Q:

In 2001, a Gallup poll surveyed 1016 households in the U.S. about their pets. Of those surveyed, 590 said the had at least one dog or cat as a pet. Construct and interpret a 99% interval for the proportion of households who had at least one dog and one cat.

Accepted Solution

A:
The confidence interval is 
[tex]0.58\pm0.04[/tex].  

This means that if we take repeated samples, 99% of the intervals would contain the population proportion.

To construct this interval, we use
[tex]p\pm z*\sigma_p[/tex], 
where 
[tex]sigma_p=\sqrt{\frac{p(1-p)}{N}}[/tex]

Since 590/1016 said they had a cat and a dog, p=0.581 and N=1016:
[tex]\sigma_p=\sqrt{\frac{0.581(1-0.581)}{1016}}=\sqrt{\frac{0.581(0.419)}{1016}}=0.015.[/tex]

We need the z-score associated with this confidence level:
Convert 99% to a decimal:  99/100 = 0.99
Subtract from 1:  1-0.95 = 0.01
Divide by 2:  0.01/2 = 0.005
Subtract from 1:  1-0.005 = 0.995

Using a z-table, we see that this value is equally distant from z=2.57 and z=2.58, so we will use z=2.575:
[tex]p\pm 2.575(0.015)=0.58\pm0.04[/tex]