MTH 119-04 (9:30am)

Section 6.1: Confidence Intervals

Example: Salary of Workers

Load the data set workers:

workers<-read.file("/home/emesekennedy/Data/Ch5/workers.txt")

## Reading data with read.table()

The mean salary for the entire population of 1000 workers is

mu<-mean(~salary, data=workers)
mu

## [1] 23600.19

The standard deviation of salary for the entire population of 1000 workers is

s<-sd(~salary,data=workers)
s

## [1] 19269.82

Create 1000 simple random samples of size 100, and record the mean \(\bar{x}\) for each of the samples:

samples<-do(1000)*mean(~salary,data=sample(workers, 100, replace=T))

Rename the column name in samples:

names(samples)[1]<-"xbar"

Find \(z^*\) for the 95% confidence interval

qdist("norm", mean=0, sd=1, c(.025, .975))

## [1] -1.959964  1.959964

So \(z^*\) is approximately 1.96. Compute the margin of error:

m<-1.96*s/10
m

## [1] 3776.885

Add a column to samples that has the lower bound for the 95% confidence intervals for each of the 1000 samples:

samples<-transform(samples, lower=xbar-m)

Add a column to samples that has the upper bound for the 95% confidence intervals for each of the 1000 samples:

samples<-transform(samples, upper=xbar+m)

Create a subset with the samples for which the population mean \(\mu\) lies within the computed 95% confidence intervals:

samples2<-subset(samples, lower<mu & upper>mu)

Output the number of observations in this subset using the command

nrow(samples2)

## [1] 955

So 955 out of the 1000 samples, or 95.5% of the samples had the population mean lie in the 95% confidence interval.