| title: “Confidence Intervals for a Proportion - DATA 606 2020 Fall” |
| author: “Evan McLaughlin” |
date: “10.14.2020” |
knitr::opts_chunk$set(eval = TRUE, results = FALSE, fig.show = "show", message = FALSE)
library(tidyverse)
library(openintro)
library(ggplot2)
library(dplyr)
library(infer)
library(knitr)
library(png)Exercise 5.7 on page 187
In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”. However, this value was based on a sample, so it may not be a perfect estimate for the population parameter of interest on its own. The study reported a standard error of about 1.2%, and a normal model may reasonably be used in this setting.
Create a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions.
The formula for calculating confidence intervals is as follows:
## point estimate ± z⋆ × SE
Z Score Table
## 95% of the data is within 1.96 standard deviations of the mean. The three pertinent values are
## point estimate = 45
## z⋆ = 1.96
## SE = 1.2
45 + (1.96 * 1.2)
45 - (1.96 * 1.2)