MATH SOLVE

4 months ago

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]

[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]