Question

In a survey of 529 travelers, 386 said that location was very important and 323 said that room quality was very important in choosing a hotel.

1. Construct a 90% confidence interval estimate for the population proportion of travelers who said that location was very important for choosing a hotel.
2. Construct a 90% confidence interval estimate for the population proportion of travelers who said that room quality was very important for choosing a hotel.

Answer 1

Solution :

Given that,

n = 529

x = 386 location

a)

= x / n = 386/529 = 0.730

1 - = 1 - 0.730 = 0.27

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * 0.730 * 0.27 / 529

= 0.318

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.730 - 0.318 < p < 0.730 + 0.318

0.412 < p < 1.048

(0.412,1.048)

The 90% confidence interval from 0.412 to 1.048

b) x = 323 Quality

= x / n = 323/529 = 0.611

1 - = 1 - 0.611 = 0.389

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * 0.611 * 0.389 / 529

= 0.035

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.611 - 0.035 < p < 0.611 + 0.035

0.576 < p < 0.646

(0.576,0.646)

The 90% confidence interval from 0.576 to 0.646