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A researcher wishes to estimate the proportion of​ left-handed people among a certain population. In a random sample of 820820 people from the​ population, 22.722.7​% are​ left-handed. Find the margin of error for the​ 95% confidence interval for the population proportion. Round to four decimal places.

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Answer:

The margin of error for the​ 95% confidence interval for the population proportion is of 0.0287.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

The margin of error is:


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

95% confidence level

So
\alpha = 0.05, z is the value of Z that has a pvalue of
1 - (0.05)/(2) = 0.975, so
Z = 1.96.

Find the margin of error for the​ 95% confidence interval for the population proportion.

We have that
n = 820, p = 0.227

So


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


M = 1.96\sqrt{(0.227*0.773)/(820)} = 0.0287

The margin of error for the​ 95% confidence interval for the population proportion is of 0.0287.

User Raymond Tau
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