Question

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.

Answer 1

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.