Question

A simple random sample of 64 8th graders at a large suburban middle school indicated that 88% of them are involved with some type of after school activity. find the margin of error associated with a 98% confidence interval that estimates the proportion of them that are involved in an after school activity.

Answer 1

The margin of error of a 98% confidence interval is given by


$ME=2.33\left(\sqrt{ \frac{p(1-p)}{n} }\right) \\ \\ =2.33\left(\sqrt{\frac{0.88(1-0.88)}{64}}\right) \\ \\ =2.33\left(\frac{0.1056}{64}\right)=2.33(0.0406) \\ \\ =0.0946$