Suppose a study is conducted to test a claim at a 0.01 significance level. That less than 75% of college stude
610539
Try finishing the question.
814472
p stands for probability and it is less than 0.1%
188534
zero is real. I'd go with a 1 however just because it doesn't give people the option to "opt out" of thinking about thier answer. Although it may be woth it just to give choice types such as "very affected" or "Strongly agree". They will create a "universal" standard instead of having each person come up with a revised scale in thier mind (for example some people never put 10s and 1s on surveys for fear of looking bad)
759589
It will show up in less than an hour.\n\nIt isn't likely that they will be able to tell how much unless it is a very high-end test.
809295
18/360 *100=5%\nthe claims regenuine,not tall claims
834919
Probably between 6-8 percent, just like the general population. I don't know where you got the 1% from, but it's wrong.
629585
not C or B, so it come to A, or D, a pvalue between 0.01-0.05 is generally considered strong evidence for alternetive hupothothes and against null hypotheses, A says same thing as choice C so D is the right anwer
628073
please modify your question and add a study hypothesis of what
890026
it depends on the test results
826746
It's not completely to do with the weighting efficiency itself.  It depends how large your actual sample size is. If you only had 3 people in the sample then it wouldn't be very reliable regardless of the weighting; if you have 3 million people in the sample then it'd still probably be as reliable as you could hope for even with the heaviest weighting.\n\nI think (I'm not sure of this, but fairly sure) that your effective sample size is your achieved sample size multiplied by the weighting efficiency, so multiply your sample size by 0.58\n\nEg. if you had 1,000 in your sample, then your results have an effective sample size of 580, so your answers would be as reliable as if you'd asked 580 people, so use this base size in significance tests rather than the full 1,000.  ("reliability" in this case would rely on the % confidence you want to rely on, the actual percentages that you're comparing, and the effective sample sizes of the groups you're comparing).
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There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
983117
C 49.6%\n\n(or A with rounding to 0 decimal places/1 significant figure)
758530
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.
835646
haha. 71% is around a C-. to me, that is bad, but it's average.
748810
76.5% (159 million) -2001 ARIS study
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/
172997
ZERO\n\nand ironically not everything is true, you have to study it and find out for yourself