Question

2. 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

a) x = 386

Point estimate = sample proportion = = x / n = 386 / 529 = 0.730

1 - = 1 - 0.730 = 0.270

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645(((0.730 * 0.270) / 529)

= 0.032

A 90% confidence interval for population proportion p is ,

± E   

= 0.730 ± 0.032

= ( 0.698, 0.762 )

b) x = 323

Point estimate = sample proportion = = x / n = 323 / 529 = 0.611

1 - = 1 - 0.611 = 0.389

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 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   

= 0.611 ± 0.035

= ( 0.576, 0.646 )