Question

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)

Answer 1

• 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 $\pi$ in a sample of size n, with a confidence level of $1-\alpha$, is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

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

The confidence interval is:

$\pi \pm M$

In this problem:

• Poll of 1012 people, thus $n = 1012$.
• 32% keep a dog, thus $\pi = 0.32$

95% confidence level

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

The margin of error is:

$M = 1.96\sqrt{\frac{0.32(0.68)}{1012}} = 0.029$

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:

$32 - 2.9 = 29.1$

$32 + 2.9 = 34.9$

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

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