Joan conducted a study to see how common binge drinking is on her college campus. She defined “frequent binge drinking” as having five or more drinks in a row three or more times in the past two weeks. Out of 593 students who replied to her survey, 64 fit this criterion. What is the 99% confidence interval for this population proportion? Answer choices are rounded to the hundredths place.

I used the function 1-proportion z-interval on the calculator, where I inputted

Successes(x) = 64Sample size(n) = 593Confidence Level(C level) = 0.99

It would result in zInterval_1Prop 64,593,0.99: stat.results, where the values are shown below as:

Lower bound (CLower) = 0.075105Upper bound (CUpper) = 0.140747test statistic (p^) = 0.107926Margin of error (ME) = 0.032821

Therefore, the 99% confidence interval would be around 0.11 ± 0.03 or range from 0.08 to 0.14.

Note: not sure if this is correct O_o

Iwas in a argument about a certain class grade with my teacher. it was a ‘a-’when i felt that i deserved a ‘a’. i showed him (my teacher) why i felt i derserved so. i questioned what i got wrong and showed him what i learned. i look back to that teacher with nothing but love and support because that’s the class that i actually learned the most. the moral of the story is to see where the other person is coming from and what they have done. also, when in a conflict, never attack the persons character. this will just make the argument end poorly.

Where are the startments?