Stat Question. Should be easy, please help?
408294
in statistics what you do is take a small sample, do the tests on the small sample and then generalize the results to the whole population.\n\nHere you took a small sample, found the mean to be 0.68\n\nBut the mean in the whole population will not be always equal to 0.68 . It can be 0.68, but it can lie anywhere around it to. \n\nHence, we have our confidence intervals around 0.68\n\nSo, when we say that the confidence intervals are (.63692, .72308), it means that we are 95% sure that the whole population mean will lie between 0.63692 seconds and .72308 seconds. \n\nSince 0.75 is not in that range, the chance of the population mean being 0.75 is less than 5% or negligible. \n\nI hope that helps.
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Start from www.statistics.com or search "statistics books" in Yahoo Answers. Greetings.
847480
Q = m * Cp * DeltaT.\n\nPlug the numbers in.  Get Cp from your book.
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Try this website for help with your statistics homework:\n\nhttp://www.statsoft.com/textbook/stathome.html
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Answer is 10.2% of the sample(25) have a mean score within 3 points of the population mean score of 220.\n\nSD=10 --> 34% (normal distribution)\n   3 from mean = 3/10 x 34% = 10.2%.
742241
The second problem is easier than the first problem, so I'll tackle it first.\n\nLet's imagine that the trip is 120 miles. The distance doesn't matter if the average rates are as given above.\n\nIt takes 3 hours to travel 120 miles at 40 miles per hour.\nIt takes 2 hours to travel 120 miles at 60 miles per hour.\n\nSo it took 5 hours to travel a total of 240 miles.\n\nThis is an average rate of 48 miles per hour.\n\nI'll work on the first problem and hopefully I'll get back to you.\n\nUPDATE\n\nI think I have the first problem solved as well. It involves an infinite series! Call the people involved A and B.\n\nThe probability that A wins on the first toss is 1/2.\n\nIf A doesn't win on the first toss, then the probability that B wins on the second toss is also 1/2. But this only happens in 1/2 of all games. (In the other half, A won on the first toss.)\n\nIf B doesn't win on the second toss, the the probability that A wins on the third toss is also 1/2. But this only happens in 1/4 of all games (In half the other games, A won on the first toss. In half the remaining games - 1/4 of all games, B won on the second toss.\n\nFollowing this logic, A wins:\n\n1/2 + 1/4(1/2) + 1/16(1/2) + 1/64(1/2) + ...\n\nof all games.\n\nB wins:\n\n1/2(1/2) + 1/8(1/2) + 1/32(1/2) + ...\n\nof all games.\n\nLooking again at A, we have:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...)\n\nThe term in parentheses is an infinite series having initial term 1 and ratio 1/4.\n\nThe sum of an infinite series having ratio whose absolute value is less than 1 is equal to:\n\na/(1 - r)\n\nWith a = 1, and r = 1/4, the sum = 4/3. So:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...) = 1/2 * 4/3 = 2/3\n\nLooking again at B, we have:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...)\n\nThis sum should be half of the sum of the infinite series in A.\n\nWith a = 1/2 and r = 1/4, a/(1 -r) = 2/3\n\nSo:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...) = 1/2 * 2/3 = 1/3.\n\nA should win 2/3 of the time, and B should win 1/3 of the time. There is an advantage to going first!
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Sorry no one will be able to help you as we do not the data in #3.\n\nYou will need to figure this one out yourself...
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50%\n\nThe mean is 200 with a deviation of 50 either way.  That means that the chances are 50 percent that the rating will be higher than 200 and 50 percen the rating will be lower than 200 because 200 is the mean.
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This is not the best place to ask for help. Meet your faculty advisor, or best - meet your Stats professor. Personally I think the best approach is to just take a book, and solve the problems in it to ace Stats.
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what type of statistics?
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The correct answer is 68%. Good luck on the game!
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Find yourself a z-statistic table and look up critical value in the table.  Check your stats book on how to interpret the table.\nhttp://www.math2.org/math/stat/distributions/z-dist.htm
151556
Ratio is the correct answer.
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Just google it dear..very simple..calculate mean and average the sum of square of variation of each score from the mean.\nhttp://davidmlane.com/hyperstat/A16252.html
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There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
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tehre's a formula for it...\nit depends how much you run druing the game
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mathematically, you CAN take an average of averages. \n\nBut statistically, it won't be meaningful, because you will be combining information from non-identical studies.