## 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.

1. 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?

2. 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?

3. anonymous

@nincompoop @dan815 @jim_thompson5910 @ganeshie8

4. anonymous

@campbell_st

5. 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

6. jim_thompson5910

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

7. anonymous

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

8. jim_thompson5910

do you have a calculator?

9. anonymous

yes

10. jim_thompson5910

a TI calculator?

11. anonymous

yes

12. jim_thompson5910

hit 2nd, then the vars key

13. jim_thompson5910

scroll down to invNorm

14. jim_thompson5910

|dw:1435127861838:dw|

15. anonymous

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

16. jim_thompson5910

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

17. jim_thompson5910

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

18. jim_thompson5910

so do you see how the area under the curve to the left of z = k is 0.99 ? |dw:1435127996415:dw|

19. anonymous

alright, I see the curve

20. jim_thompson5910

you said it's an online calculator?

21. anonymous

well, the 0.99

22. anonymous

yes?

23. jim_thompson5910

24. anonymous
25. jim_thompson5910

ok that calculator won't work

26. jim_thompson5910

27. jim_thompson5910

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

28. jim_thompson5910

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

29. 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?

30. jim_thompson5910

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

31. anonymous

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

32. anonymous

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

33. jim_thompson5910

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

34. anonymous

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

35. 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 $\Large L = \overline{x} - z_c*\frac{\sigma}{\sqrt{n}}$ $\Large U = \overline{x} + z_c*\frac{\sigma}{\sqrt{n}}$

36. anonymous

thank you c:

37. jim_thompson5910

no problem

38. anonymous

I got 48.60 and 51.40 (for future users)

39. 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?

40. jim_thompson5910

either way is fine

41. 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?

42. 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

43. jim_thompson5910

also, n = sample size

44. jim_thompson5910

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

45. jim_thompson5910

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

46. 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%

47. 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

48. anonymous

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

49. anonymous

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

50. jim_thompson5910

The only difference I see is 1.78% and 1.8%

51. jim_thompson5910

well they wanted the smallest margin of error

52. jim_thompson5910

sorry I forgot to mention that

53. anonymous

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