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StClowers

  • 10 months ago

In order to estimate the mean amount of time computer users spend on the internet each month, how many computer users must be surveyed in order to be 95% confident that your sample mean is within 15 minutes of the population mean? Assume that the standard deviation of the population of monthly time spent on the internet is 192 min. The minimum sample size required is ___ computer users. What is a major obstacle to getting a good estimate of the population mean? a. The data does not provide information on what the computer users did while on the internet. b. There may not be 630 computer users to survey c. It is difficult to precisely measure the amount of time spent on the internet, invalidating some data values. d. There are no obstacles to getting a good estimate of the population mean.

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  1. mathmale
    • 10 months ago
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    This topic is closely related to confidence intervals and uses most or all of the same parameters (variables). Pls bring me up to date: What type of references do you have available? Do you have a textbook, a list of statistics formulas, skill in looking things up on the 'Net, or ... ?

  2. mathmale
    • 10 months ago
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    This web site is a bit too advanced, but does present the formula you need for minimum sample size: http://www.itl.nist.gov/div898/handbook/prc/section2/prc222.htm About 3/4 of the way from the top of the first page, there's a formula that begins with N. That's the one we need to consider here.

  3. StClowers
    • 10 months ago
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    I have a Ti84 calc.....

  4. StClowers
    • 10 months ago
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    I see several formulas that begin with N lol

  5. StClowers
    • 10 months ago
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    @mathmale ...will the formula be N\[\ge \left(\begin{matrix}15 \\ 192\end{matrix}\right)^{2}\delta ^{2}\]

  6. StClowers
    • 10 months ago
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    I dunno how to solve it though...can you tell me which number goes where

  7. mathmale
    • 10 months ago
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    Glad you're at least trying something. If y ou'll look at the formula in that web site again, you'll see several quantities: 1.96 is called the z-critical value and is connected with a 95% confidence interval; the quantity in the denominator is the tolerance (which in this problem is 15 min) and that sigma is the std. deviation. I'd like to take a lot more time to explain this in more depth, but it does appear to me that you have all the info you need to calculate N.

  8. StClowers
    • 10 months ago
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    im trying to enter it in my Ti84 and not sure how to do it...

  9. StClowers
    • 10 months ago
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    @mathmale

  10. mathmale
    • 10 months ago
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    \[(\frac{ 1.96 }{ 15 })^{2}(192)^{2}\] is waht y ou want to calculate. Are you comfortable using parentheses on your calculator?

  11. StClowers
    • 10 months ago
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    yes i think so...i will find out lol

  12. StClowers
    • 10 months ago
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    so I got 2.27 Is that right?

  13. mathmale
    • 10 months ago
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    I haven't actually done it myself, but if you wouldn't mind, I'm going to lead you thru this calculation again. Type this into your calc.: (1.96/15) (in precisely that order) and then press the 2 x key on the left side of your calculato0r keyboard. What do you get?

  14. mathmale
    • 10 months ago
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    Note that the 2 x key, when pressed, will square the quantity within parentheses for you after you've pressed the ENTER key.

  15. StClowers
    • 10 months ago
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    when i typed that in, I typed 2 then the 2x key and it put 2^2, is that right?

  16. StClowers
    • 10 months ago
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    nevermind, got it...lol i got .0170737778

  17. mathmale
    • 10 months ago
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    sorry, i didn't see what y ou'd typed and was waiting. Plese clear the calculator.

  18. mathmale
    • 10 months ago
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    Type in ( 1.96 / 15 ) x^2 and ENTER in that order. you should obtain 0.01707.

  19. StClowers
    • 10 months ago
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    did you see what i got...i think we typed at the same time....

  20. StClowers
    • 10 months ago
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    yes i got that same answer

  21. mathmale
    • 10 months ago
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    Ooops. Do you have 0.017... still displayed? if so, pls don't erase it.

  22. StClowers
    • 10 months ago
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    yes i do

  23. StClowers
    • 10 months ago
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    so then i would type (192)^2 ?

  24. mathmale
    • 10 months ago
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    Great. Now we're going to do the last part of the calculation. keeping that 0.01707 as it is, type in the following: 2 x 192 x ENTER in that order. The first x stands for multiplication and is a key by itself on the right side of your calc, and the 2 x is the same command key you used earllier.

  25. mathmale
    • 10 months ago
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    I got the result 141616.7, which I rounded up to 141617.

  26. StClowers
    • 10 months ago
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    i got 629.407 ...?!?!

  27. mathmale
    • 10 months ago
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    Let me check my own work. Actually, your result looks more reasonable than does mine!!

  28. mathmale
    • 10 months ago
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    How extremely embarrassing. You're right; I'm wrong. You must now round up (not down) to the nearest integer number N. What's your N?

  29. StClowers
    • 10 months ago
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    my N is 629.408 ? is that right? or maybe 629.41

  30. StClowers
    • 10 months ago
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    oh wait....actually that would be 629

  31. StClowers
    • 10 months ago
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    since its asking for number of computer users

  32. mathmale
    • 10 months ago
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    What I meant was that you must eliminate all decimal fractions in your result by rounding up appropriately. 629.4077 has that decimal fraction 0.4077, which we do NOT want. MUST roud up to the nearest integer. Try again.

  33. StClowers
    • 10 months ago
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    629

  34. mathmale
    • 10 months ago
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    Obviously we don't want to see 0.4077 computer user. Bloody. :)

  35. StClowers
    • 10 months ago
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    LOL

  36. mathmale
    • 10 months ago
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    You've rounded down. We MUST round UP to be conservative in our estimate.

  37. mathmale
    • 10 months ago
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    round 629.4077 UP to the nearest integer.

  38. StClowers
    • 10 months ago
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    oh yeah...sorry bout that....it would be 630?

  39. mathmale
    • 10 months ago
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    Right. Perfect. You went beyond what the question asked you for; you have enuf data to choose the fourth possible answer as the correct answer, and you went thru the actual calculations to find N=630, the number of computer users you must survey (poll) in order to be no more than 15 min off the true population mean of the total time the pop of computer users spends on the Internet/month.

  40. mathmale
    • 10 months ago
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    Very, very good. Thanks for your persistence and positive attitude.

  41. StClowers
    • 10 months ago
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    Thank you so much!!! You are the best helper on here indeed...its so true...not many take these steps and time to explain. :)

  42. mathmale
    • 10 months ago
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    While this in itself is not sufficient practice, you've made huge, positive strides towards learning this material!

  43. mathmale
    • 10 months ago
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    thank you! I've been on the Internet nearly 12 hours today, so need to log off; I urge you to post each new part of this same problem and see if someone can help you. I may be on the 'Net tomorrow (but probably not).

  44. StClowers
    • 10 months ago
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    ok thank you so much!

  45. mathmale
    • 10 months ago
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    Great pleasure! Bye.

  46. StClowers
    • 10 months ago
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    gnight

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