anonymous
  • anonymous
I've got 2 questions that I'd appreciate some help with. I will grant a medal to those who help.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
In a sample of 100 households, the mean number of hours spent on social networking sites during the month of January was 50 hours. In a much larger study, the standard deviation was determined to be 6 hours. Assume the population standard deviation is the same. What is the 98% confidence interval for the mean hours devoted to social networking in January?
anonymous
  • anonymous
and Two studies were completed in Florida. One study in northern Florida involved 2,000 patients; 64% of them experienced flu-like symptoms during the month of December. The other study, in southern Florida, involved 3,000 patients; 54% of them experienced flu-like symptoms during the same month. Which study has the smallest margin of error for a 95% confidence interval?
anonymous
  • anonymous
@nincompoop @dan815 @jim_thompson5910 @ganeshie8

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anonymous
  • anonymous
@campbell_st
jim_thompson5910
  • jim_thompson5910
In a sample of 100 households, the mean number of hours spent on social networking sites during the month of January was 50 hours. In a much larger study, the standard deviation was determined to be 6 hours. Assume the population standard deviation is the same. What is the 98% confidence interval for the mean hours devoted to social networking in January? from that we can find n = 100 xbar = 50 sigma = 6
jim_thompson5910
  • jim_thompson5910
if the confidence level is 98%, then what is the critical value?
anonymous
  • anonymous
I'm not exactly sure honestly. If I were to be looking for critical value, how would I find it?
jim_thompson5910
  • jim_thompson5910
do you have a calculator?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
a TI calculator?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
hit 2nd, then the vars key
jim_thompson5910
  • jim_thompson5910
scroll down to invNorm
jim_thompson5910
  • jim_thompson5910
|dw:1435127861838:dw|
anonymous
  • anonymous
I just get x, y, z, t, and s. It's an online calculator.
jim_thompson5910
  • jim_thompson5910
98% confidence means that the area under the curve (shown below) is 0.98 |dw:1435127873474:dw|
jim_thompson5910
  • jim_thompson5910
1 - 0.98 = 0.02 0.02/2 = 0.01 is in the left tail |dw:1435127929596:dw|
jim_thompson5910
  • jim_thompson5910
so do you see how the area under the curve to the left of z = k is 0.99 ? |dw:1435127996415:dw|
anonymous
  • anonymous
alright, I see the curve
jim_thompson5910
  • jim_thompson5910
you said it's an online calculator?
anonymous
  • anonymous
well, the 0.99
anonymous
  • anonymous
yes?
jim_thompson5910
  • jim_thompson5910
what's the link to it
anonymous
  • anonymous
http://web2.0calc.com/
jim_thompson5910
  • jim_thompson5910
ok that calculator won't work
jim_thompson5910
  • jim_thompson5910
use this instead http://onlinestatbook.com/2/calculators/inverse_normal_dist.html
jim_thompson5910
  • jim_thompson5910
well I mean use it to find the value of k in that drawing
jim_thompson5910
  • jim_thompson5910
type in the area of 0.98 then click "between" leave the mean and std dev as they are
anonymous
  • anonymous
Thanks, I've been looking around you asked if I had a TI Calculator. I have a normal one but I guess that won't do. Alright, I've set it to .98. Now just set between to 6 and 50?
jim_thompson5910
  • jim_thompson5910
if you have TI83 or higher, then you can use that
anonymous
  • anonymous
no, it's just the basic calculator, nothing like a scientific one.
anonymous
  • anonymous
I couldn't find anything to emulate TI83 or anything else
jim_thompson5910
  • jim_thompson5910
ok, anyways, use that link I gave you to get what you see in the attached image
anonymous
  • anonymous
alright, so 2.327 is the critical value, where would this come in?
jim_thompson5910
  • jim_thompson5910
\[\large n = 100\] \[\large \overline{x} = 50\] \[\large \sigma = 6\] \[\large z_c = 2.327\] you will use the values above to construct the confidence interval (L,U) L = lower bound U = upper bound \[\Large L = \overline{x} - z_c*\frac{\sigma}{\sqrt{n}}\] \[\Large U = \overline{x} + z_c*\frac{\sigma}{\sqrt{n}}\]
anonymous
  • anonymous
thank you c:
jim_thompson5910
  • jim_thompson5910
no problem
anonymous
  • anonymous
I got 48.60 and 51.40 (for future users)
anonymous
  • anonymous
alright, as for the other question, would you like me to post it in a different question or would you like to help on this thread?
jim_thompson5910
  • jim_thompson5910
either way is fine
anonymous
  • anonymous
alright, so that you won't have to go back up or anything : Two studies were completed in Florida. One study in northern Florida involved 2,000 patients; 64% of them experienced flu-like symptoms during the month of December. The other study, in southern Florida, involved 3,000 patients; 54% of them experienced flu-like symptoms during the same month. Which study has the smallest margin of error for a 95% confidence interval?
jim_thompson5910
  • jim_thompson5910
Margin of Error (of a proportion): \[\Large z_c*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] \(\Large z_c\) is the critical value (based on the confidence level) \(\Large \hat{p} \) is the sample proportion
jim_thompson5910
  • jim_thompson5910
also, n = sample size
jim_thompson5910
  • jim_thompson5910
use the link I gave you to find zc (now at 95% confidence)
jim_thompson5910
  • jim_thompson5910
the p-hat values are the percentages of people who got sick
anonymous
  • anonymous
alright so critical is at 1.96, for northern Florida I get 2.1% and southern Florida is 1.8%. Out of the options, both "The southern Florida study with a margin of error of 1.8%." and "The northern Florida study with a margin of error of 2.1%." although southern Florida was 1.78%
jim_thompson5910
  • jim_thompson5910
Northern Florida Margin of Error \[\Large \large E = z_c*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] \[\Large \large E = 1.96*\sqrt{\frac{0.64(1-0.64)}{2000}}\] \[\Large \large E \approx 0.02103692753231 \approx 2.10\%\] ------------------------------------------------------------- Southern Florida Margin of Error \[\Large \large E = z_c*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] \[\Large \large E = 1.96*\sqrt{\frac{0.54(1-0.54)}{3000}}\] \[\Large \large E \approx 0.0178349230444 \approx 1.78\%\] ------------------------------------------------------------- so you have them both correct
anonymous
  • anonymous
well, 50/50 chance of getting this right, thank you for your help again c:
anonymous
  • anonymous
gah, it wasn't northern florida with 2.1%, I guess it was southern florida with 1.8% ._.
jim_thompson5910
  • jim_thompson5910
The only difference I see is 1.78% and 1.8%
jim_thompson5910
  • jim_thompson5910
well they wanted the smallest margin of error
jim_thompson5910
  • jim_thompson5910
sorry I forgot to mention that
anonymous
  • anonymous
it's alright, I got 80% of the questions completed correctly c:

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