Hi, I need help on these stats problems.

A random sample of 16 flights offered by a certain national air carrier is selected. For each flight chosen. the delay time On minutes) was observed. The "delay time" is defined as the difference between the time the plane scheduled to pull away from the jet way and the actual time the plane pulls away from the jet way, with positive values indicating that the flight is late. For now, assume the flight-delay variable is normally distributed. The data is shown below. Copy and paste this dataset into R. x=c(0, 1, -6, -6, 157, -3, 178, -3, -10, 42, -2, 120, 5, 59, 0, -2) (a) Compute the sample mean and the sample standard deviation. 5 = E (use as many decimals as R prints) {A II E (use as many decimals as R prints) (b) Find the t-multiplier needed for the t-version of the confidence interval for the population mean )1. Recall that the t-multiplier is the T(%,n_1) portion of the following formula: 5 i an1) i To find this value using R/RStudio, copy the following code into your R console: qt(0.975,df=16-1) . Next, use this value to help calculate a 95% confidence interval for p. 95% confidence interval: 5 p, g I Ii! I ll! -1_/-r1_/ (0) Now consider a 99% bootstrap confidence for u. Before doing any computation, do you expect this confidence interval to be wider, narrower, or have the same width as the 95% confidence interval you computed in (b) ? I expect the 99% confidence interval for p to be ? 3 compared to the 95% confidence interval for [1. (cl) Copy and paste the following code into R/RStudio to bootstrap the sample: library(mosaic) RNGkindlsample.kind="Reiection"): set.seed(1);#so you will get the "same" random sampling as me B=do(1000) * mean(resample(c(0, 1, -6, -6, 157, -3, 178, -3, -10, 42, -2, 120, 5, 59, 0, -2), 16)); Now compute a 99% bootstrap confidence interval for u by copying the following code into R/RStudio quantile(B$mean,0.005); quantile(B$mean,O.995); The 99% confidence interval for p is: E S u S I ll! Not sure how to perform some of the tasks? Click to see similar problem In a cluster sample of 109 students taking Statistics 213, each student was asked if they support differential tuition fees. If a student did support differential fees, the response was coded with a "1". A "non support" was coded with a "0". The data is as follows: 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1 (a) The number of Statistics 213 students in this sample who "support" differential tuition fees - let's call it the random variable X - can be modeled with a binomial distribution. To find the value of X, rather than manually counting the number of '1's in our dataset, copy and paste the following R code into R/Rstudio: x=c(0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1); sum(x); The sum(x) function will sum all the values in our x dataset. Because our values are all either 0 or 1, this is the same as counting the number of 1's observed. This is our value of X. X = (b) Using the formula p + Za/2V P(1-P) n compute the 97% confidence interval for p, the proportion of all Statistics 213 students who support differential tuition fees. Use the code below to compute the interval bounds and report all decimals that R prints. 55/109-qnorm(0.985)*((55/109)*(1-55/109)/109)^.5 55/109+qnorm(0.985)*((55/109)*(1-55/109)/109)^.5 The 97% confidence interval for p is:(c) Now use R to compute a 97% bootstrap confidence interval. Copy and paste the following code into R/RStudio to bootstrap the sample: library(mosaic) RNGkind(sample.kind="Rejection"); set.seed(1);#so you will get the "same" random sampling as me B=do(1 000) * mean(resample(c(rep(1 .55),rep(0,109-55)). 109)): Compute a 97% bootstrapped confidence interval for p: Copy and paste the following code into R/Rstudio to compute the interval bounds: quantile(B$mean,0.015); quantile(B$mean,0.985); (d) Recently, an executive of the Student's Union (3U) was quoted saying that 30% of all U of C students support differential tuition fees. From your finding in part (b), is this figure supported? Complete the statement below. ? , the SU executive's claim ? c because the 97% confidence interval ? the value of Not sure how to perform some of the tasks? Click to see similar