Question

car insurance company suspects that the younger the driver is, the more reckless a driver he/she is. They take a survey and group their respondents based on age. Of 217 respondents between the ages of 16-25 (Group 1), 183 claimed to wear a seat belt at all times. Of 398 respondents who were 26+ years old (Group 2), 322 claimed to wear a seat belt at all times. Find a 95% confidence interval for the difference in proportions. Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group? Enter the confidence interval - round to 3 decimal places.

Answer 1

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Here, , n1 = 217 , n2 = 398
p1cap = 0.8433 , p2cap = 0.809

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.8433 * (1-0.8433)/217 + 0.809*(1-0.809)/398)
SE = 0.0316

For 0.99 CI, z-value = 2.58
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.8433 - 0.809 - 2.58*0.0316, 0.8433 - 0.809 + 2.58*0.0316)
CI = (-0.047 , 0.116)

As 0 is included in the above CI, there is no difference in the proportion of the two groups.