p value and n value in statistics?
478561
I don't remember N value, but I know that P value is the probability that something is done by "chance".  In other words, I cannot "prove" that I am a psychic, but I can do things that point farther and farther away from me 'cheating" or "just guessing".  The farther one gets from something being "chance" the smaller the P value is.  In Statistics, P-value is only considered "significant" if it is below .05 (or less than possibly happening one in 20 times).
662741
In basic statistical hypothesis testing, the P-value represents the probability of encountering a result more extreme than the value that we've observed in our testing.\n\nAs an example - suppose we had conducted a properly designed experiment, and discovered that the mean score of our sample was 103.5. We would evaluate this finding of 103.5 by comparing it to a known "sampling distribution" - a distribution of sample means of size n that was randomly drawn from a population with (for the sake of this example) a mean of 100 and a standard deviation of 1.75.\n\nWhat we want to do is evaluate the likelihood that a sample of size n - randomly drawn from the population - would have a mean as extreme as 103.5.\n\nWe begin with the assumption that the null hypothesis is true - that our sample belongs to the population - and that the deviation of 3.5 from the population mean of 100 is simply due to random error.\n\nIn this situation, the appropriate statistical tool for evaluating the probability of observing a sample mean as extreme as 103.5 is the t-test. We begin by "standardizing" the raw score of 103.5 by converting it to a deviation score and then dividing that deviation score by the standard deviation of the sampling distribution.\n\nThe formula is simply  (103.5 - 100.0) / 1.75  =  2.0\n\nThis means that our sample mean of 103.5 is 2.0 standard deviation units above the mean of the sampling distribution. Now we need to find out the probability of encountering a sample that is 2 standard deviation units from the mean. \n\nBecause the sampling distribution is assumed to be approximately normally distributed, we can refer to the z-table (reproduced in any statistic text), and discover that only 2.28 percent of all the scores in a normal distribution fall above 2 standard deviation units above the mean. The proper interpretation of this result is that the probability of observing a sample this extreme from a population with a mean of 100 and a standard deviation of 1.75 is 2.28% or less.\n\nP-values are used in conjunction with critical values - pre-determined probabilities that we feel are so unlikely to have occurred by chance, that we are comfortable concluding that results this extreme could not have been selected from the population (thereby rejecting the null hypothesis that the sample was selected from the population).\n\nThe common critical values are .05 and .01, meaning, in each case, that we'll reject the null hypothesis if our finding is so extreme that it would only be expected to happen by chance less than 5 percent of the time - or in the case of .01, less than 1 percent of the time.\n\nReferring back to our finding - we concluded that a result of 103.5 would only happen 2.28 percent of the time or less - if our pre-selected critical value had been .05 (5 percent), we would reject the null hypothesis, and conclude that our sample was not selected from the population. If the critical value had been 1 percent, because our finding happens 2.28 percent of the time, we would fail to reject the null hypothesis.\n\nMan, I hope this made some sense!
1047890
class C\n{\npublic:\nenum { NUM_VALUES = 10 };\n\nprivate:\ndouble m_values[NUM_VALUES];\n\npublic:\n// Sets the number stored at index i in m_values to v\nvoid SetValue(int i, double v) throw( std::exception )\n{\nif ( i < 0 || i >= NUM_VALUES ) throw std::exception("Index out of range");\nm_values[i] = v;\n}\n\n// Gets the number stored at index i in m_values\ndouble GetValue(int i) const throw( std::exception )\n{\nif ( i < 0 || i >= NUM_VALUES ) throw std::exception("Index out of range");\nreturn m_values[i];\n}\n\n    // Returns the mean of the numbers stored in m_values\n    double GetMean(void) const \n    {\n        double sum = 0;\n        for ( int i = 0; i < NUM_VALUES; i ++ ) sum += m_values[i];\n        return sum / NUM_VALUES;\n    }\n\n    // Returns the sample standard deviation of the numbers stored in m_values\n    double GetStandardDeviation(void) const \n    {\n        double mean = GetMean();\n        double sum = 0;\n        for ( int i = 0; i < NUM_VALUES; i ++ ) sum += (m_values[i]-mean)*(m_values[i]-mean);\n        return sqrt(sum/(NUM_VALUES-1));\n    }\n};
319924
It depends on the distribution.  Some probability distributions have multiple parameters (such as the mean and standard deviation for the Normal distribution).  Phi2 would be one parameter of the particular distribution used in the problem.
47438
Take each of your values, and then subtract the mean from them.  E.g., 70 - 56 = 14, 60 - 56 = 4, etc.\n\nNow take each of these answers and square it.  Half of the starting numbers will be negative (eg 30 - 56 = -26) but once you square all of these numbers, they should all be positive.\n\nNow you'll have 7 positive numbers. Add them all together and divide the answer by 7.  Finally, take the square root of that value, and that's the standard deviation.\n\nSo 14, 4, -6, 4, 4, 9, and -26\nbecome 196, 16, 36, 16, 16, 81, and 676\nSummed together gives 1037\ndivided by 7 is ~148.14\nSquare-rooted is ~12.2\n\nThe standard deviation is 12.2.\n\nNOTE: the rt11guru answer that follows is based on a technique used in some larger sample sets where 1 is subtracted from the number of entries, ie the sum is divided by 7-1=6 rather than 7.  That is called the *sample standard deviation*, as shown in the link.
919154
It depends on whether you want the mean number of heads, the total number of heads, etc.\nIf you want the total, multiply the standard deviation by sqrt(n), where you have n tosses.\nIf you want the average, divide by sqrt(n).\n\n(Note, the formula talked about below is for the *margin* of error, not the standard error. And thats only for means.)
189338
You have to say the starting value for n.  \nThe first value for n has to be given; it could be basically anything.
849629
Alpha, as related to level of significance, is purely up to the statistician.  It simply denotes how sure you are of your findings.  The p-value is more descriptive.  The p-value is basically the probability that the statistician will reject the null hypothesis when it in fact should not be rejected.  If the p-value is less than the chosen alpha level, the null hypothesis is rejected.
273970
No problem \n\ngo to google and search under "Z table" +statistics. \n\nYou'll find a lot.
714755
A.  Since the values range from 5 to 55, and if we believe that almost all of the values are within 2 sd's of the mean, then that would mean that there are a total of 4 sd's between 5 and 55.  So we can approximate this standard deviation by saying\n\n(55-5)/4 = 12.5\n\nB.  Since you are taking a sample of 400, then the standard error of the mean is 12.5/sqrt(400) = 0.625.  Since the sample size is large, then the sample mean is approximately normal.  To be within 2 units, then you would have to be within mu-2 and mu+2.  So we can standardize.\n\nz = (mu-2-mu)/0.625 = -3.2\nz = mu+2-mu)/0.625 = 3.2\n\nThe area between these two values would then be approximately 0.9986.\n\nC. In this case, our hypothesis test would look like this.\n\nH0: mu = 35\nHA: mu < 35\n\nOur test statistic would be (31.75-35)/0.625 = -5.2\n\nWe would then reject the null hypothesis.  So we would conclude that the scientists are correct.
837986
http://www.statsoft.com/textbook/stathome.html
282608
If you can provide more details then I can help you. p-value has to be compared with the level of significance. z-table and t-tables are different and u'll requir specific table for that. are you solving a question of hypothesis testing??\n\nCritical value for 't' requires the degree of freedom.
806818
1. If you know the variances you'd know the mean, because you need to calculate the mean to find the variance. Newer stat texts have finally started to get smart and quit putting stupid topics like this in there!\n\n2. If you suspect that the population variances are equal, you can combine the samples to get one estimate of that variance. That's called 'pooling.' When variances are equal use the formula that combines those variances for the calculation.\n\n3. The p-value comes from this calculation so you don't use it to determine which formula to use. Your background knowledge is what you would use to determine whether variances are likely to be equal.\n\n4. It's never wrong not to pool. Pooling when appropriate does give a test more power, but not pooling is always OK.
953152
The binomial distribution is given by:\n\n    N!\n----------- x (p)^n x (1-p)^(N-n)\nn!(N-n!)\n\nWhere N is the number of trials, n is the number of successes and p is the probability of success\n\nIn this question, n, the "number of successes" is the number of failures caused by operator error. p, the "probability of success" is the probability that the failure is caused by operator error (i.e. 30%).\n\nSo, in the questions you need to substitute the following for each variable:\n\ni) N=20, p=0.3\n\nAs this question uses the phrase "at least", you will need to use the equation more than once and add all the results. Repeat the calucation for each n for n=10 to n=20 and add the results.\n\nI got the following values for each n (rounded to 8 d.p)\n\nn=10, 0.03081708\nn=11, 0.01200665\nn=12, 0.00385928\nn=13, 0.00101783\nn=14, 0.00021811\nn=15, 0.00003739\nn=16, 0.00000501\nn=17, 0.0000005\nn=18, 0.00000004\nn=19, 0.00000000\nn=20, 0.00000000\n\nTotal: 0.0479619 or 4.80% chance that at least 10 are caused by operator error\n\nii) N=20, p=0.3\n\nThis question is similar to q1, however you need to repeat the calculation for each n for n=0 to n=4 and add the results.\n\nI got the following values for each n (rounded to 8 d.p)\n\nn=0, 0.00079792\nn=1, 0.00683934\nn=2, 0.02784587\nn=3, 0.07160367\nn=4, 0.13042097\n\nTotal: 0.23750778 or 23.75% chance that no more than 4 are caused by operator error\n\niii) In the random sample, only 5 are errors. Work out the probability of this being the case using N=20, n=5, p=0.3. If it seems reasonable, then the 30% figure is probably right, and the sample just contains a low number of failures due to operation error. If the probability is impossibly small, then the 30% is not correct.\n\nThe probability I calculated for n=5 was 0.178863051. This indicates a 17.89% chance that there would be 5 errors in the sample. This is not so outrageous to suggest that 30% is an incorrect percentage of operator errors. It's just that in this sample, the number of actual errors was less than the number of expected errors (which is 30% x 20, or 6 errors).
795924
Binomial distribution, (a) p=0.5, n=12; (b) p=0.5, n=13.  You can do it!
885975
Good lord. I had to teach myself statistics, and I don't remember a lot of it, but I'll do what I can.\n\na) The null hypothesis is that the two learning environments have no impact on the grades of the course. The alternative hypothesis is that they will be different.\nb) Two-tailed, since you do not know which environment will be better or worse.\nc) The time spent in one vs. the other, and the quality of the teacher (as measured as a variable.)\nd) That there would be a standard margin for disparity.\ne) I have no clue. I never got that far.\n\n2) Don't know.
712323
p is a probability. So you're almost right - the question is, can you have a p value of 0 or 1?\nAnd the answer is yes. Although nearly every single statistical test you will use will never give 0 or 1, it is theoretically possible.