Question

Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying within their school. They select a random sample of 240 students who are asked to complete a survey anonymously. Of these students, 54 report that they have experienced bullying. Use this information to find a 95% ???? ‑confidence interval for ???? , the true proportion of students at this school who have experienced bullying.

Answer 1

Answer:   $(0.273,\ 0.783)$

Step-by-step explanation:

Given : Total number of students who are asked to complete a survey anonymously : n= 240

Number of students

The proportion report that they have experienced bullying =54

Then , the proportion that students have experienced bullying =$p=\dfrac{54}{240}=0.225$

Significance interval : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

The confidence interval for the true proportion is given by :-

$p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=0.225\pm (1.96)\sqrt{\dfrac{(0.225)(1-0.225)}{240}}\\\\\approx0.225\pm0.0528\\\\=(0.255-0.528,\ 0.255+0.528)\\\\=(0.273,\ 0.783)$