Question

In a survey of 5,000 travelers, 2000 said the location was very important in choosing an airline. Construct a 99% confidence interval estimate for the population proportion of travelers who said the reputation was very important in choosing an airline.

Answer 1

Solution :

Given that,

n = 5000

x = 2000

Point estimate = sample proportion = = x / n = 2000/5000 = 0.4

1 - = 1 - 0.4 = 0.6

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 (((0.4 * 0.6) /5000 )

= 0.0178

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.4 - 0.0178< p < 0.4+0.0178

0.3822< p < 0.4178

The 99% confidence interval for the population proportion p is : (0.3822 , 0.4178)