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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 鈥攁ll of which would be classified as bullying. Do these data provide convincing evidence at the =0.05significance level that the media report鈥檚 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 =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

Interpreting a P-value A student performs a test of H0:p=0.3H0:p=0.3versus Ha:p<0.3Ha:p<0.3and gets a P-value of 0.22The student says, "This means there is about a22%chance that the null hypothesis is true." Explain why the student's explanation is wrong.

After once again losing a football game to the archrival, a college鈥檚 alumni association conducted a survey to see if alumni were in favor of firing the coach. An SRS of 100 alumni from the population of all living alumni was taken, and 64 of the alumni in the sample were in favor of firing the coach. Suppose you wish to see if a majority of all living alumni is in favor of firing the coach. The appropriate standardized test statistic is

(a)z=0.64-0.50.64(0.36)100z=0.64-0.50.64(0.36)100

role="math" localid="1654432946823" (b)t=0.64-0.50.64(0.36)100t=0.64-0.50.64(0.36)100

(c)z=0.64-0.50.5(0.5)100z=0.64-0.50.5(0.5)100

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Do you Tweet? The Pew Internet and American Life Project asked a random sample of U.S. adults, 鈥淒o you ever 鈥 use Twitter or another service to share updates about yourself or to see updates about others?鈥 According to Pew, the resulting 95% confidence interval is (0.123, 0.177).11 Based on the confidence interval, is there convincing evidence that the true proportion of U.S. adults who would say they use Twitter or another service to share updates differs from 0.17? Explain your reasoning.

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observations from a Normal population. The t statistic is t=2.25

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b. falls between 0.02 and 0.04.

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A random sample of 100 likely voters in a small city produced 59 voters in favor of Candidate A. The observed value of the standardized test statistic for performing a test of H0:p=0.5H0:p=0.5versus Ha:p>0.5Ha:p>0.5

is which of the following?

a)z=0.59-0.50.59(0.41)100

b)z=0.59-0.50.5(0.5)100

c)z=0.5-0.590.59(0.41)100

d)z=0.5-0.590.5(0.5)100

e)z=0.59-0.5100

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