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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.

Respuesta :

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]
Cricetus

The confidence interval will be "[tex]0.58 \pm 0.03489[/tex]".

Given values are:

  • [tex]X = 590[/tex]
  • [tex]N = 1016[/tex]

Now,

[tex]P = \frac{X}{N}[/tex]

   [tex]= \frac{590}{1016}[/tex]

   [tex]= 0.581[/tex]

→ [tex]\sigma p = \sqrt{\frac{\hat{p}(1- \hat{p})}{N} }[/tex]

By substituting the values, we get

→      [tex]= \frac{0.581(0.419)}{1016}[/tex]

→      [tex]= 0.015[/tex]

  • c = 0.98
  • [tex]z_{\frac{e}{2} } = 2.326[/tex]  

The confidence interval is:

= [tex]\hat{p} \pm 2.326(0.015)[/tex]

= [tex]0.58 \pm 0.03489[/tex]

Learn more:

https://brainly.com/question/19808090

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