one sample test : binomial test
126684
Can you please elaborate on your question?
404878
Here is what i found.  I'm not sure how right this is, but you can check it out.  Good Luck with your testing.
565431
2x<1 , x<0.5\n -1<x\n\n     -1< x<0.5
441869
sample space=8\nnumber of outcomes with at\nleast 2 heads=4\n\nP(at least 2 heads)=4/8=1/2\n\nodds:even money\n\ni hope that this helps
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.
130845
the correct answer should be # 2 because the main point of the experiment is that "the same steps don't always produce the same results"
413399
Binomial distribution, (a) p=0.5, n=12; (b) p=0.5, n=13.  You can do it!
428931
Your hypothesis seems to be correct.  Also, this would be a single-tailed distribution since there is only one open end.
254561
It is actually binomial with parameter p=0.5 and n=400. Since 400 is quite a large number, u can approximate this by the Normal distribution with the same mean and variance, which are, np=400 and np(1-p)=100.\n\nX ~ N(200,100)\n\nu wanna test that if 216 is within 95% interval.\n\nVariance=100\nSDev=10\n\n216-200=16,\n16/stdev=1.6\n\nso, 216 is 1.6 stdev from the mean 200. which is within +/-1.96(95% confidence interval. hence, the null hy is accepted, with 95% significance level.\n\n\n\nbtw, drbinsomething's solution is equivalent to mine. he directly test P as a random variable while I tested mean nP as the random variable. He didn't mention it, but he took P to be asymptotically Normal as I did. '1.96' only applies when the distribution is Gaussian/Normal.
172387
This is a simple two tailed test of the equality of means.\n\nLook this up in a stats book, but the Hypothesis that you are testing is Mean1=Mean2.
304866
P(At least one is white) = 1-P(none are white)\n= 1-P(BBB)\n= 1-13/36)(12/35)(11/34)\nP(One is white) = P(BBW) + P(BWB) + P(WBB)\n= (13/36)(12/35)(23/34) + (13/36)(23/35)(12/34) + (23/36)(13/35)(12/34)\nP(Two are white) = P(BWW) + P(WBW) + P(WWB)\n= (13/36)(23/35)(22/34) + (23/36)(13/35)(22/34) + (23/36)(22/35)(13/34)\nP(Three are white) = P(WWW)\n= (23/36)(22/35)(21/34)
545934
a) i +(1/3)(j-i)\n\n  b)  \n1-12 =>1-4 & 5-12\n\n... 1-4=>1 & 3-4\n.......3-4=>3 & 4\n\n...5-12=> 5-6 & 7-12\n......5-6=> 5 & 6\n......7-12=>7-8 & 9-12\n..........7-8=>7 & 8\n..........9-12=> 9 & 10-12\n...............10-12=>10 & 11-12\n....................11-12=>11 & 12\n\n c) 1/3 of the computations lead to a subset of the previous set\nwhich is 1/2 the size of the other subset which is lead to twice\nas often. So it averages out to the same number of\ncomputations as a conventional binary search.
676257
In this case 'N', which is the sample size, is 120.  The probability of an event occuring is equal to the number of times it occurred in the sample divided by N, so:\n\nprobability of X=0:  20/120 = 1/6 ~ 0.1667\nprobability of X=1:  65/120 = 13/24 ~ 0.5417\nprobability of X=2:  25/120 = 5/24 ~ 0.2083\nprobability of X=3:  10/120 = 1/12 ~ 0.0833\n\nAlso the average value of X would be 1.2082 using this probability distribution through an infinite number of line checks.
363162
samples or sample size; usually designated "n"
116687
did a sample question on a very inportant test
545904
Okay, keep in mind that this problem is asking the probablility that AT LEAST 8 were married - that means the probability that 8 were married plus the probability that 9 were married plus the probablility that 10 were married.\n\nSo let's start with the probability that 8 were married:\n\nIf the probability that someone was never married is .15, that means the probability that they were married is 1 minus .15, or .85. And if 8 people in the sample were married, that means that 2 were not. So what you're going to do is multiply .85^8 (for married people) x .15^2 (for unmarried). \n\nThe principle is the same for 9 married people and ten married people: \n\nFor 9 married, you multiply .85^9 x .15^1\n\nFor 10 married, you multiply .85^10 x .15^0\n\nFinally, add all three answers together, which gives you the probability that either 8, 9, or 10 people in the sample were married, which is what the question was asking! Hope that helped!
489958
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.