Question

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.\ calculate and interpret the standard error of \\(\hat{p}\\) where \\(\hat{p}\\) = the sample proportion of students who answered \yes.\
\\(se_{\hat{p}} = \\) \\(\boxed{\text{incorrect answer}}\\) (round to 4 decimal places.)
in \\(\boxed{\text{srss}}\\) of size \\(n = 172\\), the sample proportion of students at this university who would answer \yes\ will typically vary from \\(\boxed{\text{the true proportion}}\\) by about \\(\boxed{0.0239}\\).

Explanation:

Step1: Calculate sample proportion

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

Step2: Apply SE formula for proportion

$SE_{\hat{p}} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.1105(1-0.1105)}{172}}$

Step3: Compute the final value

$\sqrt{\frac{0.1105 \times 0.8895}{172}} = \sqrt{\frac{0.0983}{172}} \approx \sqrt{0.0005715} \approx 0.0239$

In simple random samples (SRSs) of size $n=172$, the sample proportion of students at this university who would answer "Yes" will typically vary from the true proportion by about 0.0239. This standard error quantifies the typical amount of sampling variability in the sample proportion.

Answer:

$SE_{\hat{p}} = 0.0239$