Confidence Intervals

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.

cps04 <- read.csv("cps04.csv", header = T, sep = ",")

# Average Hourly Earnings
ahe.resp = na.omit(cps04$ahe)        
n = length(ahe.resp)                  
sigma = sd(ahe.resp)                  
sem = sigma/sqrt(n)                   

xbar <- mean(ahe.resp); xbar

## [1] 16.7712

xbar + c(-E, E)

## [1] 16.7212 16.8212

# Marital Status
mar.response = na.omit(cps04$bachelor)
n = length(mar.response)
m = sum(mar.response == "1"); m

## [1] 3640

sigma = sd(mar.response)
SE=sigma/sqrt(n)
E= qnorm(0.975)*SE; E

## [1] 0.01092388

xbar <- mean(mar.response); xbar

## [1] 0.4557976

xbar+c(-E,E)

## [1] 0.4448738 0.4667215

# Female
female.response = na.omit(cps04$female)
n = length(female.response)
f = sum(female.response == "1"); f

## [1] 3313

sigma = sd(female.response)
SE=sigma/sqrt(n)
E= qnorm(0.975)*SE; E

## [1] 0.01080662

xbar <- mean(female.response); xbar

## [1] 0.414851

xbar+c(-E,E)

## [1] 0.4040444 0.4256576

# Age
age.resp = na.omit(cps04$age)         
n = length(age.resp)                  
sigma = sd(age.resp)                  
sem = sigma/sqrt(n)                   
E = qnorm(.975)*sem ;E

## [1] 0.06340892

xbar <- mean(age.resp); xbar

## [1] 29.75445

xbar + c(-E, E)

## [1] 29.69104 29.81785

Please explain something from your exercise result.

4 Case Study

Assume you have access to data on an entire population, say the size of every house in all residential home sales in Ames, Iowa between 2006 and 2010 it’s straight forward to answer questions like,

• How big is the typical house in Ames?
• How much variation is there in sizes of houses?.
• How much is the average price of house in Ames?
• How much is the confidence interval price of house in Ames?

But, If you have access to only a sample of the population, as is often the case, the task becomes more complicated. What is your best guess for the typical size if you only know the sizes of several dozen houses? This sort of situation requires that you use your sample to make inference on what your population looks like.

4.1 Collect Data

To access the data in R, type the following code:

download.file("http://www.openintro.org/stat/data/ames.RData", destfile = "ames.RData")
load("ames.RData")

In this case study we’ll start with a simple random sample of size 60 from the population. Specifically, this is a simple random sample of size 60. Note that the data set has information on many housing variables, but for the first portion of the lab we’ll focus on the size of the house, represented by the variable Gr.Liv.Area.

#randomly set seed to fix outputs in this assignment
set.seed(0)
population <- ames$Gr.Liv.Area
samp <- sample(population, 60)
samp

##  [1] 2200 2093 1040 2233 1523 1660 1555 1102  848 1136 2061 1122  960 1092 2610
## [16] 2217 1959 2334 1660 1576  848 2004  988 1500  874 1340 1800 1069 1456  784
## [31]  985 1928  882 1124 1639 1214 1434 1150 1544 1812 1511 1949 1077 1248 1480
## [46] 1320 1717 1367  928 2552 1953  693 2690 2276 1173 1258 2582 1558  672 1488

4.2 Visualization

As usual, before you begin to analyze more about your data. It’s important to visualize the data in advance. Here, we use a random sample of size 60 from the population.

# Histogram
library(moments)
hist(samp, breaks = 20, col = 'pink')

# Make a histogram of your sample
hist(samp, main ="Distribution fo Samp", 
     col = "deeppink3", 
     xlim = c(200, 3500), 
     freq = F,
     xlab = "Samp")
# ...and add a density curve
curve(dnorm(x, 
            mean=mean(samp), 
            sd=sd(samp)), add=T, 
            col="blue", lwd=2)

Your Challenge:

• Describe the distribution of your sample. What would you say is the “typical” size within your sample? Also state precisely what you interpreted “typical” to mean.

hist(samp, breaks = 10)

summary(samp)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##     672    1100    1484    1514    1933    2690

mean(population)

## [1] 1499.69

# The distribution in this sample case seems to be right skewed. I would not consider this is a “typical” sample size while in this case if we compared to our original population mean of 1499.6904437 we noticed that it is “considerable” near. A “typical” mean is the value that is near the most concurrent values in a distribution.

• Would you expect another student’s distribution to be identical to yours? Would you expect it to be similar? Why or why not?

No, this is a random sample of 60 randomly selected observations. The question here will be how similar is similar? or how near is near?, then if the values are near then yes, it could happen.

4.3 Confidence Intervals

One of the most common ways to describe the typical or central value of a distribution is to use the mean. In this case we can calculate the mean of the sample using,

sample_mean <- mean(samp)
sample_mean

## [1] 1514.133

Return for a moment to the question that first motivated this lab: based on this sample, what can we infer about the population? Based only on this single sample, the best estimate of the average living area of houses sold in Ames would be the sample mean, usually denoted as \(\bar{x}\) (here we’re calling it sample_mean). That serves as a good point estimate but it would be useful to also communicate how uncertain we are of that estimate. This can be captured by using a confidence interval.

We can calculate a 95% confidence interval for a sample mean by adding and subtracting 1.96 standard errors to the point estimate (I assume that you have been familiar with this formula).

se <- sd(samp) / sqrt(60)
lower <- sample_mean - 1.96 * se
upper <- sample_mean + 1.96 * se
c(lower, upper)

## [1] 1381.351 1646.915

This is an important inference that we’ve just made: even though we don’t know what the full population looks like, we’re 95% confident that the true average size of houses in Ames lies between the values lower and upper. There are a few conditions that must be met for this interval to be valid.

Your Challenge:

• For the confidence interval to be valid, the sample mean must be normally distributed and have standard error $ $. What conditions must be met for this to be true?

The sample must consists of at least 30 independent observations and the data should not be strongly skewed, then the distribution of the sample mean is well approximated by a normal model.

• What does “95% confidence” mean?

Is the confidence interval level for the normal model with standard error SE. The confidence interval for the population parameter is point estimate ± z.SE where corresponds to the confidence level selected.

• Does your confidence interval capture the true average size of houses in Ames? If you are working on this case study, does your classmate’s interval capture this value?

Yes, the above confidence interval capture the true average size of houses in Ames. I would expect my classmate to capture the mean value on their lab as well.

4.4 Simulation

let’s simulate a scenario of confidence interval in classroom to capture the true average size of houses in Ames. Suppose we have 100 students in the classroom.

count = 0
for (i in 1:100) {
  samp <- sample(population,60)
  samp_mean<- mean(samp)
  se <- sd(samp)/sqrt(60)
  lower <- samp_mean-1.96*se
  upper <- samp_mean+1.96*se
  if ((lower <= 1499.69) & (upper >= 1499.69)) {
    count = count+1
  }  
}
count

## [1] 97

Using R, we’re going to recreate many samples to learn more about how sample means and confidence intervals vary from one sample to another. Loops come in handy here (If you are unfamiliar with loops, review the Sampling Distribution Lab).

Here is the rough outline:

• Obtain a random sample.
• Calculate and store the sample’s mean and standard deviation.
• Repeat steps (1) and (2) 50 times.
• Use these stored statistics to calculate many confidence intervals.

But before we do all of this, we need to first create empty vectors where we can save the means and standard deviations that will be calculated from each sample. And while we’re at it, let’s also store the desired sample size as \(n\).

samp_mean <- rep(NA, 50)
samp_sd <- rep(NA, 50)
n <- 60
n

## [1] 60

Now we’re ready for the loop where we calculate the means and standard deviations of 50 random samples.

for(i in 1:50){
  samp <- sample(population, n) # obtain a sample of size n = 60 from the population
  samp_mean[i] <- mean(samp)    # save sample mean in ith element of samp_mean
  samp_sd[i] <- sd(samp)        # save sample sd in ith element of samp_sd
}
samp

##  [1] 2206 3672 2270 1786 1041 2614 1655 1378 1250 1884 1358  764 1176 1595 1419
## [16] 1620 1299 1097 1073 1647 1220 1086 1928 1412 1091 2263 1968 1261 1538  793
## [31] 1337 1768 1604 1609 1479  980  480  816  951 1069 1709 1742 2237 1458  864
## [46] 1665 1778 1949 1040 1414  954 1142 1614 1368 5642 1383 1242  816 2082 1728

Lastly, we construct the confidence intervals.

lower_vector <- samp_mean - 1.96 * samp_sd / sqrt(n) 
upper_vector <- samp_mean + 1.96 * samp_sd / sqrt(n)

Lower bounds of these 50 confidence intervals are stored in lower_vector, and the upper bounds are in upper_vector. Let’s view the first interval.

c(lower_vector[1], upper_vector[1])

## [1] 1400.415 1718.352

# confidential interval visualization
plot_ci(lower_vector, upper_vector, mean(population))

# For a 95% confidence interval, the critical value is -1.959964 and 1.959964.
qnorm((1-0.95)/2)

## [1] -1.959964

qnorm((1+0.95)/2)

## [1] 1.959964

Your Challenge:

• What proportion of your confidence intervals include the true population mean? Is this proportion exactly equal to the confidence level? If not, explain why.

Vector <- data.frame(lower_vector, upper_vector)
meanp <- mean(population)

left <- sum(Vector$upper_vector < meanp)
right <- sum(Vector$lower_vector > meanp)

noMeanIncluded <- left + right
noMeanIncluded

## [1] 2

proportion <- round(noMeanIncluded/n ,2)
proportion

## [1] 0.03

In this case only 97% include the population mean. This proportion is not necesarily the same as our confidence level but very near approximation of it.

• Pick a confidence level of your choosing, provided it is 99%. What is the appropriate critical value?

Let’s pick 99%. For this, the critical value will be 0.99.

• Calculate 50 confidence intervals at the confidence level you chose in the previous question. You do not need to obtain new samples, simply calculate new intervals based on the sample means and standard deviations you have already collected. Using the plot_ci function, plot all intervals and calculate the proportion of intervals that include the true population mean. How does this percentage compare to the confidence level selected for the intervals?

lower_vector <- samp_mean - 0.99 * samp_sd / sqrt(n) 
upper_vector <- samp_mean + 0.99 * samp_sd / sqrt(n)

plot_ci(lower_vector, upper_vector, mean(population))

Vector <- data.frame(lower_vector, upper_vector)
meanp <- mean(population)

left <- sum(Vector$upper_vector < meanp)
right <- sum(Vector$lower_vector > meanp)

noMeanIncluded <- left + right
noMeanIncluded

## [1] 17

proportion <- round(noMeanIncluded/n ,2)
proportion

## [1] 0.28

In this case only 99% include the population mean. This proportion is not necesarily the same as our confidence level but very near approximation of it.