A survey of 1010 college seniors working towards an undergraduate degree was conducted. Each student wasasked, ʺAre you planning or not planning to pursue a graduate degree?ʺ Of the 1010 surveyed, 658 stated thatthey were planning to pursue a graduate degree. Construct and interpret a 98% confidence interval for theproportion of college seniors who are planning to pursue a graduate degree.
A) $( 0.616,0.686 )$; we are $98 \%$ confident that the proportion of college seniors who are planning to pursue a graduate degree is between $0.616$ and $0.686$.
B) $( 0.620,0.682 )$; we are $98 \%$ confident that the proportion of college seniors who are planning to pursue a graduate degree is between $0.620$ and $0.682$.
C) $( 0.621,0.680 )$; we are $98 \%$ confident that the proportion of college seniors who are planning to pursue a graduate degree is between $0.621$ and $0.680$.
D) $( 0.612,0.690 )$; we are $98 \%$ confident that the proportion of college seniors who are planning to pursue a graduate degree is between $0.612$ and $0.690$. Construct a 95% Z-interval or a 95% t-interval about the population mean. Assume the data come from a population thatis approximately normal with no outliers.

Question

A survey of 1010 college seniors working towards an undergraduate degree was conducted. Each student wasasked, ʺAre you planning or not planning to pursue a graduate degree?ʺ Of the 1010 surveyed, 658 stated thatthey were planning to pursue a graduate degree. Construct and interpret a 98% confidence interval for theproportion of college seniors who are planning to pursue a graduate degree. 
A) $( 0.616,0.686 )$; we are $98 \%$ confident that the proportion of college seniors who are planning to pursue a graduate degree is between $0.616$ and $0.686$.
B) $( 0.620,0.682 )$; we are $98 \%$ confident that the proportion of college seniors who are planning to pursue a graduate degree is between $0.620$ and $0.682$.
C) $( 0.621,0.680 )$; we are $98 \%$ confident that the proportion of college seniors who are planning to pursue a graduate degree is between $0.621$ and $0.680$.
D) $( 0.612,0.690 )$; we are $98 \%$ confident that the proportion of college seniors who are planning to pursue a graduate degree is between $0.612$ and $0.690$. Construct a 95% Z-interval or a 95% t-interval about the population mean. Assume the data come from a population thatis approximately normal with no outliers.

Quizplus · Accepted Answer

The sample proportion is $ \hat{p} = \frac{658}{1010} \approx 0.651 $. Using the formula for a confidence interval for a proportion and a 98% confidence level, the interval is calculated as $ \hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $, where $ Z \approx 2.33 $ for 98% confidence. This results in the interval $(0.616, 0.686)$.

A Survey of 1010 College Seniors Working Towards an Undergraduate (0.616,0.686)( 0.616,0.686 )(0.616,0.686)

A Survey of 1010 College Seniors Working Towards an Undergraduate $( 0.616,0.686 )$