Answer in R. # 1. Simulate $n=50$ observations from a normal distribution with mean 3, standard deviation .80, using: ANSWER: ```{r} set.seed(87612345) # add your own R code below this line ``` Calculate a 98% t-confidence interval for the mean of 'data' ANSWER: ```{r} # add your own R code below this line ``` Carry out the following bootstrap sampling scheme. Draw $B=2000$ bootstrap samples. For each sample, calculate the sample average, using the R built-in function "mean". You can use a for loop to do this, where the loop index $i$ runs from 1 to $B$. Inside the loop you can do the bootstrap sampling with "bdata=sample(data,n,replace=T)", where $n$ is the number of sample in your dataset. and then you need something similar to: "bootstrapmeans=c(bootstrapmeans,mean(bdata))". ANSWER: ```{r} set.seed(87612345) # add your own R code below this line # use a variable called B for the number of bootstrap sample (2000) # use a variable called n for the number of sample ``` + (1a) make a histrogram of the bootstrapmeans with 20 bars, using something like: "hist(bootstrapmeans,nclass=20)". ANSWER: ```{r} # add your own R code below this line ``` Does the histogram look like it has a normal shape? ANSWER: + (1b) calculate the 1'th and 99'th percentiles of the bootstrapped means using "quantile(bootstrapmeans,probs=c(.01,.99))". ANSWER: ```{r} # add your own R code below this line ``` This is a valid 98% bootstrap confidence interval for the unknown population mean, which does not depend on the assumption of normality. This interval uses the so-called "bootstrap percentile method". (7 points: 5 points for histogram, 2 points for percentiles) # 2. Repeat problem 1, but using a sample of size $n=60$ from a chi-square distribution with $0.2$ degree of freedom (note that this does not need to be an integer). Use $B=1500$ bootstrap samples. You will need first to generate the data, and then to write a loop for producing bootstrap samples from this data. ANSWER: ```{r} set.seed(117426) # add your own R code below this line ``` (2a) make the histogram of the bootstrap means with 25 bars. ANSWER: ```{r} # add your own R code below this line ``` Does the histogram look a close to normal as it did when you sampled from a normal distribution? ANSWER: (5 points for histogram) (2b) make a 96% confidence interval for the population mean using the percentile method ANSWER: ```{r} # add your own R code below this line ``` (2 points for the interval) (2c) calculate the usual 96% t-interval assuming normality, using the t-test procedure. ANSWER: ```{r} # add your own R code below this line ``` Compare the percentile (bootstrap method) confidence interval obtained in 2b and and t-confidence intervals obtained in 2c. Do you expect the two methods to give similar results or not for this distribution (which is not normal)? ANSWER: (2 points for extracting the confidence interval from the t.test output, or for calculating using first principles.) # 3. The following code draws a sample of size $n=30$ from a chi-square distribution with 15 degrees of freedom, and then puts $B=200$ bootstrap samples into a matrix M. After that, the 'apply' function is used to calculate the median of each bootstrap sample, giving a bootstrap sample of 200 medians.  ```{r} set.seed(117888) data=rchisq(30,15) M=matrix(rep(0,30*200),byrow=T,ncol=30) for (i in 1:200)M[i,]=sample(data,30,replace=T) bootstrapmedians=apply(M,1,median) ``` (3a) Use the 'var' command to calculate the variance of the bootstrapped medians. (This provides a valid estimate of the variance of the median of a random variable.) ANSWER: ```{r} # add your own R code below this line ``` (1 point for variance) (3b) Use the apply command to calculate bootstrapped means rather than bootstrapped medians, and then use 'var' to calculate an estimate of the variance of the mean (hint: use 'apply' on the 'mean') . ANSWER: ```{r} # add your own R code below this line ``` (3 points for the variance of the bootstrapped means. Note: if you run the sampling again, yournumerical result may change. This will not affect your mark if your code is correct). (3c) The theoretical variance of a chi-square random variable with 15 degrees of freedom is 2(15)=30, so the theoretical variance of the mean is 30/30=1 To show this we have applied the formula: $$Var(\bar{x}) = Var (X)/n$$ where n is the number of items that are averaged. So we know that the true variance of the mean is $1$. If we had an infinite number of samples, the estimated variance of the mean would be $1$. But we only have $n=30$ samples. Did the bootstrap do a reasonable job of estimating the variance of the mean? ANSWER: # 4. Following is a recursive function which takes as input (argument) a vector $x$ and returns a scalar. Either by looking carefully at the function, or by running the function with a few simple input vectors and observing the returned values in each case, determine what the function is doing. ```{r} myfun=function(x){ if(length(x)==1){return(x) } else return(x[1]+myfun(x[-1])) } myfun(c(1:1000)) #here's what happens when it is applied to 1:1000 ``` ANSWER (4 points) (Describe in precise words what the function does) Modify the function slightly so that it returns the value: $\sum_{i=1}^n e^{x_i}$ In other words, we want to write a function that calculates the sum of the exponential of each component of the input vector $x$. Test your function with the argument vector: x=c(2,1,3,0.4). You should get an answer approximately equal to 31.6847 ANSWER (6 points): ```{r} # add your own R code below this line ```