Please explain easily the Student T Test and another tests used in Biostatistics, thanks?
104149
T-tests, orginally designed by Gossett for Guiness Brewery is a test of significance.  It is used when there are two groups and one wants to see if chance could account for differenes between the two groups.
578047
This depends on the focus of your study.  What or who are you trying to assess the risk of?  Biostatistics has several methods for assessing risk when it comes to diseases.  Quality Assurance methods assess the risk of malfunction and ways of reducing the likelihood of it happening.  I'm sure there are many other forms of assessing risk as well, so I'm afraid your question needs to be a little more specific.
249970
Statistics is applicable in every field of science which involves the data analysis.\nIf You wanna know the reliability of a medicine to patient, statistics can predict the results in some level of confidence.\nThe pills and medicines to stop pregnancy is tested in this way.
242094
In the experiment of testing for the number of students out of 37 who consider statistics exciting\n\nNo. of trials made \n==> No. of students tested = 37\n==>n = 37\n\nLet the event of a student tested being intersted in the subject of statistics be a success\n\nProbability of success in each trial\n= probability that a student tested is interested in the subject of statistics\n==> p = 0.75\n[Given that 75% of the population of all business statistics 1 students consider statistics exciting]\n\nTherefore, \nq = 1 - p ==> q = 1 - 0.75 ==> q = 0.25\n\nLet "x" indicate the number of successes in "n" trials\nsince the number of students who are interested in statistics can be either 0, 1, 2, .... 37\nThe values "x" can carry are 0, 1, 2, .... 37\n==> "X" is a discrete random variate with range = {0, 1, 2, .... 37}\n\nThe probability distribution of "x" is a binomial distribution..\n\n[Since there a number of trials all of which are identical and independent with only two possible outcomes in each trial either a success i.e. the student being interested or a failure i.e. the student not being interested, with the probability of success in each trial remaining the same all throught]\n\nThe probability function of a binomial distribution\n==> P(X=x) = nCx * p^x * q^(n-x)\n\nTherefore, for the above distribution\nP(X=x) = 37Cx * (0.75)^x * (0.25)^(37-x)\n==>P(X=x) = 37Cx * (0.75)^x * (0.25)^(37) * (0.25)^(-x)\n==>P(X=x) = 37Cx * (0.75)^x * (0.25)^(37) * 1/(0.25)^(x)\n==>P(X=x) = 37Cx * (0.75/0.25)^x * (0.25)^(37)\n==>P(X=x) = 37Cx * (3)^x * (0.25)^(37)\n\nTherefore, \nProbability of observiing 36 or more of the students who consider statistics to be an exciting subject\n==> P(X >= 36) = P(X=36) + P(X=37)\n\n\nP(X = 36) = C(37,36) * (3)^36 * (0.25)^(37)\nP(X = 37) = C(37,37) * (3)^37 * (0.25)^(37)\n\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
870154
Here is the web site for A.P. statistics:\nhttp://www.collegeboard.com/student/testing/ap/sub_stats.html?stats\n\ns
1094318
It is not exactly clear what you need, but maybe you can use the Kolmogorov-Smirnov test (KS-test) for your problem. The KS-test tries to determine if two datasets differ significantly. The KS-test has the advantage of making no assumption about the distribution of data.  Note however, that this generality comes at some cost: other tests (for example Student's t-test) may be more sensitive if the data meet the requirements of the test.
24979
The simple answer is no.  The K-S (Kolmogorov-Smirnov) test is actually used to test if two data sets have the same distribution, so the previous answer is not correct.  \n\nA T-Test is used when the data comes from an underlying normal population (you can assess that with K-S), and assumes continuous data, and a common unknown variance for both samples.  \n\nWhat exactly are you trying to test?
861521
you would need to look at table for t values.\nyou are looking for a 95% confidence level, so you would divide .5 (which is 1-.95) by 2 and get .025 that is the column you find the value for t.\nyour n number, the row for the t number, is 15-1 or 14 and 12-1 or 11.\nLook at the table IV in colum t=0,025 and row 14 write down the number. that number is the t value for the first equation.\nLook at table IV in column t=0.025 and row 11, that is your t value for the second equation.\nNow that you have the t values you can plug them into the formula for confidence interval.\n\nGood luck
892708
Probably a t-test. \n\nTell us what hypothesis you are testing.
242624
http://en.wikipedia.org/wiki/Mean_%28statistics%29\n\nThis is a link from Wikipedia. It shows lots of examples along with explanations.
878741
There are certain steps to perform a hypothesis test.\n\n1. Construct a null hypothesis Ho .\n2. Construct a alternate hypothesis H1.\n3. Decide test statistics and parameters [this includes degrees of freedom, significance value] [inmost cases its denoted by Z or t].\n4. Compute the p-value.\n5. Check with the available table.\n6. make inferences wether the hypothesis is accepted or not.\n\nThere is no fixed fomulae for this, but the choice of null hypothesis and test statistics depends on the problem. \n\nThe type of null hypothesis depends on the probelm(wether certain parameters are given or certain distributions are followed by given data etc) and the test statistics depends on parameters.\n\nI hope now you know it.\n\nAll the best
884290
Ho: mean=40 hours/week\nHa: mean > 40 hours/week\nalpha: 95%\nn = 30\ndf = 30-1\n\nOne sample t-test:\n\nt=(mean-assumed mean)/(sample standard deviation/square root of n)\n\nt=(40-45)/(7.5/sqrt 30)\nt=-5/1.369\nt=-3.65\n\nFrom the tables with alpha = 0.05 and df = 29 the t value is 2.045.\n\nSince our calculated t-value is outside of the t value from the table, there is little statistical significance to conclude that the average work week has increased.
830991
I'm wondering why you created the data 1-3500?  I would think you would use the set of data available and figure the standard deviation and average.  With that information, you would be able to predict tomorrows value at a certain confidence level.  I forget is that the p test or the t test.  its been a while since I took statistics.
579675
Yes, you have to understand a lot of Biostatistics.  In a nutshell Biostatistics is the application of statistical techniques to scientific research in health-related fields, including medicine, biology, and public health, and the development of new tools to study these areas. Since the beginning of the twentieth century, the field of biostatistics has become an indispensable tool in improving health and reducing illness. Biostatisticians play essential roles in designing studies, analyzing data and creating methods to attack research problems such as :\n\n* the determination of major risk factors for heart disease, lung disease and cancer \n* the testing of new drugs to combat AIDS \n* the evaluation of potential environmental factors harmful to human health, such as tobacco smoke, asbestos or pollutants, etc
481083
It depends what you want to predict.  For example, you could want to know how many people at a school go out for lunch instead of eat at the cafeteria.  Instead of asking everyone, you could choose to ask 30 people at random in the school and use that sample to predict the total school population who go out for lunch.\n\nFor example, if 12 out of the 30 people go out to eat lunch and there are 800 people in the school, then you could predict the total number of persons going out to lunch by taking 12/30 and multiplying that to the total population.\n\n(12/30)*800=320\n\nYou use the sample to predict the behavior of the larger group.  The bigger the sample, usually the more accurate the prediction is.
108115
What are the main topics of Biostatistics? ANOVA?
898945
Considering the data you had shown,\nyou must use a one-tailed test.\nyour null hypothesis would be:\nHo: There is no significant difference between the mean amount of time devoted to patient care per week by doctors using conventional methods and alternative methods.\nHo: mean(conventional) = mean(alternative)\n\nyour alternative hypothesis must be:\nHa: The mean amount of time devoted to patient care per week by doctors using conventional methods exceeds that of doctors using alternative methods\nHa: mean(conventional) > mean(alternative)\n\nNow that you have the hypotheses, next is to set the level of significance but it's all up to you depending on how confident are that you will not commit any kind of error.\n\nAnd your asking about the appropriate test?\nFor me, the choice depends on the given.\nBut you can only choose from the second and fourth test since you have to use a directional test (one-tailed test).\n\nUse Z-test if the standard deviation or variance of the population is known.\nUse t-test if the standard deviation or variance of the population is unknown.\nThat's the general rule of choosing the appropriate test.\nBut there's another consideration in the selection of test.\nyou must also take into consideration the number of your sample, whether it exceeds 30 or not.\n\nThat's all. I hope I've been a great help to you. Please inform me if there is something wrong with my lecture here.\nthank you.