Homework 6

Due Thurs 4 Nov in class

Assignment:

This assignment contains homework problems out of the text as well as a lab portion.

1. As usual, I recommend going thru the conceptual exercises in Chapter 5.
2. Chapter 4 Exercises: 26, 27, 30.
3. Note that in Exercise 30, subjects should ideally be able to tolerate 5 times as much sunlight when wearing SPF 5 sunscreen. This means we might expect the effect of the treatment to be multiplicative. Think about what transformation would be most appropriate.
4. Also, use the bootstrap to obtain a 95% confidence interval for the multiplicative effect in Exercise 30. You can do this by computing the ratio of (treatment duration)/(pretreatment duration). This number gives the multiplicative effect for each subject. Now take these 13 ratios and compute a confidence interval using both the mean and the median as the statistic of interest.
5. In addition to the above exercises, compute the p-value for testing whether the urban site shows a higher level of pollution than the downwind site for the pollution data ( pollution.asc ) Go ahead and do this using a permutation test for paired differences. Compare this p-value to that of the rank sum test.
   Bootstrap for estimating the effect in a paired differences experiment
   
   Below is an example of using Splus commands to carry out a bootstrap analysis for paired differences. This may prove helpful for the supplemental problem to Exercise 30.
   
   We'll use the pollution dataset ( pollution.asc ) which was used in lecture.
   • Read the dataset into a dataframe called pollution.
   • Now create a vector d1 to hold the differences by day. I'm going to remove the NA's to make life easier on me.

     d1_pollution$city - pollution$rural
     bad_is.na(d1)
     d1_d1[!bad]
     

     Above, bad is a vector of T's or F's - T where d1 is NA, and F where d1 is a valid number. I can delete these NA's by selecting the components of d1 that not T's. The ! operator reverses the T's and F's in bad.
   • Now I go ahead and compute the bootstrap samples and get the mean of each of the samples. I'll hold the bootstrapped means in the vector bootmean. The vector ib holds the indicies of the bootstrap sample and d1[ib] is the actual bootstrapped sample.

     n_length(d1)
     bootmean_rep(NA,1000)
     for(i in 1:1000){
       ib_sample(1:n,n,replace=T)
       bootmean[i]_mean(d1[ib])
     }
     par(mfrow=c(1,1))
     hist(bootmean,xlab="bootstrapped means")
     points(mean(d1),0,cex=2,pch=16)
     ci95_quantile(bootmean,c(.025,.975))
     text(ci95,c(0,0),c("[","]"))
     

     Now ci95 is a vector of length 2 holding the upper and lower ends of the confidence interval.
   
   Permutation test for a paired differences experiment
   
   Below is an example of using Splus commands to carry out a permutation test for paired differences. This may prove helpful for the supplemental problem on the pollution data.
   
   We'll use the shoe dataset ( shoe.asc ) from the midterm:
   A shoe company has devoloped a new material A which they hope will wear out less quickly than the standard material B. The data below are measurements of the amount of wear of the soles of shoes worn by 10 boys -- a high score means high wear. The soles were made of the two different materials, A and B. Each boy wore a special pair of shoes, the sole of one shoe having been made with A and the sole of the other with B. The decision as to whether the left or the right sole was made with A or B was determined by a flip of a coin.
   
   
   • First read the data into a dataframe called shoe. Since the data are obviously paired (each person wears a pair of shoes where the shoes have one of each sole), we can take the differences and put them into a vector called d1

     d1_shoe$B - shoe$A
     n_length(d1)
     

   • Now sample many permutations and put the mean into the vector pdif. You can generate the random signs for the test by using the sample command below.

     pdif_rep(NA,1000)
     for(i in 1:1000){
       isgn_sample(c(1,-1),n,replace=T)
       pdif[i]_mean(isgn*abs(d1))
     }
     

   • Now go ahead and plot the results...

     par(mfrow=c(1,1))
     hist(pdif,xlab="permutation means")
     points(mean(d1),0,cex=2,pch=16)
     text(0.2,154,paste("pval =",round(mean(pdif>=mean(d1)),2)),cex=2)