Question

Bullies in middle school A media report claims that more than $75\%$of

middle school students engage in bullying behavior. A University of Illinois study on aggressive behavior surveyed a random sample of $558$middle school students. When asked to describe their behavior in the last $30$days, $445$students admitted that they had engaged in physical aggression, social ridicule, teasing, name-calling, and issuing threats —all of which would be classified as bullying. Do these data provide convincing evidence at the $\alpha =0.05$significance level that the media report’s claim is correct?

Short answer

There is enough convincing evidence that media report's claim is correct.

Step-by-step solution

Given Information

Given that $\alpha =0.05$

$n=558$

$x=445$

Explanation and Calculation

The claim can be null or alternative hypothesis.

Null: Population proportion is equal to value given in claim. If this is claim, alternate hypothesis is opposite of this.

$H_0:p=75\%=0.75$

$H_1:p>0.75$

The three conditions are:

Random: Sample is a random sample, this is met.

Independent: $558<10\%$of population of all students. It is also met.

Normal: $n{p}_0=558(0.75)=418.5$and $n\left(1-p_0\right)=558(1-0.75)=558(0.25)=139.5$both are greater than $10$.

All conditions are satisfied. Hypothesis test can be performed.

Hypothesis test

Sample proportion, $\hat{p}=\frac{x}{n}=\frac{445}{558}=0.7975$

Test Static: $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}=\frac{0.7975-0.75}{\sqrt{\frac{0.75(1-0.75)}{558}}}=2.59$

$P$is calculated as: $P=P(z>2.59)$

$=1-P(Z<2.59)$

$=1-0.9952$

$=0.0048$

$P<0.05\Rightarrow \text{Reject}{H}_0$

There is enough convincing evidence that claim by media is correct.