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Bullies in middle school 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 said their behavior included physical aggression, social ridicule, teasing, name-calling, and issuing threats. This behavior was not defined as bullying in the questionnaire. Is this evidence that more than three-quarters of the students at that middle school engage in bullying behavior? To find out, Maurice decides to perform a significance test. Unfortunately, he made a few errors along the way. Your job is to spot the mistakes and correct them.

$\begin{array}{r}{H}_0:p=0.75\\ H_a:\stackrel{-}{p}>0.797\end{array}$

where p= the true mean proportion of middle school students who engaged in bullying.

- A random sample of 558 middle school students was surveyed.

- 558(0.797)=444.73 is at least 10.

$z=\frac{0.75−0.797}{\sqrt{\frac{0.797\left(0.203\right)}{445}}}=−2.46;\text{P-value}=2\left(0.0069\right)=0.0138$

The probability that the null hypothesis is true is only 0.0138, so we reject $H_0$. This proves that more than three-quarters of the school engaged in bullying behavior.

Step by step solution

01

Introduction

The scientific technique requires that one can test it. Scientists for the most part base scientific speculations on previous observations that can't satisfactorily be explained with the available scientific theories

02

Explanation Part (1)

The sample size is n= $558$

Behaviour included physical aggression, social ridicule, teasing, name-calling, and issuing threats is x = $445$

$\stackrel{-}{p}=\frac{x}{n}=\frac{445}{558}=0.797$

Calculating the null and alternative hypotheses,

$\begin{array}{r}{H}_0:p=0.75\\ H_a:p>0.75\end{array}$

On a random selection of students,

$n{p}_0=558×0.75=418$

$n{p}_0â‰\yen 10$

$n\left(1−p_0\right)=558×(1−0.75)=140$

$n\left(1−p_0\right)â‰\yen 10$

03

Explanation Part (2)

Using,

$Z=\frac{\stackrel{-}{p}−p_0}{\sqrt{\frac{p_0\left(1−p_0\right)}{n}}}$

$Z=\frac{0.797−0.75}{\sqrt{\frac{0.75(1−0.75)}{558}}}=\frac{0.047}{0.018}=2.6$

The p-value is $2.6$

$P(Z<2.6)=0.9953$

$P(Zâ‰\yen 2.6)=1−P(Z<2.6)$

$=1−0.9953=0.0047$

Hence it is less than the significance level $0.05$.

More than three-quarters of the school engaged in bullying behavior from above

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