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

Answer:

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

Step-by-step 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.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

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

This means that $n = 6458, \pi = \frac{754}{6458} = 0.1168$

95% confidence level

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

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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