Is there help or advice for the following project on

Classical Probability and Statistics Problems and Bootstrap Estimation.

R studio project attached

upload must have both the html (produced by knitting the Rmd file) and the Rmd files!

--title: "STAT270 - Project 2 - Classical Probability and Statistics Problems and Bootstrap Estimation" author: "YOUR GROUP # LIST of studends who worked" date: "March 1, 2020" output: html_document --```{r setup, include=FALSE} knitr::opts_chunk$set(echo = TRUE) ``` ```{r packages, echo = FALSE, message = FALSE} # load the packages for graphing and data wrangling library(ggplot2) library(PASWR2) library(dplyr) library(lattice) library(boot) library(MASS) ``` Note: If you `Rmd` file submission knits you will receive total of **(5 points)** ### 1. PROBLEM SOLVING PART **Problem 1 (5 pts) - similar to Pr. 5 / page 238 in text** A box contains 20 consecutive balls numbered 1 to 20. If four numbers are drawn at random, how many ways are there for the largest number to be a 13 and the smallest number to be 3 ? Use the Fundamental Principle of Counting! **Solution:** YOUR CODE HERE: ```{r} ``` ################################################################################### ###################### **Problem 2 (5 pts) - similar to Pr. 14 / page 239 in text** On a multiple-choice exam with **four** possible answers for each of the five questions, what is the probability that a student would get four or more correct answers just by guessing? Hint: Use the fact that $P(E F) = P(E) P(F)$ for two independent event (generalized for more than events) Getting one answer correct is independent of another. Also $$P(at least 4) = P(exactly 4) + P(exactly 5)$$, the last events are mutually exclusive so $P(A \\cup B) = P(A) + P(B)$ YOUR CODE HERE: ```{r} ``` Out of 200 students who adopt this test taking approach how many are expected to get at least 4 correct? **Hint:** Use Binomial experiment settings to answer this questions. #### Solution: ################################################################################### ############# **Problem 3 (5 pts) - similar to Pr. 44 page 243 in text** Consider tossing **four** fair coins. There are 16 possible outcomes, e.g `HHHH,HHHT,HHTH,HTHH,THHH, ...` possible outcomes. Define `X` as the random variable \"number of heads showing when four coins are tossed.\" Obtain the mean and the variance of X. Simulate tossing three fair coins 10,000 times. Compute the simulated mean and variance of X. Are the simulated values within 2% of the theoretical answers? Hint: to find the theoretical values use `dbinom (x= , size = , prob = )` **Solution:** YOUR CODE HERE: ```{r} library(MASS) ``` **Problem 4 (5 pts) - similar to Pr. 11 / page 307 in text** Traffic volume is an important factor for determining the most cost-effective method to surface a road. Suppose that the average number of vehicles passing a certain point on a road is 5 every 60 seconds. (a) Find the probability that more than 12 cars will pass the point in a period of 3 minutes. (b) What is the probability that more than 100 cars pass the point in half an hour? Hint: Adjust the Poisson parameter $\\lambda = 5$ every 60 seconds to `#` per every 3 minutes YOUR CODE HERE: ```{r pr 11} ``` ################################################################################### ################ **Problem 5 (5 pts) - similar to Pr. 18 / page 308 in text** Suppose the percentage of drinks sold from a vending machine are 70% and 30% for soft drinks and bottled water, respectively. (a) What is the probability that on a randomly selected day, the first soft drink is the fourth drink sold? (b) Find the probability that exactly 4 out of 10 drinks sold is a soft drink. Hint: Let `X` = number of waters (failures) purchased before the first soft drink is purchased. Then, $X \\sim Geo(0.70)$. **Solution:** YOUR CODE HERE: (a) ```{r pr. 18a} ``` (b) Let `X` = number of soft drinks sold. Then, $X \\sim Bin(10, 0.70)$ and $P(X = 1) = 0$ since one has: ```{r pr. 18b} ``` ################################################################################### ################ **Problem 6 (10 pts) - Exponential Distribution: Light Bulbs** If the life of a certain type of light bulb has an exponential distribution with a mean of 10 months, find (a) The months. (b) The How `0.05`? (c) The than 25 probability that a randomly selected light bulb lasts between 6 and 15 95th percentile of the distribution. the probability of the light bulb to last more than 36 months compare to probability that a light bulb that has lasted for 10 months will last more months. **Solution:** YOUR CODE HERE: ```{r Ex. 4.17} ``` ################################################################################### ################ **Problem 7 (10 pts) - Pr. 8 page 400 ** A population has the following elements: 2, 5, 8, 12, 13 (**finite population**). (a) Enumerate all the samples of size 2 that can be drawn with and without replacement. (Hint: Use the `srs()` function from the `PASWR2` package) (b) Calculate the mean of the population. (c) Calculate the variance of the population. (d) Calculate the standard deviation of the population. (e) Calculate the mean of the sample mean, $E[X]$. (f) Calculate the variance of the sampled mean, $Var(\\bar X)$ (g) Calculate the standard deviation of the sample mean. (h) Calculate the mean of the sample variance, $E[S^2]$ (i) Is the variance of $\\bar X$ larger when sampling with or without replacement? Explain your answer **Solution:** YOUR CODE HERE: ```{r} ``` ################################################################################### ####################### **Problem 8 (10 pts) - similar to Pr. 10 / page 400 in text** Use the data frame `WHEATUSA2004` from the `PASWR2` package; draw all samples of sizes 4, 5, and 6; and calculate the mean of the means. What size provides the best approximation to the population mean? What is the variance of these means? **Solution:** The mean of the means for samples of size two, three, and four are all equal to `1148.7333` (the population mean). Consequently, all samples provide equal approximations to the population mean when taking all possible samples of a given sample size. The variance of the means, however, decreases with increasing sample size from `645755.872` for samples of size four, to `496720.646` for samples of size five, to `397374.397` for samples of size six. **Hint:** Use the `srs()` function from the `PASWR2` package to draw simple random samples. YOUR CODE HERE: ```{r} ``` ################################################################################### ######################## **Problem 9 (10 pts) - Pr. 16 / page 513 in text** A group of engineers working with physicians in a research hospital is developing a new device to measure blood glucose levels. Based on measurements taken from patients in a previous study, the physicians assert that the new device provides blood glucose levels slightly higher than those provided by the old device. To corroborate their suspicion, `15` diabetic patients were randomly selected, and their blood glucose levels were measured with both the new and the old devices. The measurements, in `mg/100 ml`, appear in data frame `GLUCOSE` from the `PASWR2` package: (a) Are the samples independent? Why or why not? (b) If the blood glucose level is a normally distributed random variable, compute a 95% confidence interval for the mean differences of the population. (c) Use the results in (b) to decide whether or not the two devices give the same results. **Solution:** (a) (b) **Hint:** look at the QQ-plot to see if normality is reasonable. YOUR CODE HERE: ```{r} GLUCOSE % mutate(DIFF = old-new) # create variable DIFF for the difference between old and new level head(GLUCOSE) ggplot(data = GLUCOSE, aes(sample = DIFF)) + stat_qq() + theme_bw() CI Look for the answer with the following : **Empirical bootstrap confidence interval for the mean.** For the data contained in a vector `x = c(36,30,37,43,42,43,43,46,40,42)` Estimate the mean `` of the underlying distribution and give an `90%` bootstrap confidence interval. ```{r} x = c(36,30,37,43,42,43,43,46,40,42) qqnorm(x) # see if the data appear to follow normal distribution ``` > Data is not quite normal as it appears on the Q-Q PLOT! ```{r} n = length(x) set.seed(25) # sample mean xbar = mean(x) cat("data mean = ",xbar,'\ ') nboot = 10000 # number of bootstrap samples # Generate 10000 bootstrap samples, i.e. an n x 10000 array of random resamples from x. tmpdata = sample(x,n*nboot, replace=TRUE) # sample with replacement bootstrapsample = matrix(tmpdata, nrow=n, ncol=nboot) # Compute the sample mean xbar for each bootstrap sample xbarstar = colMeans(bootstrapsample) # Compute delta* for each bootstrap sample (difference between the original mean and the bootstrap sample mean) deltastar = xbarstar - xbar # Find the 0.1 and 0.9 quantiles for deltastar d = quantile(deltastar,c(0.05,0.95)) # for the 905 CI we need 90% area between the quantiles # Calculate the 90\\% confidence interval for the mean. ci = xbar - c(d[2],d[1]) ci ``` > We can achieve the same CI with the functions `boot()` and `boot.ci()` from the package `boot` ```{r} MEAN Review the problem below. **Pr. 30 / page 695 in text** The \"Wisconsin Card Sorting Test\" is widely used by psychiatrists, neurologists, and neuropsychologists with patients who have a brain injury, neurodegenerative disease, or a mental illness such as schizophrenia. Patients with any sort of frontal lobe lesion generally do poorly on the test. The data frame `WCST` and the following table contain the test scores from a group of 50 patients from the Virgen del Camino Hospital (Pamplona, Spain). (a) Use the function `eda()` from the `PASWR2` package to explore the data and decide if normality can be assumed. For details type in console `?eda` (b) What assumption(s) must be made to compute a `95%` confidence interval for the population mean? (c) Compute the confidence interval from (b). (d) Compute a 95% BCa bootstrap confidence interval for the mean test score. (e) Should you use the confidence interval reported in (c) or the confidence interval reported in (d)? **Solution:** (a) Assuming the variable score has a normal distribution is not reasonable. See the results of EDA analysis below: ```{r} data(WCST) WCST %>% pull(score) %>% eda() ``` (b) In order to construct a `95%` confidence interval for the population mean, one assumes that the values in the variable score are taken from a **normal distribution**. Although this is not a reasonable assumption, the sample size might be sufficiently large to overcome the skewness in the parent population. Consequently, one might appeal to the **Central Limit Theorem** and claim that the sampling distribution of `X` is **approximately normal** due to the sample size `(50)`. In this problem, the skewness is quite severe, and one should not be overly confident in the final interval. (c) If we assume normality one has: ```{r} # CI % pull(score) %>% t.test() %>% .$conf CI ``` (d) Use the `boot.ci()` nonparametric bootstrap CIs functions to obtain the bootstrap CI. ```{r} library(boot) # use boot package MEAN