Question

By law, all new cars must be equipped with airbags on both the driver’s side and the passenger’s side. There is concern, however, whether the airbags pose a danger to children seated on the passenger side. In a study of 55 deadly accidents among the people killed by the explosive force of airbags, 35 were children seated on the passenger side. Let p be the actual proportion of fatal motor vehicle accidents involving children seated on the passenger side. Construct a 99% confidence interval for p.

A) [0.509; 0.763] B) [0.197; 1.531] C) [0.991; 1.034] D) [0.469; 0.803] E) [0.981; 1.019]

Answer 1

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E Z/2 *(( * (1 - )) / n)

= 1.645 *((0.) / )

E =

A 90% confidence interval for population proportion p is ,

- E < p < + E

< p <

< p <

The 90% confidence interval for the population proportion p is :

Solution :

Given that,

n = 55

x = 35

Point estimate = sample proportion = = x / n = 35/55=0.636

1 -   = 1- 0.636 =0.364

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

  Margin of error = E = Z/2   * (((( * (1 - )) / n)

= 2.576* (((0.636*0.364) / 55)

E = 0.1671

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.636-0.1671 < p < 0.636+0.1671

0.469< p < 0.803

The 99% confidence interval for the population proportion p is : 0.469, 0.803