anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The 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
  • anonymous
@DanJS
anonymous
  • anonymous
@Hero

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@Loser66
anonymous
  • anonymous
Loser I dont get this that much
anonymous
  • anonymous
@vera_ewing
anonymous
  • anonymous
I think its southern california
anonymous
  • anonymous
Just write something and I will give you a medal @Loser66
anonymous
  • anonymous
What ever you want
Loser66
  • Loser66
hehhehe... I don't want medal, just interesting on the field I never know before. Here is the answer: the first one.
Loser66
  • Loser66
I made a short research about it. I learn how to find the answer out, not guess!
anonymous
  • anonymous
Help me though
Loser66
  • Loser66
if 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
  • Loser66
Apply the formula margin error = \(z*\sqrt{\dfrac{p(1-p)}{population}}=2.33*\sqrt\dfrac{0.74(1-0.74}{1000}\ )
Loser66
  • Loser66
=3.23% the net is so bad.
anonymous
  • anonymous
Thanks

Looking for something else?

Not the answer you are looking for? Search for more explanations.