Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up.

1. What is the sample statistic and what is its distribution? Include the parameters of this distribution.
2. Construct a 97% confidence interval the population percent of drivers who always buckle up before riding in a car.
3. Interpret the above confidence interval
4. What is the minimum number you would need to survey to be 95% confident that the population proportion is estimated to within 0.03?

Answer 1

SolutionA:

sample proportion=p^=x/n=320/400=0.8=80%

here n=400 .large sample follows normal distribution with mean=p^=0.8

and standard deviation=sqrt(p^*(1-p^)/n

=sqrt(0.8*(1-0.8)/400)

=0.02

=2%

parameters are mean=80% and standard deviation=2%

Solution-b:

z critical in R

qnorm(0.985)=2.17009

97% confidence interval the population percent of drivers who always buckle up before riding in a car.

=p^-z*sqrt(p^*(1-p^)/n,p^+z*sqrt(p^*(1-p^)/n

=0.8-2.17009*sqrt(0.8*(1-0.8)/400),0.8+2.17009*sqrt(0.8*(1-0.8)/400)

=0.7565982,0.8434018

=0.7565982*100,0.8434018*100

97% lower limit for p=0.7565982=75.66%

97% upper limit for p=0.8434018=84.34%

Solution-c:

we are 97% confident that the true population percent of drivers who always buckle up before riding in a car. lies in between

75.66% and 84.34%

Solutiond:

p^=0.

z crit for 95%=196

E=0.03

required sample szie=n

nz^2*p*(1-p)/E^2

=(1.96^2*0.8*(1-0.8)/0.03^2

n=682.95

n=683

minimum number you would need to survey =683