Question

Suppose a national survey intends to identify the children experiencing physical bullying in a population of junior high school students in Texas. Out of a sample of 6,458, a total of 754 reported that they have experienced physical bullying. Calculate 95% confidence intervals for the proportion of the sample who have been bullied.

Answer 1

The 95% confidence intervals is (0.109, 0.125).

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)`.

Out of a sample of 6,458, a total of 754 reported that they have experienced physical bullying.

This means that
`n = 6458, \pi = (754)/(6458) = 0.1168`

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`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.1168 - 1.96\sqrt{(0.1168*0.8832)/(6458)} = 0.109`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.1168 + 1.96\sqrt{(0.1168*0.8832)/(6458)} = 0.125`

The 95% confidence intervals is (0.109, 0.125).