Question

1. Please generalize ci95.t to ci.t so that we can calculate confidence interval at any confidence level given a sample using the KORGER.csv data

The function prototype looks like,

ci.t <- function(n, mu, sd_mu, conf)

{
            ##your code goes here

}

The sample outputs need to be replicated are,

setwd("…")

kg <- read.csv("KORGER.csv")

kd <- split(kg$Distance, kg$Team)$KOREA

n <- length(kd); m <- mean(kd); s <- sd(kd)/sqrt(n);

>ci.t(n, m, s, 0.95)

95%CILower 95%CIUpper

  8448.693 11307.852

>ci.t(n,m,s,0.90)

90%CILower 90%CIUpper

  8715.393 11041.152

>ci.t(n,m,s,0.99)

99%CILower 99%CIUpper

  7844.859 11911.686

I've got this so far.....

kg <- read.csv("KORGER.csv")
kd <- split(kg$Distance, kg$Team)$KOREA

n <- length(kd); m <- mean(kd); s <- sd(kd)/sqrt(n);

ci.t <- function(n, mu, sd_mu)
{
   lower <- round(mu - qt(0.975, n-1)*sd_mu, 3);
   upper <- round(mu + qt(0.975, n-1)*sd_mu, 3);
   ci <- c(1,lower,1,upper)
   names(ci) <- c("%CILower ","%CIUpper")
   ci
  
}
ci.t(n,m,s,0.95)

Answer 1

The Central Limit Theorem states that the sampling distribution of the sample means approaches a normal distribution the sample size gets larger — no matter what the shape of the population distribution. This fact holds especially true for sample sizes over 30. All this is saying is that as you take more samples, especially large ones, your graph of the sample means will look more like a normal distribution.

An essential component of the Central Limit Theorem is that the average of your sample means will be the population mean. Similarly, if you find the average of all of the standard deviations in your sample, you’ll find the actual standard deviation for your population.

Thus

Using the CLT, the required distribution is approximately Normally distributed. Furthermore,

• The mean of the distribution is = 54 months
• The standard deviation is

b) Assuming that the scientist’s claim is true, what the probability the audit group’s sample has a mean life of 52 or fewer months?

If the population standard deviation is known, we can transform the sample mean to an approximately standard normal variable, z score, Z:

Thus, the probability that the audit group will observe the sample mean life of 52 or less is 0.0094. If the 50 tested animals do have a mean of 52 or fewer months, the audit group will have strong evidence that the scientist's claim is untrue because the chance of that happening is very small.

R code for plotting and calculating probability

mean=54; sd=0.85
ub = 52;
x <- seq(-4,4,length=36)*sd + mean
hx <- dnorm(x,mean,sd)
plot(x, hx, type="n", xlab="Lifespan", ylab="",
main="Normal Distribution", axes=FALSE)
i <- x >= lb & x <= ub
lines(x, hx)
polygon(c(lb,x[i],ub), c(0,hx[i],0), col="red")

area <- pnorm(ub, mean, sd)
result <- paste("P("," X <=",ub,") =",
signif(area, digits=3))
mtext(result,3)
axis(1, at=seq(51.45, 56.55, 0.85), pos=0)