Construct a 94% confidence interval for the true proportion of the students that approve the way the president is handling the job.

1. Procedure: Select an answer One mean T procedure One variance χ² procedure One mean Z procedure One proportion Z procedure
2. Assumptions: (select everything that applies)
   • Population standard deviation is unknown
   • Normal population
   • Simple random sample
   • The number of positive and negative responses are both greater than 10
   • Population standard deviation is known
   • Sample size is greater than 30
3. Unknown parameter: Select an answer σ², population variance p, population proportion μ, population mean  
4. Point estimate: Select an answer sample proportion, p̂ sample mean, x̄ sample variance, s² = (Round the answer to 3 decimal places)
5. Confidence level % and $\displaystyle \alpha=$ , also
   •  $\displaystyle \frac{\alpha}{{2}}=$ , and $\displaystyle {1}-\frac{\alpha}{{2}}=$
   • Critical values: (Round the answer to 3 decimal places)
     • left= right=
6. Margin of error (if applicable): (Round the answer to 3 decimal places)
7. Lower bound: (Round the answer to 3 decimal places)
8. Upper bound: (Round the answer to 3 decimal places)
9. Confidence interval:(, )
10. Interpretation: We are % confident that the true population proportion is between % and %.