Question

An insurance company collects data on​ seat-belt use among drivers in a country. Of

2000 drivers

30 -39 years​ old,

23​%

said that they buckle​ up, whereas

482

of

1800

drivers

55-64

years old said that they did. Find a

90​%

confidence interval for the difference between the proportions of​ seat-belt users for drivers in the age groups

30-39

years and

55-64

years.

Construct a

90​%

confidence interval.

Answer 1

solution:-

given that n1 = 2000 ,

p1 = 23% = 0.23

and

n2 = 1800 here x2 = 482

then p2 = x2/n2 = 482/1800 = 0.27

the value of 90% confidence from z table is 1.645

confidence interval formula for the difference between the proportions

=> (p1-p2) +/-z * sqrt(p1(1-p1)/n1 + p2(1-p2)/n2)

=> (0.23-0.27) +/- 1.645 * sqrt(0.23*(1-0.23)/2000 + (0.27*(1-0.27)/1800))

=> (-0.0632 to -0.0168)

=> (-0.063 to -0.017) round to three decimal places