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 558middle school students. When asked to describe their behavior in the last 30days, 445students 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 α=0.05significance level that the media report’s claim is correct?

Short Answer

Expert verified

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

Step by step solution

01

Given Information

Given that α=0.05

n=558

x=445

02

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.

H0:p=75%=0.75

H1: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: np0=558(0.75)=418.5and n1-p0=558(1-0.75)=558(0.25)=139.5both are greater than 10.

All conditions are satisfied. Hypothesis test can be performed.

03

Hypothesis test

Sample proportion, p^=xn=445558=0.7975

Test Static: z=p^-p0p01-p0n=0.7975-0.750.75(1-0.75)558=2.59

Pis calculated as: P=P(z>2.59)

=1-P(Z<2.59)

=1-0.9952

=0.0048

P<0.05RejectH0

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

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Most popular questions from this chapter

Home computers Jason reads a report that says 80%of U.S. high school

students have a computer at home. He believes the proportion is smaller than 0.80at his large rural high school. Jason chooses an SRS of 60students and finds that 41have a computer at home. He would like to carry out a test at the α=0.05significance level of H0:p=0.80versus Ha:p<0.80, where p= the true

proportion of all students at Jason’s high school who have a computer at home. Check if the conditions for performing the significance test are met.

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a. Construct and interpret a 95% confidence interval for the true proportion p of all teens in the state who passed their driving test on the first attempt. Assume that the conditions for inference are met.

b. Explain why the interval in part (a) provides more information than the test in Exercise 51.

Which of the following 95%confidence intervals would lead us to reject H0:p=0.30in favor of Ha:pnotequalto0.30at the 5%significance level?

a. (0.19,0.27)

b.(0.24,0.30)

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d. (0.29,0.31)

e. None of these

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