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 304 students who are asked to complete a survey anonymously. Of these students, l 14 report that they have experienced bullying. Use this information to find a 90% z-confidence interval for p, the true proportion of students at this school who have experienced bullying. 

Give the limits (bounds) of the confidence interval as a proportion, precise to three decimal places. 

lower limit=_______ 

upper limit=_______ 

Answer 1

$\mathbf{given:}\;x=\mathbf{114},\;n=\mathbf{304}$

$\mathbf{for\; 90\%\; confidence \;level,\; z_{\frac{\alpha }{2}}=1.645}$

$\mathbf{sample \;proportion\;(\hat p)}=\frac{x}{n}=\frac{114}{304}=\mathbf{0.375}$

$\mathbf{formula:}\;Confidence \;interval \;for \;population\; proportion \;p$

$=\hat p \pm z_{\frac{\alpha }{2}}*\left \{ \sqrt{\frac{\hat p*(1-\hat p)}{n}} \right \}$

$=0.375 \pm 1.645*\left \{ \sqrt{\frac{0.375 *(1-0.375 )}{304}} \right \}$

$=0.375 \pm 0.046$

$\mathbf{lower\;limit}=0.375 - 0.046=\mathbf{{\color{Red} 0.329}}$

$\mathbf{upper\;limit}=0.375+ 0.046=\mathbf{{\color{Red} 0.421}}$