An IQ test is designed so that the mean is 100 and the standard deviation is 19 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 95% confidence that the sample mean is within 2 IQ points of the true mean. Assume that Ơ = 19 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation. (a) The required sample size is ____. (Round up to one integer as needed) (b) Is this sample size reasonable?

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An IQ test is designed so that the mean is 100 and the standard deviation is 19 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 95% confidence that the sample mean is within 2 IQ points of the true mean. Assume that Ơ = 19 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation. (a) The required sample size is ____. (Round up to one integer as needed) (b) Is this sample size reasonable?

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You'll find a formula for required sample size in the same chapter that discusses confidence intervals. You're given all the info necessary to calculate this minimum sample size: confidence level (90%), standard deviation (17) and acceptable margin of error (5). Please look for this formula.
But i have no idea how to calculate this
Your formula is basically OK, but you must square the quantity within parentheses. also, you're dividing that (z-score)(sigma) by E.

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z, as you've written it, is called the "z critical value" and is, I believe, 1.645 for 90% confidence.
This also might help - http://www.wolframalpha.com/input/?i=confidence+interval&a=*C.confidence+interval-_*Formula.dflt-&f2=0.90&f=TInterval.c_0.90&f3=40&f=TInterval.n%5Cu005f40&f4=17&f=TInterval.s_17&f5=100&f=TInterval.xbar_100
Please be sure you know how to find that; if you don't, either someone else or I can explain where that comes from. Forgive me, but I need to get off my computer; if by the time I come back on, someonehas not helped you complete this problem already, I certainly will. Thanks, ybarrap.
Im so lost in statistics....I have no idea....can you please help me compute this problem?
Use the students t-value: $$ \large{ \bar x\pm\cfrac{t_{(1-c)/2}}{\sqrt n}\\ } $$ Where c = 0.90 so (1-c)/2 = .05 Now let's find \(t_{.05}\), the two-tail student's t at the .05 significance level. Can you find that? Then you want: $$ \cfrac{t_{(1-c)/2}}{\sqrt n}=5 $$ Now you are left with an equation that you can solve for n. Does this help?
Ont he calculator you gave me, when i plugged in the numbers...then entered my answer, it was wrong. ugh
I forgot to include the sample variance, s: $$ \large{ \bar x\pm\cfrac{t_{(1-c)/2}s}{\sqrt n}\\ } $$ So then find $$ \cfrac{t_{(1-c)/2}s}{\sqrt n}=5 $$
You need the student's t-value at the .05 significance level. Get that and you should be done. The calculator is just a validation-- you need to know how to look up this value.
no clue. I think the answer is 32 but have no idea how to get that
I'm back. The formula (you typed earlier in this conversation) for the minimum number of samples is (after a correction or two): \[n=(\frac{ [z critical value][standard deviation] }{ E })^{2}\] Not that this is essentially the same as yours, except that I've indicated division by a horiz. line and have squared the quantity within parentheses.
z, the z-critical value corresponding to a 90% confidence interval, is 1.28 (not 1.645, as I typed in earlier). The standard deviation given in the problem statement is 17. E is the "margin of error;" we're hoping that we can estimate the "true mean" to within E, or 5. Putting this all together gives us \[n=(\frac{ (1.28)(17) }{ 5 })^{2}.\]
I've calculated this and have obtained as result n = 18.94. This must be rounded UP, to 19, which represents the minimum number of samples you must work with in order not to be further than 5 IQ points of the true mean IQ.
is that the answer then? I am so beyond frustrated that I cannot figure this stuff out.
First, please read this thru and decide whether or not the process is clear to you. Have you calculated (or looked up) z-critical values before? For a 99% confidence interval, the z-critical value is 2.33; for 95%, it's 1.645, and for 90% (with which you're dealing here), it is 1.28. That's where the 1.28 in the formula for n came from. Please see: http://easycalculation.com/statistics/z-critical-value.php If you type in 90%, this calculator will return 1.28.
I'd certainly like for you to feel comfortable and confident in dealing with this sort of problem. Can you think of any questions to ask that might help you come to a clearer understanding of what's happening here? Where did you obtain that "32" figure that you thought was the answer?
thats what it said when i typed that into the box where i put the answer. I have cried over trying to figure this out...sorry Im difficult. I have no idea what the answer is or how to get it. I dont even know what to say to answer your question
This homeowrk is due now and i dont have a clue.
What kind of reference(s) (if any) do you have available to you? Have you been able to locate the formula you typed out for me 'way back? Where did you get it? Except for not being squared, your expression looked OK.
thank you for helping. I am really trying but Im not understanding these formulas.
my work is out of the mymathlabs and it was a formula given in the example...
its all foreign language to me
Although I can't stay online for more than a few minutes longer, perhaps I could help you tomorrow or later to understand those formulas. What deadlines are you working with?
its due tomorrow but i have 5 more questions. bleh
I don't have reason to doubt the accuracy of the problem we've supposedly just completed. I'd type in "19" for the minimum number of samples you'd need. Pick one more of those questions and post it, as quickly as you can.
Please post it as an entirely new question. (This assumes, of course, that you want to and have time to do so now.)
I will. thank you. they are already posted but noone is answering them - it wont let me post again
An IQ test is designed so that the mean is 100 and the standard deviation is 19 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 95% confidence that the sample mean is within 2 IQ points of the true mean. Assume that Ơ = 19 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation. (a) The required sample size is ____. (Round up to one integer as needed) (b) Is this sample size reasonable?

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