Question

In a survey of 900 US individuals regarding the more dangerous drivers, it was found that teenagers account for 63%, followed by people over 75 years old who account for 33% and those who express no opinion account for 4%. Construct a 99% confidence interval for the proportion of individuals who think that teenagers are the more dangerous drivers

Answer 1

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

n = 900

P = x/n = 0.63

Confidence level = 99%

Critical Z value = 2.5758

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.63 ± 2.5758* sqrt(0.63*(1 – 0.63)/900)

Confidence Interval = 0.63 ± 2.5758*0.0161

Confidence Interval = 0.63 ± 0.0415

Lower limit = 0.63 - 0.0415 =0.5885

Upper limit = 0.63 + 0.0415 = 0.6715

Confidence interval = (0.5885, 0.6715)