anonymous
  • anonymous
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.
Probability
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmale
  • mathmale
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 ... ?
mathmale
  • mathmale
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.
anonymous
  • anonymous
I have a Ti84 calc.....

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anonymous
  • anonymous
I see several formulas that begin with N lol
anonymous
  • anonymous
@mathmale ...will the formula be N\[\ge \left(\begin{matrix}15 \\ 192\end{matrix}\right)^{2}\delta ^{2}\]
anonymous
  • anonymous
I dunno how to solve it though...can you tell me which number goes where
mathmale
  • mathmale
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.
anonymous
  • anonymous
im trying to enter it in my Ti84 and not sure how to do it...
anonymous
  • anonymous
@mathmale
mathmale
  • mathmale
\[(\frac{ 1.96 }{ 15 })^{2}(192)^{2}\] is waht y ou want to calculate. Are you comfortable using parentheses on your calculator?
anonymous
  • anonymous
yes i think so...i will find out lol
anonymous
  • anonymous
so I got 2.27 Is that right?
mathmale
  • mathmale
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?
mathmale
  • mathmale
Note that the 2 x key, when pressed, will square the quantity within parentheses for you after you've pressed the ENTER key.
anonymous
  • anonymous
when i typed that in, I typed 2 then the 2x key and it put 2^2, is that right?
anonymous
  • anonymous
nevermind, got it...lol i got .0170737778
mathmale
  • mathmale
sorry, i didn't see what y ou'd typed and was waiting. Plese clear the calculator.
mathmale
  • mathmale
Type in ( 1.96 / 15 ) x^2 and ENTER in that order. you should obtain 0.01707.
anonymous
  • anonymous
did you see what i got...i think we typed at the same time....
anonymous
  • anonymous
yes i got that same answer
mathmale
  • mathmale
Ooops. Do you have 0.017... still displayed? if so, pls don't erase it.
anonymous
  • anonymous
yes i do
anonymous
  • anonymous
so then i would type (192)^2 ?
mathmale
  • mathmale
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.
mathmale
  • mathmale
I got the result 141616.7, which I rounded up to 141617.
anonymous
  • anonymous
i got 629.407 ...?!?!
mathmale
  • mathmale
Let me check my own work. Actually, your result looks more reasonable than does mine!!
mathmale
  • mathmale
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?
anonymous
  • anonymous
my N is 629.408 ? is that right? or maybe 629.41
anonymous
  • anonymous
oh wait....actually that would be 629
anonymous
  • anonymous
since its asking for number of computer users
mathmale
  • mathmale
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.
anonymous
  • anonymous
629
mathmale
  • mathmale
Obviously we don't want to see 0.4077 computer user. Bloody. :)
anonymous
  • anonymous
LOL
mathmale
  • mathmale
You've rounded down. We MUST round UP to be conservative in our estimate.
mathmale
  • mathmale
round 629.4077 UP to the nearest integer.
anonymous
  • anonymous
oh yeah...sorry bout that....it would be 630?
mathmale
  • mathmale
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.
mathmale
  • mathmale
Very, very good. Thanks for your persistence and positive attitude.
anonymous
  • anonymous
Thank you so much!!! You are the best helper on here indeed...its so true...not many take these steps and time to explain. :)
mathmale
  • mathmale
While this in itself is not sufficient practice, you've made huge, positive strides towards learning this material!
mathmale
  • mathmale
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).
anonymous
  • anonymous
ok thank you so much!
mathmale
  • mathmale
Great pleasure! Bye.
anonymous
  • anonymous
gnight

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