Question

Reporting cheating What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of $172$$172$undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” Only $19$answered “Yes.” Assume the conditions for inference are met.

a. Determine the critical value$Z^{*}$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

a. The critical value is${\mathrm{z}}_{\mathrm{c}}={\mathrm{z}}_{\mathrm{a}/2}\text{critical value}=2.04$

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

c. True population proportion is between$0.0617\text{and}0.1593$

Step-by-step solution

Given Information

It is given that $x=19$

$n=172$

Confidence Level $=0.98$

Level of significance$a=0.04$

Critical Value

Using Excel Formula

Using EXCEL FORMULA$=\mathrm{ABS}\left(\mathrm{NORMSINV}\right(0.04/2\left)\right)$${\mathrm{Z}}_{\mathrm{C}}={\mathrm{Z}}_{\mathrm{a}/2}\text{critical value}=2.04$

Constructing Confidence Interval

Sample Proportion $\hat{p}=\frac{x}{n}$

$\hat{p}=\frac{19}{172}=0.1105$

Margin of Error $E=Z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$E=2.04\times \sqrt{\frac{0.1105(1-0.1105)}{172}}=0.0488$

The confidence interval is $(\hat{p}-E,\hat{p}+E)$

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

The $96\%$confidence interval for population proportion is$50.0617<p<0.1593$

Interpreting Confidence Interval

There is $0.0617<p<0.1593$confidence that correct population of all UG at this university going to professor lie between $0.0617\text{and}0.1593$