a) contstruct a 95% confidence interval estimate of the population standard deviation of pulse rates of males.\nb) contstruct a 95% confidence interval estimate of the population standard deviation of pulse rates of females.\nc) compare and interpret the results. Does the variation appear to be different for the two groups?
205322
To obtain 95% confidence intervals for the population variance we use the result:\n\n[(n-1) S²] ÷ (sigma²) ~ chi-square(n-1)\n\nWhere sigma² is the population variance\nand S² is the sample variance\n\nFor the males we have n = 40, so using the result we have a chi-square(39).\n\nFor a 95% confidence interval we want the values that contain 95% of the distribution between them (the lower limit will have 2½% below it and the upper limit will have 2½% above it).\n\nUnfortunately my tables don't have a chi-square 39 listed - so you will have to look these up in your tables - I will use the chi-square 40 results just to show you the method.  The chi-square 40 results are 24.43 and 59.34.\n\nNow we have these values (called "critical values") - we stick them into our thrilling result:\n\n[39 × 11.3²]/(sigma²) = 24.43\n\nand:\n\n[39 × 11.3²]/(sigma²) = 59.34\n\nRearranging each of these gives us the sigma²:\n\n(sigma²) = [39 × 11.3²]/24.43 = 203.844\n\n(sigma²) = [39 × 11.3²]/59.34 = 83.922\n\nNow you want the confidence interval for the standard deviation (not the variance), so we need to root these:\n\nsigma = 14.3\n\nsigma = 9.16\n\nSo our 95% confidence interval is (9.16, 14.3).\n\nYou can do a very similar thing for the females and then can compare these two limits to see if there is any overlap between them.\n\nHope this helps.
470808
OK - I will try and explain it, but forgive me if I get myself confused...  hehehe\n\nFirst step - make a bell curve.  Put a mark in the center and label that "7 hours" - that's your mean.  Next, put three marks on either side of that middle point - those will be your standard deviation points.  For the points on the right, ADD 5.3 to the mean for the first one, then add 5.3 again for the 2nd, etc...  Points on the left side of the mean have the 5.3 SUBTRACTED.  (Am I still making sense?  lol)\n\nNow - the 68-95-99.7 rule says that 68% of the freshmen will fall into the area between the first line on EITHER side of the mean, 95% between the second lines and 99.7% between the third lines on either side of the mean.\n\nI think you should be able to figure it from there - I hope so...  Right now I am so confused I think I gave myself a headache!  lol\n\nGood luck!
463322
Is that really the question?  Prob of a normal variable being greater than the mean?  It's just 0.5.  What's the real question, and what has Z to do with it?
488179
The width of the confidence interval is 40, so the error about the mean is 20. You have a sample of 25 batteries, so the number of degrees of freedom is 24. The critical t-score for 24 degrees of freedom and a confidence level of .95 is approximately 2.063899.\n\nWe can use the t-score to approximate the sample standard deviation:\n\nt = error / (s/sqrt(n))\nSolving for s we get \ns = error*sqrt(n)/t, \ns = 20*sqrt(25)/2.063899\ns = 48.45199
268968
First of all, if the population is normal, the distribution of sample means is normal regardless of sample size.  So I. is true.\n\nSecond, if the population is not normal, the distribution of sample means is approximately normal if the sample size is large.  So III. is also true.\n\nII. cannot be determined to be true unless we know what the population looks like.\n\nSo I would go with C.\n\nUPDATE: Apparently ghakh hasn't heard of the Central Limit Theorem.  You know, the "Fundamental Theorem of Statistics"?  In layman's terms, it states that if the sample size is large enough, then the sample means will be approximately normal if the sample size is large enough REGARDLESS OF WHAT THE POPULATION LOOKS LIKE!!!!!!  All you need is a finite variance, which for elementary statistics students is not something they need to be concerned with.\n\nMaybe you should take my statistics class, ghakh.  You might learn something!
458459
This would be a binomial.  Assuming that p=1/2 , equal probability of male or female.\n\nThen it's like 80 trials with success (male) or failure (female)\n\nAt least 30 male means that you have to add probs of 30, 31, 32, ...,80.  It's easier to add probs of 1,2,...,29 and subtract from 1.\n\n80 is a large enough number to use the normal approximation.\n\nFor the normal, the mean would be 80/2 = 40.  So, at least 30 is\n\n1-P(X<less than 30) \n\nconvert X to Z to use standard tables.
67223
For timed events, you can look at the average, median, standard deviation, etc of the time.  Similar for distances on things like javelin or shotput, or accuracy rates for archery.\n\nJust was watching the biathalon (cross country skiing with a few sessions of marksmanship with a rifle along the way), and there's some interesting math there.  When they stop to shoot at targets, they have to do a penalty lap for each target they miss.  Which brings up questions like, is it more important to be accurate or fast?  If an athelete can spend a month training, what mix of target practice and conditioning should they do?  If you know the speed and accuracy of each competitor, can you pick the favorites?\n\nFor the last one, obviously if someone is both faster and a better shot than anyone else, they're a favorite.  But in reality, the atheletes have strong and weak spots, so which combination of skills will come out on top?
89417
I am pondering this question a bit. I might have a better answer for you, but here is a start:\n- - - - - - - - - - - -  - \n\nYour data is not exactly amenable to standard quantitative analysis. You can't really form an average, for example. \n\nYou can find the mode. The most common answer is, diet. So the mode of your data is "diet". \n\nThe data categories, as they stand, are not represented by ordinal numbers. So it is a bit hard to even describe the distribution, let alone the dispersion of the distribution. What you could do, is say that 35% preferred diet, and 65% did not. Then you could try to estimate the error bounds on these figures, given some assumptions. Or do a robust estimate using a jackknife or bootstrap.\n\n- - - - - -  - - \n\nHere is another idea. You might want to know the average number of people that are in a given category. That is, you might want to know how many people like a given type of soda. You have 5 categories, so the average is \n\n(20)/5=4\n\nThere are an average of 4 people in each category.\n\nNow you can find the standard deviation.\n\nFirst find an estimate of the variance V using the standard formula:\n\nV=(1/4)*[(1-4)^2 + (3-4)^2 +(3-4)^2 +(7-4)^2 +(6-4)^2 ]\nV=(1/4)*[9 + 1 + 1 + 9 +4 ]\nV=(1/4)*[24]\nV=6\n\nThen take the square root\n\nS=sqrt(V)= sqrt(6)\nS=sqrt(6) = 2.44948974\n\nThe average soft drink is liked by 4 people, but there is a standard deviation of about 2.4.\n\nSo one can say a soft drink is liked by 4 += 2.4 people (if we set the error bounds to be 1 standard deviation, but some people set it at 2).\n\nYou could even convert this to percentages. Then you could say that a soft drink is liked by 20% of the people, +- 12%.\n\nThe standard deviation assumes that the distribution is normal, that is a bell-shaped curve (aka Gaussian). However, it probably is not. A robust estimate of the dispersion of the dispersion can be formed just resampling methods, like the bootstrap or the jackknife. I have placed a link to these if you are interested, but I think it might be too advanced for your needs.\n\nAnother thing to do is to look at the median, and the median average deviation, or the median average deviation away from the median. A reference discusses these and other measures of dispersion if you are interested.\n\nI still feel that one might be able to do more with this. I will think some more and if I have any good ideas I will post them. I am not a statistician, so it is likely that one of them might have a much better suggestion right away.
479987
The bigger the sample size, the more accurate the estimate.\n\nSay I was estimating the percentage of red m&m's in a pack.\n\nIf I sampled 10 m&m's, I might I might find that the sample contains 1 red one. This could lead me to think that a pack contained 10% red m&m's.\n\nHowever, if I sampled 100 m&m's, I might find that the sample contains 15 red m&m's, leading me to think that a pack contains 15% red m&m's. This answer is impossible using the sample size of 10, as you can't get 1.5 m&m's of a particular colour.\n\nIf I sampled 1000 m&m's I might find that I have 143 red ones, leading me to think that the pack contains 14.3% red m&m's.\n\nSo, as you can see, the bigger the sample size, the closer the estimation comes to the actual answer.
110776
To answer that question you would have to know if the differences between "having a variation" and "not having the variation" are normally distributed.  If so, you could make an assumption on the number to look at based on normal distribution and 3 standard deviations.\n\nIf there is no normal distribution to the data, then you would have to determine your own "statistics" for the odds that the deviation would occur, and calculate the number based on the possible number of "variations" that could occur, versus the odds that any one could occur.
203120
your confidence interval of .4873 to .5526 indicates the range of the true mean with 95% confidence.  Since values above and below .50 are in this range, the poll is a tossup.
434906
OK, here goes:\n\nThe data could be\n7 9 10 12 12 13 14\n\nThe median (center number) is 12\nThe mode (most repeated number) is 12\nThe mean (average) is 11.\n\nThe point is that you need to look at all three measures to get a good idea of the data.\n\nHere's another set:\n1 3 10 12 12 19 20\nThis has the same measurements.\n\nIt is only when we look at things like the variance and std deviation that we can really see the difference between those two data sets.\n\nGood luck!
481229
Why should it say >> male << if it meant researchers ... dnt think we actually have that large number of ' brilliant male' health professionals ...;) \ni think its talking about the subjects of the study \nbut the paragraph has a weird structure though
180924
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.
170100
The problem is incompletely or improperly described.\n\nEach member of the population has some characteristic value.  The mean is the average of this value taken over the entire population, which, assuming each member is an equal contributor, is M = sum(values)/N, where N is the population.  If each member has the same value, say 515, then mean would be 515 and the standard deviation would be zero.  If each even-numbered member has value 1030, and each odd-number member has value 0, the mean is 515 and the standard deviation is (approximately) 515.\n\nSo, you have provided enough information to solve the problem satisfactorily.
93073
Here's how to figure this out.\n\nMean: 75% times 250\n\nThe other two are a little trickier, and there's more than one way to do it.  To get the variance, you take the sum of the squares of the individual deviations and divide by the sample size.  Because 0.75 of the sample passes, each pass gives a deviation of 0.25 because 1 pass - 0.75 is 0.25.  For the fails, the deviation is -0.75.  So that should be an expression like this:\n\n((0.25)^2 x M + (-0.75)^2 x (250 - M))/250\nwhere M is the mean you calculated above.  The ^2 is the exponent 2 for squaring something.\n\nThe standard deviation is the square root of the variance.
385640
I think the answer would be 1), though it is a stupid question. The right way to frame the question would be around a confidence interval.