Komputasi Statistik

1 Brief Introduction

Please watching this video, to get some ideas about Confidence Intervals (CI)

2 CI in Business

This video guide you, how can you apply Confidence Intervals in Business.

3 Your Exercise

In this section, your expected to get familiar with confidential intervals exercise:

3.1 Exercise 1

Find a point estimate of average university student Age with the sample data from survey!

# Survey data from MASS package
library(MASS)

# The point estimate of average university student `Age`
mean(survey$Age)

## [1] 20.37451

From MASS package we get the survey data. After that, we use mean function to get the point estimate of age student.

3.2 Exercise 2

Assume the population standard deviation \(\sigma\) of the student Age in data survey is 7. Find the margin of error and interval estimate at 95% confidence level.

Age.resp = na.omit(survey$Age)        # filter out missing values in Age
n = length(Age.resp)                  # assign the length of response
sigma = 7                             # population standard deviation 
sem = sigma/sqrt(n)                   # standard error of the mean
E = qnorm(.975)*sem ;E                # margin of error (upper tail 95% of CI)

## [1] 0.8911934

xbar <- mean(Age.resp); xbar

## [1] 20.37451

xbar + c(-E, E)

## [1] 19.48332 21.26571

First we have to use na.omit function to filter out missing values in Age. After that, we can find the standard error of the mean from sigma/sqrt(n). Finally we can get the error value by multiplying 1.96 or qnorm(0.975) with standard error of the mean .

3.3 Exercise 3

Without assuming the population standard deviation \(\sigma\) of the student Age in survey, find the margin of error and interval estimate at 95% confidence level.

Age.resp = na.omit(survey$Age)        # filter out missing values in Age
n = length(Age.resp)                  # assign the length of response
s = 7                                 # sample standard deviation 
SE = s/sqrt(n)                        # standard error estimate
E = qt(.975, df=n-1)*SE; E            # margin of error (upper tail 95% of CI)

## [1] 0.8957872

xbar <- mean(Age.resp); xbar

## [1] 20.37451

xbar + c(-E, E)

## [1] 19.47873 21.27030

First we have to use na.omit function to filter out missing values in Age. After that, we can find the standard error estimate from s/sqrt(n). Now we can get the margin of error by multiplying qt(.975, df=n-1) with standard error estimate.

3.4 Exercise 4

Improve the quality of a sample survey by increasing the sample size with unknown standard deviation \(\sigma\)!.

Please explain something from your exercise result.

3.5 Exercise 5

Assume you don’t have planned proportion estimate, find the sample size needed to achieve 5% margin of error for the male student survey at 95% confidence level!

gender.response = na.omit(survey$Sex)
n = length(gender.response)
male = sum(gender.response == "Male"); male

## [1] 118

pbar  = male/n;pbar

## [1] 0.5

zstar = qnorm(0.975)
p=0.5

# Margin or error
E = 0.05
zstar^2*p*(1-p)/E^2

## [1] 384.1459

Please explain something from your exercise result.

3.6 Exercise 6

Perform confidence intervals analysis on this data set from 2004 that includes data on average hourly earnings, marital status, gender, and age for thousands of people.