A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Two studies were completed in California. One study in northern California involved 1,000 patients; 74% of them experienced flulike symptoms during the month of December. The other study, in southern California, involved 500 patients; 34% of them experienced flulike symptoms during the same month. Which study has the smallest margin of error for a 98% confidence interval?
anonymous
 one year ago
Two studies were completed in California. One study in northern California involved 1,000 patients; 74% of them experienced flulike symptoms during the month of December. The other study, in southern California, involved 500 patients; 34% of them experienced flulike symptoms during the same month. Which study has the smallest margin of error for a 98% confidence interval?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The northern California study with a margin of error of 3.2%. The southern California study with a margin of error of 3.2%. The northern California study with a margin of error of 4.9%. The southern California study with a margin of error of 4.9%.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Loser I dont get this that much

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think its southern california

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just write something and I will give you a medal @Loser66

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1hehhehe... I don't want medal, just interesting on the field I never know before. Here is the answer: the first one.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I made a short research about it. I learn how to find the answer out, not guess!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1if the confidence is 98% , then z =2.33 for the NOrth one the sample is 74%*1000= 740 hence p = 740/1000=0.74

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Apply the formula margin error = \(z*\sqrt{\dfrac{p(1p)}{population}}=2.33*\sqrt\dfrac{0.74(10.74}{1000}\ )

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1=3.23% the net is so bad.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.