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Chapter 8: Q 36. (page 522)

Reporting cheating What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172172undergraduates at a large university: 鈥淵ou witness two students cheating on a quiz. Do you go to the professor?鈥� Only 19answered 鈥淵es.鈥� Assume the conditions for inference are met.

a. Determine the critical valueZ*for a 96%confidence interval for a proportion.

b. Construct a 96%confidence interval for the proportion of all undergraduates at this university who would go to the professor.

c. Interpret the interval from part (b).

Short Answer

Expert verified

a. The critical value iszc=za/2critical value=2.04

b. 96%confidence interval for proportion is 0.0617<p<0.1593

c. True population proportion is between0.0617and0.1593

Step by step solution

01

Given Information

It is given that x=19

n=172

Confidence Level =0.98

Level of significancea=0.04

02

Critical Value

Using Excel Formula

Using EXCEL FORMULA=ABS(NORMSINV(0.04/2))ZC=Za/2critical value=2.04

03

Constructing Confidence Interval

Sample Proportion p^=xn

p^=19172=0.1105

Margin of Error E=Z伪/2脳p^(1-p^)n

E=2.04脳0.1105(1-0.1105)172=0.0488

The confidence interval is (p^-E,p^+E)

(0.1105-0.0488,0.1105+0.0488)=(0.0617,0.1593)

The 96%confidence interval for population proportion is50.0617<p<0.1593

04

Interpreting Confidence Interval

There is 0.0617<p<0.1593confidence that correct population of all UG at this university going to professor lie between 0.0617and0.1593

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