Description

Ravish Kamath: 213893664
Question 1
In the without replacement sampling example of Table 3.2, demonstrate that an unbiased estimate of the population size, N, is provided by
n
X
wi i=1
Solution
This is re-creating our samples, denoted as S, from a population of N = 4, and having each Si contain n = 2 elements without replacement
u = c(1,2,3,4)
N = 4 n = 2
sample = t((combn(1:N, n))) sample
## [,1] [,2] ## [1,] 1 2 ## [2,] 1 3 ## [3,] 1 4
## [4,] 2 3
## [5,] 2 4
## [6,] 3 4
We now calculate the probabilities, denoted as P(S), for each of the elemnts of S. Take note that our population elements have unequal probability of selection, represented by δi, as shown below. So in order to accommodate the δi, we take the inclusion probability, denoted by πi, for each element of our population.
delta = c(0.1, 0.1, 0.4, 0.4)
prob = round(delta[sample[,1]]*delta[sample[,2]]/(1-delta[sample[,1]])+ delta[sample[,2]]*delta[sample[,1]]/(1-delta[sample[,2]]),4)
pi = rep(0, length(u)) for(i in 1:length(u)){ pi[i] = sum(prob[sample[,1]== i]) + sum(prob[sample[,2] == i])
}
data.frame(sample,prob)
## X1 X2 prob ## 1 1 2 0.0222 ## 2 1 3 0.1111 ## 3 1 4 0.1111 ## 4 2 3 0.1111 ## 5 2 4 0.1111 ## 6 3 4 0.5333
In order to get the weights, wi, we must reciprocate our πi. Next, we sum up the wi for each element per sample.
wi = 1/pi wi
## [1] 4.091653 4.091653 1.323627 1.323627
sumWeights = rep(0, length(sample[,1])) sumWeights = wi[sample[,1]] + wi[sample[,2]] data.frame(sample,prob,sumWeights)
## X1 X2 prob sumWeights ## 1 1 2 0.0222 8.183306 ## 2 1 3 0.1111 5.415280 ## 3 1 4 0.1111 5.415280 ## 4 2 3 0.1111 5.415280 ## 5 2 4 0.1111 5.415280 ## 6 3 4 0.5333 2.647253
Finally, to demonstrate that this is an unbiased estimate of the population size N, which in our case N = 4, we take the expectation of the Pni=1 wi for each sample. As shown below the E(Pni=1 wi) gives us our true population size, N.
sum(sumWeights*prob)
## [1] 4
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Question 2
Following textbook exercise 3.17, again u = 1,2,3,4 δi = (.1,.1,.4,.4) but this time, take a sample of size n = 3 without replacement. Use R to show that an unbiased estimate of the population size N = 4 is provided by P3i=1 wi
Solution
A similar approach to Question 1, we will now take a sample of n = 3 from a population of N = 4 elements.
u = c(1,2,3,4)
N = 4 n = 3
sample = t((combn(1:N, n))) sample
## [,1] [,2] [,3]
## [1,] 1 2 3 ## [2,] 1 2 4 ## [3,] 1 3 4 ## [4,] 2 3 4
We now calculate the probabilities for each of the sample. Take note that our N elements have unequal probability weights, represented by δi, as shown below. So in order to accommodate the δi, we take the inclusion probability, denoted by πi, for each element of our population.
delta = c(0.1, 0.1, 0.4, 0.4)
prob = round(delta[sample[,1]]*(delta[sample[,2]]/(1-delta[sample[,1]]))*
(delta[sample[,3]]/(1-delta[sample[,1]]-delta[sample[,2]]))+ delta[sample[,1]]*(delta[sample[,3]]/(1-delta[sample[,1]]))* (delta[sample[,2]]/(1-delta[sample[,1]]-delta[sample[,3]]))+ delta[sample[,2]]*(delta[sample[,3]]/(1-delta[sample[,2]]))* (delta[sample[,1]]/(1-delta[sample[,2]]-delta[sample[,3]]))+ delta[sample[,2]]*(delta[sample[,1]]/(1-delta[sample[,2]]))* (delta[sample[,3]]/(1-delta[sample[,2]]-delta[sample[,1]]))+ delta[sample[,3]]*(delta[sample[,1]]/(1-delta[sample[,3]]))* (delta[sample[,2]]/(1-delta[sample[,3]]-delta[sample[,1]]))+ delta[sample[,3]]*(delta[sample[,2]]/(1-delta[sample[,3]]))* (delta[sample[,1]]/(1-delta[sample[,3]]-delta[sample[,2]])),4)
pi = rep(0, length(u)) for(i in 1:length(u)){
pi[i] = sum(prob[sample[,1]== i]) + sum(prob[sample[,2] == i]) + sum(prob[sample[,3]== i])
}
data.frame(sample,prob)
## X1 X2 X3 prob ## 1 1 2 3 0.0556 ## 2 1 2 4 0.0556 ## 3 1 3 4 0.4444 ## 4 2 3 4 0.4444
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In order to get the weights, represented by wi we must now reciprocate πi. Then, we sum up the wi for each element per sample.
wi = 1/pi wi
## [1] 1.799856 1.799856 1.058873 1.058873
sumWeights = rep(0, length(sample[,1]))
sumWeights = wi[sample[,1]] + wi[sample[,2]] + wi[sample[,3]] data.frame(sample,prob,sumWeights)
## X1 X2 X3 prob sumWeights ## 1 1 2 3 0.0556 4.658585 ## 2 1 2 4 0.0556 4.658585 ## 3 1 3 4 0.4444 3.917603 ## 4 2 3 4 0.4444 3.917603
Finally,to demonstrate that this is an unbiased estimate of the population size N, which in our case N = 4, we take the expectation of our weights from each sample. As shown below the E(Pni=1 wi) gives us our true population size, N.
sum(sumWeights*prob)
## [1] 4
4
Question 3
Expanding on Exercise 3.18, data on K–12 education variables and populations for all 50 states are available via links in Electronic Section 3.0. Using a sample size of n = 5, select repeated random samples without replacement from this population of states and calculate the mean number of teachers per state for each sample. Plot the sample means, thereby generating a simulated sampling distribution for the sample mean for samples of size 5.
a. Describe the shape of the simulated sampling distribution. Does it look normal? Why or why not?
b. Calculate the standard deviation for the set of generated sample means. Is it close to the theoretical value of

Solution
Part A
In this question, we are trying to calculate the sample mean, denoted by y, of the number of teachers per state in a sample of n = 5 from N = 50 states. We will repeat this process a 1000 times.
ybar = rep(0,1000) for(i in 1:1000){
n = sample(1:50, 5, replace = FALSE) teachers = schools[n,]$Teachers ybar[i] = mean(teachers)
}
hist(ybar, xlab = ‘Generated sample means’, main = ‘Sampled Means for Teachers in the US.’)
Sampled Means for Teachers in the US.

Generated sample means
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Part B
When calculating the standard deviation,denoted by s for our generated sample means, it does come quite close to the theoretical σ = 28,465.
sd(ybar)
## [1] 26954.27
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Question 4
A study to assess the attitudes of accountants toward advertising their services involved sending questionnaires to 200 accountants selected from a list of 1400 names. A total of 82 usable questionnaires were returned. The data summary for one question is shown in the accompanying table.
a. Estimate the population proportion virtually certain to advertise in the future.
b. Estimate the population proportion having at least a 50–50 chance of advertising in the future.
c. Among those who advertised in the past, estimate the population proportion some- what unlikely to advertise again.
d. Among those who advertised in the past, estimate the population proportion having at least a 50–50 chance of advertising again.
Place bounds on the errors of estimation in all cases. Do parts (c) and (d) require further assumptions over those made for parts (a) and (b)?
Solution
Let us first re-create the data from the questionnaires provided. Let it be known that our population is 1400 names, denoted by N, and our sample, S, for all respondents is n = 82, and those who have advertised in the past n′ = 46.
response = c(‘Virtual certainty’, ‘Very likely’, ‘Somewhat likely’, ‘About 50-50’,
‘Somewhat unlikely’, ‘Very unlikely’,’Absolutely not’, ‘No response’)
allRespondents = c(.22,.04,.19,.18,.06,.12,.15,.04) advertisedPast = c(.35,.05,.35,.15,.10,0,0,0)
df = data.frame(response, allRespondents, advertisedPast) df
## response allRespondents advertisedPast
## 1 Virtual certainty 0.22 0.35 ## 2 Very likely 0.04 0.05 ## 3 Somewhat likely 0.19 0.35 ## 4 About 50-50 0.18 0.15
## 5 Somewhat unlikely 0.06 0.10 ## 6 Very unlikely 0.12 0.00 ## 7 Absolutely not 0.15 0.00
## 8 No response 0.04 0.00
N = 1400 n = 82
nPrime = 46
Part A
Based on the calculations below, we can see that the estimated population proportion, denoted as pˆ, for respondents that are virtually certain to advertise in the future is 0.22 with a bound on the errors of 0.0893.
p_hat = df[1,2] q_hat = 1-p_hat p_hat
## [1] 0.22
var_hat = (1-(n/N))*((p_hat*q_hat)/(n-1))
B = 2*(sqrt(var_hat)) round(B,4)
## [1] 0.0893
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Part B
Based on the calculations below, we can see that the pˆ for respondents having at least a 50-50 chance of advertising in the future is 0.63 with a bound on the errors of 0.1041.
p_hat = sum(df[1:4,2]) q_hat = 1-p_hat p_hat
## [1] 0.63
var_hat = (1-(n/N))*((p_hat*q_hat)/(n-1))
B = 2*(sqrt(var_hat)) round(B,4)
## [1] 0.1041
Part C
Based on the calculations below, we can see that the pˆ for respondents who have advertised in the past and are unlikely to advertise again is 0.1 with a bound on the errors of 0.088.
p_hat = df[5,3] p_hat
## [1] 0.1
q_hat = 1-p_hat
var_hat = (1-(nPrime/N))*((p_hat*q_hat)/(nPrime-1))
B = 2*(sqrt(var_hat)) round(B,4)
## [1] 0.088
Part D
Based on the calculations below, we can see that the pˆ for respondents who have advertised in the past and having at least a 50-50 chance of advertise again is 0.9 with a bound on the errors of 0.088.
p_hat = sum(df[1:4,3]) p_hat
## [1] 0.9
q_hat = 1-p_hat
var_hat = (1-(nPrime/N))*((p_hat*q_hat)/(nPrime-1))
B = 2*(sqrt(var_hat)) round(B,4)
## [1] 0.088
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Solution
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Quesiton 5
Redo Exercise 3.18 on the complete data set with 50 states.
a. Sampling with Replacement
b. Sampling without Replacement
Solution
Part A
teachers = schools$Teachers states = schools$State popu = schools$Pop N = dim(schools)[1]
n = 2 sample=expand.grid(1:N,1:N)
To calculate the probabilities, P(S) for each sample, it is given in the question that the P(S) are proportional to the population of the states. Hence our P(S) for each sample will be dependent on the element’s corresponding δi value. Finally we can calculate our estimated total number of teachers for each sample, denoted by τˆ.
delta =schools$Pop/sum(schools$Pop) prob = rep(0,N)
prob = delta[sample[,1]]*delta[sample[,2]] tau_hat = 1/n*(teachers[sample[,1]]/delta[sample[,1]]
+teachers[sample[,2]]/delta[sample[,2]])
To demonstrate that our τˆ is unbiased, we take the expectation of our estimated total population.
prob%*%tau_hat
## [,1]
## [1,] 2992790
sum(schools$Teachers)
## [1] 2992790
As proved above, we come to the same total number of teachers, which proves that the E(τˆ) is the theoretical population total, τ.
Part B
Similar to Part A, we continue the same procedure, however there will be without replacement, which means our total number of S will be 1225.
sample = t(combn(1:N,n))
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Solution
To calculate the P(S) for each S, we use a similar approach to Part A, except without replacement does change how we multiply our individual probability for each element of the Si, as shown below.
prob = rep(0,N)
prob = delta[sample[,1]]*delta[sample[,2]]/(1-delta[sample[,1]]) + delta[sample[,2]]*delta[sample[,1]]/(1-delta[sample[,2]])
pi=rep(0,length(schools$Teachers)) for(i in 1:length(schools$Teachers)){ pi[i]=sum(prob[sample[,1]==i])+sum(prob[sample[,2]==i])
}
Finally we can calculate our τˆ for each S as represented by the variable .
tau_hat = rep(0,dim(sample)[1]) tau_hat = teachers[sample[,1]]/pi[sample[,1]]+teachers[sample[,2]]/pi[sample[,2]]
To demonstrate that our τˆ is unbiased, we take the expectation of our estimated τˆ.
prob%*%tau_hat
## [,1]
## [1,] 2992790
sum(schools$Teachers)
## [1] 2992790
As proved above, we come to the same total number of teachers, which proves that the E(τˆ) total is the theoretical population total, τ.
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