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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

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

I have a Ti84 calc.....

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

I see several formulas that begin with N lol

- anonymous

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

- anonymous

I dunno how to solve it though...can you tell me which number goes where

- 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

im trying to enter it in my Ti84 and not sure how to do it...

- anonymous

@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

yes i think so...i will find out lol

- anonymous

so I got 2.27 Is that right?

- 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

Note that the
2
x key, when pressed, will square the quantity within parentheses for you after you've pressed the ENTER key.

- anonymous

when i typed that in, I typed 2 then the 2x key and it put 2^2, is that right?

- anonymous

nevermind, got it...lol
i got .0170737778

- mathmale

sorry, i didn't see what y ou'd typed and was waiting.
Plese clear the calculator.

- mathmale

Type in ( 1.96 / 15 ) x^2 and ENTER
in that order. you should obtain 0.01707.

- anonymous

did you see what i got...i think we typed at the same time....

- anonymous

yes i got that same answer

- mathmale

Ooops.
Do you have 0.017... still displayed? if so, pls don't erase it.

- anonymous

yes i do

- anonymous

so then i would type (192)^2 ?

- 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

I got the result 141616.7, which I rounded up to 141617.

- anonymous

i got 629.407 ...?!?!

- mathmale

Let me check my own work. Actually, your result looks more reasonable than does mine!!

- 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

my N is 629.408 ? is that right? or maybe 629.41

- anonymous

oh wait....actually that would be 629

- anonymous

since its asking for number of computer users

- 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

629

- mathmale

Obviously we don't want to see 0.4077 computer user. Bloody. :)

- anonymous

LOL

- mathmale

You've rounded down. We MUST round UP to be conservative in our estimate.

- mathmale

round 629.4077 UP to the nearest integer.

- anonymous

oh yeah...sorry bout that....it would be 630?

- 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

Very, very good. Thanks for your persistence and positive attitude.

- 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

While this in itself is not sufficient practice, you've made huge, positive strides towards learning this material!

- 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

ok thank you so much!

- mathmale

Great pleasure! Bye.

- anonymous

gnight

Looking for something else?

Not the answer you are looking for? Search for more explanations.