 11.52, p-value = 0.008619
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -11.145  -2.022
sample estimates:
mean of x mean of y 
    20.75     27.33 

2. The amount spent on housing is an important component of the cost of living. An economist selected a sample of 20 homeowners and calculated housing costs as a percentage of monthly income, five years ago and now. Test the hypothesis the mean housing cost has decreased.

(a) Write the null and alternative hypotheses.

(b) Calculate the test statistic.

p5yrago <- c(17,20,29,43,36,43,45,19,49,49,35,16,23,33,44,44,28,29,39,22)
pnow <- c(10,39,37,27,12,41,24,26,28,26,32,32,21,12,40,42,22,19,35,12)
t.test(pnow,p5yrago,mu = 0, alternative="less", conf.level = .99, paired=TRUE)

    Paired t-test

data:  pnow and p5yrago
t = -2.258, df = 19, p-value = 0.01795
alternative hypothesis: true difference in means is less than 0
99 percent confidence interval:
   -Inf 0.7854
sample estimates:
mean of the differences 
                   -6.3 

© What is the critical value for the test? (Use the .01 level of significance).

the critical value of P for the test is 0.01795

(d) Based on your calculations, what is your conclusion?

because the p-value is greater than 0.01 we fail to reject the null hypothesis.

(e) Interpret your result.

The sample size is too samll to fit the condition for normal model for p1-p2. Therefore, I used paired t test to find the critical value of P. Because we do not want to make type 1 error a confidence interval of 99% is used. we fail to reject the the null hypothesis that there is no difference in house cost because the income level could be moving with the housing cost in a positive relation. A larger sample would be needed to evaluate the true relationship.