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anonymous
 2 years 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.
anonymous
 2 years 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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mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1This 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
 2 years ago
Best ResponseYou've already chosen the best response.1This 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
 2 years ago
Best ResponseYou've already chosen the best response.0I have a Ti84 calc.....

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I see several formulas that begin with N lol

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@mathmale ...will the formula be N\[\ge \left(\begin{matrix}15 \\ 192\end{matrix}\right)^{2}\delta ^{2}\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I dunno how to solve it though...can you tell me which number goes where

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1Glad 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 zcritical 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
 2 years ago
Best ResponseYou've already chosen the best response.0im trying to enter it in my Ti84 and not sure how to do it...

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1\[(\frac{ 1.96 }{ 15 })^{2}(192)^{2}\] is waht y ou want to calculate. Are you comfortable using parentheses on your calculator?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0yes i think so...i will find out lol

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so I got 2.27 Is that right?

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1I 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
 2 years ago
Best ResponseYou've already chosen the best response.1Note that the 2 x key, when pressed, will square the quantity within parentheses for you after you've pressed the ENTER key.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0when i typed that in, I typed 2 then the 2x key and it put 2^2, is that right?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0nevermind, got it...lol i got .0170737778

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1sorry, i didn't see what y ou'd typed and was waiting. Plese clear the calculator.

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1Type in ( 1.96 / 15 ) x^2 and ENTER in that order. you should obtain 0.01707.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0did you see what i got...i think we typed at the same time....

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0yes i got that same answer

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1Ooops. Do you have 0.017... still displayed? if so, pls don't erase it.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so then i would type (192)^2 ?

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1Great. 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
 2 years ago
Best ResponseYou've already chosen the best response.1I got the result 141616.7, which I rounded up to 141617.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i got 629.407 ...?!?!

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1Let me check my own work. Actually, your result looks more reasonable than does mine!!

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1How 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
 2 years ago
Best ResponseYou've already chosen the best response.0my N is 629.408 ? is that right? or maybe 629.41

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0oh wait....actually that would be 629

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0since its asking for number of computer users

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1What 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.

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1Obviously we don't want to see 0.4077 computer user. Bloody. :)

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1You've rounded down. We MUST round UP to be conservative in our estimate.

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1round 629.4077 UP to the nearest integer.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0oh yeah...sorry bout that....it would be 630?

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1Right. 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
 2 years ago
Best ResponseYou've already chosen the best response.1Very, very good. Thanks for your persistence and positive attitude.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you so much!!! You are the best helper on here indeed...its so true...not many take these steps and time to explain. :)

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1While this in itself is not sufficient practice, you've made huge, positive strides towards learning this material!

mathmale
 2 years ago
Best ResponseYou've already chosen the best response.1thank 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
 2 years ago
Best ResponseYou've already chosen the best response.0ok thank you so much!
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