34. Reporting cheating What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of $172$undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” Only$19$ answered “Yes.”
(a) Identify the population and parameter of interest.
(b) Check conditions for constructing a confidence interval for the parameter.
(c) Construct a $99\%$confidence interval for$p$. Show your method.
(d) Interpret the interval in context.

To identify the population and parameter of interest.

The complete set of people being investigated is referred to as the population. As a result, the overall population for the sample is made up of all undergraduate quiz participants.
The examined subject is the parameter of interest. The population proportion of all undergraduate quiz participants is the parameter of relevance here.
As a result, the total population for the sample is all undergraduate quiz participants, and the population proportion is all undergraduate quiz participants.

To check conditions for constructing a confidence interval for the parameter.

Three conditions must be met in order to find a confidence interval.
Random, Independent, and Normal are the conditions. The requirement of randomness is met because the sample was drawn at random among university students.
The requirement for independence is satisfied because the sample size is less than $10\%$ of the population size.
The poll found $19$ success stories and $153$ failure stories.
Because the odds of success and failure are both bigger than ten.
The typical condition is met.
As a result, all three confidence interval conditions are met.

To construct a $99\%$confidence interval for $p$.

Determine the sample proportion as follows:
Sample proportion$=\frac{\text{samplesize}}{\text{total populatiom}}$
$\hat{p}=\frac{x}{n}$
$=\frac{19}{172}$
$=0.1105$

Since, the confidence level is $99\%$.

Convert $99\%$into decimal as follows:

$\frac{99}{100}=0.99$

According to the table,

$z_{o/2}=z_{0.005}$

$=2.575$

Determine the margin of error as follows:

Margin of error $=\mathrm{Z}\alpha /2\times \sqrt{\frac{\text{sample size(1-samplesize)}}{\text{total population}}}$
$E=z_{q/2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
$=2.575\times \sqrt{\frac{0.1105(1-0.1105)}{172}}$
$\approx 0.0616$
The amount of margin of error might change the sample proportion.
As a result, from the smallest to the largest sample percentage, the confidence that contains true proportion varies.
$\hat{p}-E<p<\hat{p}+E$
$0.1105-0.0616<p<0.1105+0.0616$
$0.0489<p<0.1721$
As a result, the amount of confidence interval that contain the true population proportion is from $0.0489$to $0.1721$.

To interpret the interval in context. The population proportion of students likely to report cheating by other students has a confidence interval of $99$ percent.

Determine the sample proportion as follows:
Sample proportion $=\frac{\text{samplesize}}{\text{total population}}$
$\hat{p}=\frac{x}{n}$

$=\frac{19}{172}$

$=0.1105$

Since, the confidence level is $99\%$.

Then, convert $99\%$into decimal as follows:

$\frac{99}{100}=0.99$

According to the table:

$z_{\alpha /2}=z_{0.005}$

$=2.575$

Determine the margin of error as follows:
Margin of error $=\mathrm{Z}\alpha /2\times \sqrt{\frac{\text{sample size(1-samplesize)}}{\text{total population}}}$
$E=z_{q/2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
$=2.575\times \sqrt{\frac{0.1105(1-0.1105)}{172}}$
$\approx 0.0616$
The amount of margin of error might change the sample proportion.
As a result, from the smallest to the largest sample percentage, the confidence that contains true proportion varies.
$\hat{p}-E<p<\hat{p}+E$
$0.1105-0.0616<p<0.1105+0.0616$
$0.0489<p<0.1721$
As a result, the confidence interval is $99\%$, the population fraction of students likely to report cheating by other students ranges from $0.0489$to $0.1721$.