How to identify outliers in statistics?
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An outlier is a value that is more than 1.5 times the interquartile range (Q3 - Q1) away from either Q1below it) or Q3 (above it).
913451
A box and whisker plot!? No problem! \n\nThe Box is from Quartile 1(25th percentile) to Quartile 3(75th percentile),.\n\nThe Whiskers are Q1 minus 1.5 Interquartile ranges and Q3 plus 1.5 Inter Quartile ranges.\n\nIQR (Interquartile ranges) is defined as Q3-Q1\n\nWell say that Q1 is 12 and Q3 is 25 then the IQR is 13, and 1.5IQR is 19.5\n\nTherefore, Anything below -7.5 (12-19.5) and above 44.5 would be considered an outlier. \n\nAs far as drawing the damn thing, just draw a box with the outside edges being Q1 adn Q3, and then draw the whiskers 1.5IQR long.\n\nAdditional info:\n\nThe box and whisker is only 1 criterion for outliers, + or - 3 standard deviations from the mean is another common way, which can get other results.\n\nHope that helps.
334827
In data and signal analysis, an apparent trend or signal caused by random events or noise is often called spurious.
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This is an example for standard deviation:\n\nSay you have these values and need to find the deviation for them\n5, 10, 15, 20, 25 \n\nMean = 15\n\nNow, calculate the difference of the values from the mean\n\n15-5 = 10\n15-10 = 5\n15-15 = 0\n20 - 15 = -5\n25 - 15 = -10\n\nNow, square the differences and add them\n\n10^2 + 5^2 + (-5)^2 + (-10)^2 = 100 + 25 + 25 + 100 = 250\n\nStandard deviation is the squared root of this sum divided by the mean\n\nStandard Deviation = (250/15)^(1/2) \n                   = 5 * (2/3)^(1/2)\n                   = 5 * 0.816\n                   = 4.082
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most statistics follow a pattern, whether it be linear, quadratic, or whatever. Once you have enough data to establish a pattern, you can accurately predict the remainder of the line.\n\nFor instance, you have a line that looks to be half a parabola. Once it starts going back up, you can be pretty certain of where it's going to be.
286076
What kind of statistic are you looking for?
123406
Try Chi square testing, it starts out complicated but once you understand it, it is way faster than punnett squares! You can find all kinds of stuff about it on the internet, it's a statistical analysis.
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Random errors are statistical fluctuations (in either direction) in the measured data due to the precision limitations of the measurement device. Random errors usually result from the experimenter's inability to take the same measurement in exactly the same way to get exact the same number.
1038461
check the standard deviation. It represents volatility..
961030
in Statistics in theory there should zero measurements falling beyond the three standard deviations from the mean on either side. though in practical some values do fall in this range. I don't remember the exact statistical names of these values but i think they are referred to as "outliers".
706728
1.  Using your Normal Distribution table (sorry, mine is buried under my dirty laundry), find the Std Devs that represents .18 (.50 - .32).  Multiply by 30 (Std Dev).  This number represents the number of pages short of 150 (the mean).  It's probably around 140.\n\n2.  The second group (.18 below 150 pages + .25 above 150).  Repeat the process for 1 above. (about 140 to 165)\n\n3.  The third group is from .25 to .49 above the mean (about 165 to 240)\n\nGood luck.
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Approximately 95% of the data is within 2 standard deviations of the mean.\nCheck out this website\nhttp://en.wikipedia.org/wiki/Confidence_interval\nand look under the section\n"Confidence Intervals in Measurement"
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My rule of thumb is that the skewness should be less than 2 in absolute value.  Other people might have different rules of thumb.\n\nYou may find the Jarque-Bera test of interest.  It is a test for normally that uses both the sample skewness and sample kurtosis.  I have posted a link about it below.
324717
what type of statistics?
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You'll need something which measures the amplitude and frequency of the sound waves to 'see' a difference in the noises.  You can probably calculate a trigonometric function based on one particular wave within the measurement, but it won't hold true as the sounds fluctuate at random.  Possibly, you could even program a computer to calculate a random value between set parameters to account for this fluctuation.\nI'm not a math person.  Can't help you much more than that.
248449
If you suspect that scores are not the same within two groups, you must place this statement in the alternative hypothesis.\n\nHo=scores are same\nHa=scores are not the same.\n\nBy claiming this this way, you can test it using the in formation obtained from the sample data and conclude whether or not data is sufficient to claim that you can reject the null hypothesis, and decide that scores are not the same. Always, you must place a new claim, a new suspicion in the alternative hypothesis, so that you can reject the null and conclude Ha, or you can not reject and conclude that you have not enough evidence to claim Ha.
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Just enter "statistical anomaly" in the search bar on the Yahoo home page for a real answer.