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Parametric Bootstrapping

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#Problem 2 #Part 2 B=1000 sample_med_vars=numeric() CIs = character() n_values = c(10,100,1000) for (i in 1:3) { sample_meds=replicate(B, median(rnorm(n_values[i],1, 1))) sample_med_vars[i] = var(sample_meds) CIs[i] = paste(mean(sample_meds)-1.96*sd(sample_meds), mean(sample_meds)+1.96*sd(sample_meds)) } sample_med_vars #0.132826885 0.015370697 0.001603544 CIs ("0.265111943430679 1.69377252141066") #n=10 ("0.757126210069845 1.24312238725003") #n=100 ("0.922210181934582 1.07918375732663") #n=1000 #Part 3 theoretical_vars=numeric() n_values=c(10,100,1000) for (i in 1:3){ var_samplemed= (pi*(1)^2)/(2*n_values[i]) theoretical_vars[i] = var_samplemed } theoretical_vars #0.157079633 0.015707963 0.001570796 sample_mean_vars #0.132826885 0.015370697 0.001603544 *from Part 2 #The two types of variances seem to yield very similar values with very slight discrepancies. #variance of sample mean n_values=c(10,100,1000) var_sample.mean=1/n_values #0.100 0.010 0.001 #Compared to the variance of sample means, the variance of sample medians are greater in value on average. #Part 4 n=1000 sample_mean = mean(rnorm(n,1, 1)) pboot_vars=numeric() n_values = c(10,100,1000) for (i in 1:3) { pboot_vars[i] = var(replicate(B, median((rnorm(n_values[i],sample_mean, 1))))) } pboot_vars #0.131888264 0.016435213 0.001627244 sample_med_vars #0.132826885 0.015370697 0.001603544 *from Part 2

#Comparing the 2 results, we observe that #the parametric boostrap variances arevery similar to the sample median variances obtained from part 2. #Part 5 pivotalCIs = numeric() n_values = c(10,100,1000) for (i in 1:3) { pboot.meds=replicate(B, median(rnorm(n_values[i],sample_mean, 1))) pboot.percentile=quantile(pboot.meds, prob = c(.025, .975)) pboot.hat = median(rnorm(n_values[i],sample_mean,1)) pboot.CI = paste(2*pboot.hat -pboot.percentile[2], 2*pboot.hat pboot.percentile[1] ) pivotalCIs[i]=pboot.CI }

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