Statistics Question...Help?
True or False? 1. A relative frequency distribution describes the proportion of data values that fall within each class, and may be presented in a histogram form. 2.Compared to the frequency distribution, the stem-and-leaf display provides more details, since it can describe the individual data values as well as show how many are in each group, or stem. 3. A relative frequency distribution describes the proportion of data values that fall within each category.
Statistics question help!?
Ok, i'd say you're looking at a X^2 test. So the sum of all your (Observed - expected)^2/expected = you X^2 value. For A: (29-28)^2/28 = .0357 (29 is the percent of A's from the sample.) For O: (32.5-32)^2/32 = .0078 For B: (27.5-25)^2/25 = .25 For AB: (11-15)^2/15 = 1.067 1.067 + .25 + .0078 + .0357 = 1.3605 You have 3 degrees of freedom, for (row -1), 4 blood types minus 1 p(1.3605) with 3 df > .01, so the data is not significantly different from the hypothesis, meaning the data does support the examiner's hypothesis. p(1.3605) with 3 df = .715, by the way
Statistics question! HELP!?
The mean is $700. You should be able to determine that the total wage is just $700 X 12 = $8400. If the total wage bill for the inexperienced is $602 then the total for the experienced is $8400 - ($602*5) = $5390 . That distributes over the seven total workers for a mean of $5390/7 =$770 --------------------------------------... Note to Stevie, this student says he is stuck. That means he made an effort to solve the problem and cannot determine the correct procedure for solution. It is fair to ask for help at this point. If the person is actually trying get his or her homework done instead of learning, any teacher worth their salt will find that out on quizzes and tests and get the student assigned the F they clearly deserve. Assume students ask questions with integrity and help them. If a teacher is giving lots of extra points for homework, then that teacher needs to reconsider their methods, We all know that students can get outside help on homework. That's why homework only counts 5% of my grade and quizzes and tests, which they must do themselves, count 95%.
Statistics question..help me answer!?
A person picked at random has 8% chance of the illness before the test is applied. A 92% change of being healthy. We are told that with an ill patient the test is positive 96% of the time. So it is negative 4% of the time with an ill patient. I.E. 4% of a false negative. We are told it is gives a false positive in 7% of healthy patients so it gives a correct negative in 93% of healthy patients. There are 4 cases.. Ill person & correct positive => 8% * 96% => 7.68% Healthy person & false positive => 92% * 7% => 6.44% Ill person & false negative => 8% * 4% => 0.32% Healthy person & correct negative 92% * 93% => 85.56% So the answer is 7.68%. To check the result.. These 4 cases should add up to 100% 7.68 + 6.44+0.32+85.56 = 100 So the numbers are correct. -- adding and multiplying percentages can be confusing if you don't understand them. Percents are hundredths. So 80% is really a proportion of 0.8. So 80% * 20% is really 0.8*0.2 = 0.16 = 16%
Third Statistics Question - Help!?
The log likelihood of X1, ..., Xn is log(product of ((Xi / Theta^2) exp(-Xi / Theta)) = sum log((Xi / Theta^2) exp(-Xi / Theta)) = sum (log Xi - 2 log Theta + (- Xi / Theta )) = sum(log Xi) - sum(Xi) / Theta - 2n log Theta To maximize that with respect to Theta, set the first derivative with respect to Theta to zero: (Sum(Xi) / Theta^2) - 2n / Theta = 0 Sum(Xi) - 2n Theta = 0 Theta = (1/(2n)) Sum(Xi) a) the expected value of the MLE is E[MLE] = E[(1/(2n)) Sum(Xi)] = (1/(2n))E[Sum(Xi)] = (1/(2n))Sum E[Xi] = (1/(2n)) n E[X|] = (1/2) E[X] = (1/2) integral(x from 0 to +inf)(x * ((x/Theta^2)exp(-x/Theta)dx)) let u = x/Theta, du = dx/Theta, u also goes from 1 to +inf = (1/2) integral(u from 0 to +inf)(Theta u)*((Theta u) / (Theta^2)) * exp(-u) Theta dx) = (1/2) Theta integral(u from 0 to inf)(u^2 exp(-u) du) = (1/2) Theta Gamma(3) = (1/2) Theta (2!) = Theta so yes, the MLE is unbiased estimator of Theta. b) The Cramer-Rao lower bound is the reciprocal of the expected value of the square of the derivate found up above: (Sum(Xi) / Theta^2) - 2n / Theta Write that as (Sum(Xi) - 2 n Theta) / Theta^2. Its square is ((Sum(Xi))^2 - 4 n Sum(Xi) Theta + 4 n^2 Theta^2) / Theta^4. The expected value of that is computed using E[Xi Xj] = E[X^2] = 3! Theta^2 when i = j, E[Xi Xj] = E[Xi] E[Xj] = (E[X])^2 = (2! Theta)^2 when i not= j. You get (n 6 Theta^2 + n(n-1)(2 Theta)^2 - 4 n (n 2 Theta) Theta + 4n^2 Theta^2) / Theta^4 = (6n + 4n^2 - 4n - 8n^2 + 4n^2) / Theta^2 = (2n)/Theta^2. So the Cramer-Rao lower bound says that the variance of the MLE for Theta, since it is unbiased, and assuming the other qualifying assumptions, is >= the reciprocal of (2n)/Theta^2, i.e., Var(MLE) >= Theta^2 / (2n). If the actual variance achieves that lower bound, then yes, the MLE is effificent. Var(MLE) = E[MLE^2] - (E[MLE])^2 = E[MLE^2] - Theta^2. E[MLE^2] looks like it will be (1/(2n))^2 (6n Theta^2 + n(n-1)(2 Theta)^2) = Theta^2 (6n + 4n^2 - 4n)/(4n^2) = Theta^2(1 + (1/(2n))) which would make Var(MLE) = Theta^2 / (2n) = Cramer-Rao lower bound. So yes, the MLE Sum(Xi) / (2n) is both unbiased and efficient as an estimator of Theta. Dan
Statistics question please help?
The expected (mean) life of a particular type of light bulb is 1,000 hours with a standard deviation of 50 hours. The life of this bulb is normally distributed. What is the probability randomly selected bulb............. (a) Would last longer than 1150 hours? (b) Would last fewer than 1100 hours? (i must show all work to solve these questions please help)
Statistics Question! Help Please!!?
In a study of human blood types in nonhuman primates, a sample of 71 orangutans were tested and 14 were found to be blood type B. a) Construct a 95% Confidence Interval for the relative frequency of blood type B in the orangutan population b) construct a 90% Confidence Interval Thank you!
Alphabet Statistics Questions help?
1.) Because it's not a common letter. As a percentage, it doesn't occur in a lot of words. It's listed near the end of the "Letter Frequency Alphabet." (Below it are j,x,q, and z.) http://en.wikipedia.org/wiki/Letter_freq... 2.) Well, we'd survive... I only needed one of them (the q in frequency) to write this reply to you. However, I'm not sure if I agree that we would "get along fine." We need those letters too! :)
Can someone help me with statistics question #1?
remember the that for a normal distribution: 68% of observations fall within +/- 1 standard deviation 90% of observations fall within +/- 1.65 stdevs 95% of observations fall withing +/- 1.96 standard deviations and 99% of observations fall within +/- 2.58 standard deviations. a) the stdev is 100, so 700 is within 2 stdev, which is 95%, which means that 5% fall outside this range. But since it is 2 tailed, only half of them fall above, so 2.5% are above 700. b) a normal distribution is symmetrical so 1/2 are above and 1/2 are below the mean. c) 100 x 1.65 = 165 ; 500+ 165 = 665. They have to be above 665