I've got 2 questions that I'd appreciate some help with. I will grant a medal to those who help.

- anonymous

I've got 2 questions that I'd appreciate some help with. I will grant a medal to those who help.

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- 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

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

@nincompoop @dan815 @jim_thompson5910 @ganeshie8

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## More answers

- anonymous

@campbell_st

- 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

if the confidence level is 98%, then what is the critical value?

- anonymous

I'm not exactly sure honestly. If I were to be looking for critical value, how would I find it?

- jim_thompson5910

do you have a calculator?

- anonymous

yes

- jim_thompson5910

a TI calculator?

- anonymous

yes

- jim_thompson5910

hit 2nd, then the vars key

- jim_thompson5910

scroll down to invNorm

- jim_thompson5910

|dw:1435127861838:dw|

- anonymous

I just get x, y, z, t, and s. It's an online calculator.

- jim_thompson5910

98% confidence means that the area under the curve (shown below) is 0.98
|dw:1435127873474:dw|

- jim_thompson5910

1 - 0.98 = 0.02
0.02/2 = 0.01 is in the left tail
|dw:1435127929596:dw|

- 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

alright, I see the curve

- jim_thompson5910

you said it's an online calculator?

- anonymous

well, the 0.99

- anonymous

yes?

- jim_thompson5910

what's the link to it

- anonymous

http://web2.0calc.com/

- jim_thompson5910

ok that calculator won't work

- jim_thompson5910

use this instead
http://onlinestatbook.com/2/calculators/inverse_normal_dist.html

- jim_thompson5910

well I mean use it to find the value of k in that drawing

- jim_thompson5910

type in the area of 0.98
then click "between"
leave the mean and std dev as they are

- 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

if you have TI83 or higher, then you can use that

- anonymous

no, it's just the basic calculator, nothing like a scientific one.

- anonymous

I couldn't find anything to emulate TI83 or anything else

- jim_thompson5910

ok, anyways, use that link I gave you to get what you see in the attached image

##### 1 Attachment

- anonymous

alright, so 2.327 is the critical value, where would this come in?

- 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

thank you c:

- jim_thompson5910

no problem

- anonymous

I got 48.60 and 51.40 (for future users)

- 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

either way is fine

- 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

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

also, n = sample size

- jim_thompson5910

use the link I gave you to find zc (now at 95% confidence)

- jim_thompson5910

the p-hat values are the percentages of people who got sick

- 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

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

well, 50/50 chance of getting this right, thank you for your help again c:

- anonymous

gah, it wasn't northern florida with 2.1%, I guess it was southern florida with 1.8% ._.

- jim_thompson5910

The only difference I see is 1.78% and 1.8%

- jim_thompson5910

well they wanted the smallest margin of error

- jim_thompson5910

sorry I forgot to mention that

- anonymous

it's alright, I got 80% of the questions completed correctly c:

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