** Hwk-5.13_Chapter-5_Inference_Daniel_Thonn **
This Assignment requires the following:
1). R-Studio 2). R
Packages are used as shown in the Packages section
Setting up and Preparing the Environment
Packages Section Install the packages required
#Prepatory and Package installations
#install.packages(c('openintro','OIdata','devtools','ggplot2','psych','reshape2',
# 'knitr','markdown','shiny'))
#install.packages("devtools")
#devtools::install_github("jbryer/DATA606")
#require(knitr)
#library(ggplot2)
#library(utils)
# install.packages('dplyr')
#library(dplyr)
# install.packages('DATA606')
# library(DATA606)5.13 Car insurance savings. A market researcher wants to evaluate car insurance savings at a competing company. Based on past studies he is assuming that the standard deviation of savings is $100. He wants to collect data such that he can get a margin of error of no more than $10 at a 95% confidence level. How large of a sample should he collect?
Problem Summary:
Find sample size = n
ME (Margin of Error) <= 10 CI (Confidence Interval) = 95% sd (standard deviation) = 100
Formulas: ME = z* x SE
For 95% > Z* = 1.96 sd
SE = sd / ((n)^.5)
Therefore:
ME = z* x SE
SE = ME/z*
sd / ((n)^.5) = ME/z*
((n)^.5) = sd/ME/Z*
n = (sd/ME/Z*)^2
n = (100/10/1.96)^2
# ME = = z* x SE
#n <- (100/10/1.96)^2
n <- 19.6^2
# For ME <=10, then n gets larger
round(n,2) ## [1] 384.16#Ans: For ME <= 10 then n (sample size) >= 384.16 >= 385 END