Part A:
From the question, the population parameter of interest is the proportion of graduates at a mid-sized university who found a job within one year of completing their undergraduate degree.
The value of the point estimate of this parameter is given by [tex]\hat{p}= \frac{348}{400} =0.87[/tex]
Part B:
The conditions that must be met before constructing a confidence interval include:
The sample size is large.
[tex]npq=400(0.87)(1-0.87)=348(0.13)=45.24\ \textgreater \ 10[/tex]
Since, npq > 10, the conditions for constructing confidence interval are met.
Part C:
The 95%
confidence interval for the proportion of graduates who found a job
within one year of completing their undergraduate degree at this
university is given by:
[tex]95\% \ C.I.=\hat{p}\pm1.95\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ \\ =0.87\pm1.95\sqrt{\frac{0.87(1-0.87)}{400}}=0.87\pm1.95\sqrt{\frac{0.87(0.13)}{400}} \\ \\ =0.87\pm1.95\sqrt{\frac{0.1131}{400}}=0.87\pm1.95\sqrt{0.00028275}=0.87\pm1.95(0.0168)=0.87\pm0.03 \\ \\ (0.84,\ 0.90)[/tex]
Part D:
We are 95% confident that the proportion of graduates at a mid-sized university who found a job within one year of completing their undergraduate degree is between 0.84 and 0.90
Part E:
The 99%
confidence interval for the proportion of graduates who found a job
within one year of completing their undergraduate degree at this
university is given by:
[tex]99\% \
C.I.=\hat{p}\pm2.575\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ \\
=0.87\pm2.575\sqrt{\frac{0.87(1-0.87)}{400}}=0.87\pm2.575\sqrt{\frac{0.87(0.13)}{400}}
\\ \\
=0.87\pm2.575\sqrt{\frac{0.1131}{400}}=0.87\pm2.575\sqrt{0.00028275}=0.87\pm2.575(0.0168)=0.87\pm0.04
\\ \\ (0.83,\ 0.91)[/tex]
Part F:
The width of the 95% confidence interval is 0.06 while the width of the 99% confidence inteval is 0.08.
The 99% confidence interval is wider.
The wider the width of a confidence interval, the more confident we are that the required proportion is in the interval.
