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anonymous

  • one year ago

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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  1. anonymous
    • one year ago
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    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?

  2. anonymous
    • one year ago
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    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?

  3. anonymous
    • one year ago
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    @nincompoop @dan815 @jim_thompson5910 @ganeshie8

  4. anonymous
    • one year ago
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    @campbell_st

  5. jim_thompson5910
    • one year ago
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    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

  6. jim_thompson5910
    • one year ago
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    if the confidence level is 98%, then what is the critical value?

  7. anonymous
    • one year ago
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    I'm not exactly sure honestly. If I were to be looking for critical value, how would I find it?

  8. jim_thompson5910
    • one year ago
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    do you have a calculator?

  9. anonymous
    • one year ago
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    yes

  10. jim_thompson5910
    • one year ago
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    a TI calculator?

  11. anonymous
    • one year ago
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    yes

  12. jim_thompson5910
    • one year ago
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    hit 2nd, then the vars key

  13. jim_thompson5910
    • one year ago
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    scroll down to invNorm

  14. jim_thompson5910
    • one year ago
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    |dw:1435127861838:dw|

  15. anonymous
    • one year ago
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    I just get x, y, z, t, and s. It's an online calculator.

  16. jim_thompson5910
    • one year ago
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    98% confidence means that the area under the curve (shown below) is 0.98 |dw:1435127873474:dw|

  17. jim_thompson5910
    • one year ago
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    1 - 0.98 = 0.02 0.02/2 = 0.01 is in the left tail |dw:1435127929596:dw|

  18. jim_thompson5910
    • one year ago
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    so do you see how the area under the curve to the left of z = k is 0.99 ? |dw:1435127996415:dw|

  19. anonymous
    • one year ago
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    alright, I see the curve

  20. jim_thompson5910
    • one year ago
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    you said it's an online calculator?

  21. anonymous
    • one year ago
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    well, the 0.99

  22. anonymous
    • one year ago
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    yes?

  23. jim_thompson5910
    • one year ago
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    what's the link to it

  24. anonymous
    • one year ago
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    http://web2.0calc.com/

  25. jim_thompson5910
    • one year ago
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    ok that calculator won't work

  26. jim_thompson5910
    • one year ago
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    use this instead http://onlinestatbook.com/2/calculators/inverse_normal_dist.html

  27. jim_thompson5910
    • one year ago
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    well I mean use it to find the value of k in that drawing

  28. jim_thompson5910
    • one year ago
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    type in the area of 0.98 then click "between" leave the mean and std dev as they are

  29. anonymous
    • one year ago
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    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?

  30. jim_thompson5910
    • one year ago
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    if you have TI83 or higher, then you can use that

  31. anonymous
    • one year ago
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    no, it's just the basic calculator, nothing like a scientific one.

  32. anonymous
    • one year ago
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    I couldn't find anything to emulate TI83 or anything else

  33. jim_thompson5910
    • one year ago
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    ok, anyways, use that link I gave you to get what you see in the attached image

  34. anonymous
    • one year ago
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    alright, so 2.327 is the critical value, where would this come in?

  35. jim_thompson5910
    • one year ago
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    \[\large n = 100\] \[\large \overline{x} = 50\] \[\large \sigma = 6\] \[\large z_c = 2.327\] you will use the values above to construct the confidence interval (L,U) L = lower bound U = upper bound \[\Large L = \overline{x} - z_c*\frac{\sigma}{\sqrt{n}}\] \[\Large U = \overline{x} + z_c*\frac{\sigma}{\sqrt{n}}\]

  36. anonymous
    • one year ago
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    thank you c:

  37. jim_thompson5910
    • one year ago
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    no problem

  38. anonymous
    • one year ago
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    I got 48.60 and 51.40 (for future users)

  39. anonymous
    • one year ago
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    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?

  40. jim_thompson5910
    • one year ago
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    either way is fine

  41. anonymous
    • one year ago
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    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?

  42. jim_thompson5910
    • one year ago
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    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

  43. jim_thompson5910
    • one year ago
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    also, n = sample size

  44. jim_thompson5910
    • one year ago
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    use the link I gave you to find zc (now at 95% confidence)

  45. jim_thompson5910
    • one year ago
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    the p-hat values are the percentages of people who got sick

  46. anonymous
    • one year ago
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    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%

  47. jim_thompson5910
    • one year ago
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    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

  48. anonymous
    • one year ago
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    well, 50/50 chance of getting this right, thank you for your help again c:

  49. anonymous
    • one year ago
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    gah, it wasn't northern florida with 2.1%, I guess it was southern florida with 1.8% ._.

  50. jim_thompson5910
    • one year ago
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    The only difference I see is 1.78% and 1.8%

  51. jim_thompson5910
    • one year ago
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    well they wanted the smallest margin of error

  52. jim_thompson5910
    • one year ago
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    sorry I forgot to mention that

  53. anonymous
    • one year ago
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    it's alright, I got 80% of the questions completed correctly c:

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