Solved – Confidence intervals for frequency tables

bootstrapconfidence intervalr

I am analyzing the results of a survey on R. The questionnaire is a series of questions that participants answer using a Likert scale (https://en.wikipedia.org/wiki/Likert_scale).

I have obtained frequency tables for each question (i.e. what percentage of participants chose "strongly agree", what percentage chose "somewhat agree," etc.) and I am now interested in obtaining confidence intervals for those percentages.

Since I don't want to assume that my data is normally distributed, I was thinking of using bootstrapping to get the confidence intervals. However, I am not entirely sure how using bootstrapping works in this context. I am familiar with how to use bootstrapping when dealing with means, but not really when dealing with frequency tables such as these. Specifically, I am not sure how to go about coding the bootstrapping.

Thank you,
Best,

Best Answer

There are different methods for calculating confidence intervals for proportions without using bootstrapping.

For a multinomial proportion, you might try the methods in the DescTools package.

### Adapted from http://rcompanion.org/handbook/H_02.html

if(!require(DescTools)){install.packages("DescTools")}

library(DescTools)

SA = 10
A  = 9
N  = 20
D  = 5
SD = 1

observed = c(SA, A, N, D, SD)

MultinomCI(observed,
           conf.level=0.95,
           method="sisonglaz")

### Methods: "sisonglaz", "cplus1", "goodman"

   ###              est     lwr.ci    upr.ci
   ### [1,] 0.22222222 0.08888889 0.3807871
   ### [2,] 0.20000000 0.06666667 0.3585648
   ### [3,] 0.44444444 0.31111111 0.6030093
   ### [4,] 0.11111111 0.00000000 0.2696759
   ### [5,] 0.02222222 0.00000000 0.1807871

Population $R^2$

I'm firstly trying to understand the definition of the population R-squared.

Quoting your comment:

Or you could define it asymptotically as the proportion of variance explained in your sample as your sample size approaches infinity.

I think you mean this is the limit of the sample $R^2$ when one replicates the model infinitely many times (with the same predictors at each replicate).

So what is the formula for the asymptotic value of the sample $R^²$ ? Write your linear model $\boxed{Y=\mu+\sigma G}$ as in https://stats.stackexchange.com/a/58133/8402, and use the same notations as this link.
Then one can check that the sample $R^2$ goes to $\boxed{popR^2:=\dfrac{\lambda}{n+\lambda}}$ when one replicates the model $Y=\mu+\sigma G$ infinitely many times.

As example:

> ## design of the simple regression model lm(y~x0)
> n0 <- 10
> sigma <- 1
> x0 <- rnorm(n0, 1:n0, sigma)
> a <- 1; b <- 2 # intercept and slope
> params <- c(a,b)
> X <- model.matrix(~x0)
> Mu <- (X%*%params)[,1]
> 
> ## replicate this experiment k times 
> k <- 200
> y <- rep(Mu,k) + rnorm(k*n0)
> # the R-squared is:
> summary(lm(y~rep(x0,k)))$r.squared 
[1] 0.971057
> 
> # theoretical asymptotic R-squared:
> lambda0 <- crossprod(Mu-mean(Mu))/sigma^2
> lambda0/(lambda0+n0)
          [,1]
[1,] 0.9722689
> 
> # other approximation of the asymptotic R-squared for simple linear regression:
> 1-sigma^2/var(y)
[1] 0.9721834

Population $R^2$ of a submodel

Now assume the model is $\boxed{Y=\mu+\sigma G}$ with $H_1\colon\mu \in W_1$ and consider the submodel $H_0\colon \mu \in W_0$.

Then I said above that the population $R^2$ of model $H_1$ is $\boxed{popR^2_1:=\dfrac{\lambda_1}{n+\lambda_1}}$ where $\boxed{\lambda_1=\frac{{\Vert P_{Z_1} \mu\Vert}^2}{\sigma^2}}$ and $Z_1=[1]^\perp \cap W_1$ and then one simply has ${\Vert P_{Z_1} \mu\Vert}^2=\sum(\mu_i - \bar \mu)^2$.

Now do you define the population $R^2$ of the submodel $H_0$ as the asymptotic value of the $R^2$ calculated with respect to model $H_0$ but under the distributional assumption of model $H_1$ ? The asymptotic value (if there is one) seems more difficult to find.

Solved – Confidence Interval of Categorical Data with Multiple responses

If there is just one category of interest, e.g. lemons, there's no problem with "binarizing" it and extracting a CI for the population proportion who like lemons. However, if you do this for each category, you might end up with something misleading because the numbers of responses for each category are dependent and you are making multiple probability statements, and so the probability that each proportion lies in your interval could be very different than you might expect. You could report a "confidence region," or "simultaneous confidence intervals" that capture this dependency better. See Sison and Glaz.

Best Answer

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Population $R^2$

Population $R^2$ of a submodel

Solved – Confidence Interval of Categorical Data with Multiple responses

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