Question

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. ​95% confidence; nequals=​2388, xequals=1672

Answer 1

The margin of error is of 0.0184 = 1.84%.

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`.

In this question:

`\pi = (1672)/(2388) = 0.7, n = 2388`

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

`M = 1.96\sqrt{(0.7*0.3)/(2388)}`

`M = 0.0184`

The margin of error is of 0.0184 = 1.84%.