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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
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