Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car.
.25/(.03^2/1.96^2)
## [1] 1067.111
# To guarantee an error bound less than .03, we need to sample at least 1067.11 people. ie. need at least 1068 people.
\[n \geq \frac{.25}{\frac{EB^2}{z^2}} = \frac{.25 \cdot z^2}{EB^2}\]
We would need to survey more people because of the above formula.
Suppose that the insurance companies did do a survey. They randomly surveyed \(400\) drivers and found that \(320\) claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.
Explain why the conditions for inference are satisfied.
#1. Step 1: Collect known values
alpha <- .05
phat <- 320/400
n <- 400
se <- sqrt(phat*(1-phat)/n)
#2. Add and subtract the error bound to phat
EB <- qnorm(.975)*se
phat + EB
## [1] 0.8391993
phat - EB
## [1] 0.7608007
#3. Interpret results within the context of the problem
# We are 95% confident that the true proportion of drivers that always buckle up is between 76.1% and 83.9%.
We have to use \(s\), the sample standard deviation as a substitute.
The sampling distribution of sample means does not follow: \(N(\mu,\sigma/ \sqrt{n})\)
Instead we use the T-distribution.
Recall that we use \(s\) to denote the sample standard deviation.
\[T\text{-score} = \frac{\text{observation}-\text{expected}}{s}\]
Image source: OpenIntro
# Still need normality and independence conditions.