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According to a Gallup poll of 1012 people, about one-third of (32%) of Americans keep a dog for protection. Find the:Margin of Error Incorrect% (round to the tenth of a percent)The 95% Confidence Interval Incorrect% to Incorrect% (round to the tenth of a percent)

Sagot :

  • The margin of error of the proportion given is of 2.9%.
  • Applying the margin of error, the confidence interval is (29.1%, 34.9%).

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The margin of error of a confidence interval of a proportion [tex]\pi[/tex] in a sample of size n, with a confidence level of [tex]1-\alpha[/tex], is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

The confidence interval is:

[tex]\pi \pm M[/tex]

In this problem:

  • Poll of 1012 people, thus [tex]n = 1012[/tex].
  • 32% keep a dog, thus [tex]\pi = 0.32[/tex]

95% confidence level

Thus [tex]\alpha = 0.05[/tex], z is the z-score that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].  

The margin of error is:

[tex]M = 1.96\sqrt{\frac{0.32(0.68)}{1012}} = 0.029[/tex]

As a percent, 2.9%, as 0.029 x 100% = 2.9%.

Now for the confidence interval, which is percentage plus/minus margin of error, thus:

[tex]32 - 2.9 = 29.1[/tex]

[tex]32 + 2.9 = 34.9[/tex]

The confidence interval is (29.1%, 34.9%).

A similar problem is given at https://brainly.com/question/16807970