Business statistics

Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 407 drivers and find that 307 claim to always buckle up. Construct a 91% confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5] 

Answer 1

Hi Salma,

The formula for this type of confidence interval is:

CI=p-hat +/- z*SE

Let’s break this down:

p-hat=sample proportion

z*=z critical value

SE=standard error

First things first:

p-hat=307/407

p-hat=0.754

Z*=1.695 for 91% (FYI this confidence level is rare in practice)

Now, standard error has its own formula, so we need to compute that. Formula is:

SE=sqrt[(p-hat(1-p-hat)/n]

So we have:

SE=sqrt[(0.754*0.246)/407]

SE=0.021

Now, we have everything we need to compute our confidence interval:

p-hat=0.754

z*=1.695

SE=0.021

CI=0.754 +/- (1.695*0.021)

CI=[0.718, 0.790]

I hope this helps.