Data606 Lab 4b
Ritesh Lohiya
March 16, 2018
Foundations for statistical inference - Confidence intervals Sampling from Ames, Iowa If you have access to data on an entire population, say the size of every house in Ames, Iowa, it’s straight forward to answer questions like, “How big is the typical house in Ames?” and “How much variation is there in sizes of houses?”. 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.
The data In the previous lab, ``Sampling Distributions’’, we looked at the population data of houses from Ames, Iowa. Let’s start by loading that data set.
Set my working directory as C:-library.46064b
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionload("ames.RData")
head(ames)## Order PID MS.SubClass MS.Zoning Lot.Frontage Lot.Area Street Alley
## 1 1 526301100 20 RL 141 31770 Pave <NA>
## 2 2 526350040 20 RH 80 11622 Pave <NA>
## 3 3 526351010 20 RL 81 14267 Pave <NA>
## 4 4 526353030 20 RL 93 11160 Pave <NA>
## 5 5 527105010 60 RL 74 13830 Pave <NA>
## 6 6 527105030 60 RL 78 9978 Pave <NA>
## Lot.Shape Land.Contour Utilities Lot.Config Land.Slope Neighborhood
## 1 IR1 Lvl AllPub Corner Gtl NAmes
## 2 Reg Lvl AllPub Inside Gtl NAmes
## 3 IR1 Lvl AllPub Corner Gtl NAmes
## 4 Reg Lvl AllPub Corner Gtl NAmes
## 5 IR1 Lvl AllPub Inside Gtl Gilbert
## 6 IR1 Lvl AllPub Inside Gtl Gilbert
## Condition.1 Condition.2 Bldg.Type House.Style Overall.Qual Overall.Cond
## 1 Norm Norm 1Fam 1Story 6 5
## 2 Feedr Norm 1Fam 1Story 5 6
## 3 Norm Norm 1Fam 1Story 6 6
## 4 Norm Norm 1Fam 1Story 7 5
## 5 Norm Norm 1Fam 2Story 5 5
## 6 Norm Norm 1Fam 2Story 6 6
## Year.Built Year.Remod.Add Roof.Style Roof.Matl Exterior.1st Exterior.2nd
## 1 1960 1960 Hip CompShg BrkFace Plywood
## 2 1961 1961 Gable CompShg VinylSd VinylSd
## 3 1958 1958 Hip CompShg Wd Sdng Wd Sdng
## 4 1968 1968 Hip CompShg BrkFace BrkFace
## 5 1997 1998 Gable CompShg VinylSd VinylSd
## 6 1998 1998 Gable CompShg VinylSd VinylSd
## Mas.Vnr.Type Mas.Vnr.Area Exter.Qual Exter.Cond Foundation Bsmt.Qual
## 1 Stone 112 TA TA CBlock TA
## 2 None 0 TA TA CBlock TA
## 3 BrkFace 108 TA TA CBlock TA
## 4 None 0 Gd TA CBlock TA
## 5 None 0 TA TA PConc Gd
## 6 BrkFace 20 TA TA PConc TA
## Bsmt.Cond Bsmt.Exposure BsmtFin.Type.1 BsmtFin.SF.1 BsmtFin.Type.2
## 1 Gd Gd BLQ 639 Unf
## 2 TA No Rec 468 LwQ
## 3 TA No ALQ 923 Unf
## 4 TA No ALQ 1065 Unf
## 5 TA No GLQ 791 Unf
## 6 TA No GLQ 602 Unf
## BsmtFin.SF.2 Bsmt.Unf.SF Total.Bsmt.SF Heating Heating.QC Central.Air
## 1 0 441 1080 GasA Fa Y
## 2 144 270 882 GasA TA Y
## 3 0 406 1329 GasA TA Y
## 4 0 1045 2110 GasA Ex Y
## 5 0 137 928 GasA Gd Y
## 6 0 324 926 GasA Ex Y
## Electrical X1st.Flr.SF X2nd.Flr.SF Low.Qual.Fin.SF Gr.Liv.Area
## 1 SBrkr 1656 0 0 1656
## 2 SBrkr 896 0 0 896
## 3 SBrkr 1329 0 0 1329
## 4 SBrkr 2110 0 0 2110
## 5 SBrkr 928 701 0 1629
## 6 SBrkr 926 678 0 1604
## Bsmt.Full.Bath Bsmt.Half.Bath Full.Bath Half.Bath Bedroom.AbvGr
## 1 1 0 1 0 3
## 2 0 0 1 0 2
## 3 0 0 1 1 3
## 4 1 0 2 1 3
## 5 0 0 2 1 3
## 6 0 0 2 1 3
## Kitchen.AbvGr Kitchen.Qual TotRms.AbvGrd Functional Fireplaces
## 1 1 TA 7 Typ 2
## 2 1 TA 5 Typ 0
## 3 1 Gd 6 Typ 0
## 4 1 Ex 8 Typ 2
## 5 1 TA 6 Typ 1
## 6 1 Gd 7 Typ 1
## Fireplace.Qu Garage.Type Garage.Yr.Blt Garage.Finish Garage.Cars
## 1 Gd Attchd 1960 Fin 2
## 2 <NA> Attchd 1961 Unf 1
## 3 <NA> Attchd 1958 Unf 1
## 4 TA Attchd 1968 Fin 2
## 5 TA Attchd 1997 Fin 2
## 6 Gd Attchd 1998 Fin 2
## Garage.Area Garage.Qual Garage.Cond Paved.Drive Wood.Deck.SF
## 1 528 TA TA P 210
## 2 730 TA TA Y 140
## 3 312 TA TA Y 393
## 4 522 TA TA Y 0
## 5 482 TA TA Y 212
## 6 470 TA TA Y 360
## Open.Porch.SF Enclosed.Porch X3Ssn.Porch Screen.Porch Pool.Area Pool.QC
## 1 62 0 0 0 0 <NA>
## 2 0 0 0 120 0 <NA>
## 3 36 0 0 0 0 <NA>
## 4 0 0 0 0 0 <NA>
## 5 34 0 0 0 0 <NA>
## 6 36 0 0 0 0 <NA>
## Fence Misc.Feature Misc.Val Mo.Sold Yr.Sold Sale.Type Sale.Condition
## 1 <NA> <NA> 0 5 2010 WD Normal
## 2 MnPrv <NA> 0 6 2010 WD Normal
## 3 <NA> Gar2 12500 6 2010 WD Normal
## 4 <NA> <NA> 0 4 2010 WD Normal
## 5 MnPrv <NA> 0 3 2010 WD Normal
## 6 <NA> <NA> 0 6 2010 WD Normal
## SalePrice
## 1 215000
## 2 105000
## 3 172000
## 4 244000
## 5 189900
## 6 195500In this lab 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.
population <- ames$Gr.Liv.Area
samp <- sample(population, 60)Exercise 1: 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(population)
hist(samp)
Answer: The distribution is unimodal and right skewed. The typical size should be around 1500 square feet. The typical size is the mean of the sample which is a point estimate of the true population mean.
Exercise 2: Would you expect another student’s distribution to be identical to yours? Would you expect it to be similar? Why or why not?
Answer: Anothersstudent’s distribution will not be identical but similar. The distribution of a sample of size 60 is representative of the population distribution because the sample is random but it is a small sample of a large population so it will vary.
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)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 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 (See Section 4.2.3 if you are unfamiliar with this formula).
se <- sd(samp) / sqrt(60)
lower <- sample_mean - 1.96 * se
upper <- sample_mean + 1.96 * se
c(lower, upper)## [1] 1347.775 1602.792This 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.
Exercise 3: For the confidence interval to be valid, the sample mean must be normally distributed and have standard error s/n?????????. What conditions must be met for this to be true?
Answer: The sample is random(independent) and the size is greater than 30.
Confidence levels
Exercise 4: What does “95% confidence” mean? If you’re not sure, see Section 4.2.2.
Answer: A 95% confidence interval means you can be 95% certain that the true mean of a population is within the range calculated.
In this case we have the luxury of knowing the true population mean since we have data on the entire population. This value can be calculated using the following command:
mean(population)## [1] 1499.69Exercise 5: Does your confidence interval capture the true average size of houses in Ames? If you are working on this lab in a classroom, does your neighbor’s interval capture this value?
Answer: No, it does fall within the 95% confidence interval. Iam not working in a class room, so iam not sure about the other samples.
Exercise 6: Each student in your class should have gotten a slightly different confidence interval. What proportion of those intervals would you expect to capture the true population mean? Why? If you are working in this lab in a classroom, collect data on the intervals created by other students in the class and calculate the proportion of intervals that capture the true population mean.
Answer: 95% of the students will capture the true mean in their interval(values that are +/- 1.96 times away from the standard error)
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 <- 60Now 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
}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] 1333.284 1532.916On your own
- Using the following function (which was downloaded with the data set), plot all intervals. What proportion of your confidence intervals include the true population mean? Is this proportion exactly equal to the confidence level? If not, explain why.
plot_ci(lower_vector, upper_vector, mean(population))
Answer: Out of fifty confidence intervals, there are 3 intervals that do not include the true population mean. Yes this is to the confidence level.
- Pick a confidence level of your choosing, provided it is not 95%. What is the appropriate critical value?
Answer: Lets choose 84
critical84 <- qnorm(.84)
critical84## [1] 0.9944579- 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_84 <- samp_mean - critical84 * samp_sd / sqrt(n)
upper_vector_84 <- samp_mean + critical84 * samp_sd / sqrt(n)
c(lower_vector_84[1], upper_vector_84[1])## [1] 1382.456 1483.744plot_ci(lower_vector_84, upper_vector_84, mean(population))
This is a product of OpenIntro that is released under a Creative Commons Attribution-ShareAlike 3.0 Unported. This lab was written for OpenIntro by Andrew Bray and Mine Çetinkaya-Rundel.