The answer is 9. Let's first use and example, then view it conceptually. If we have scores of 3, 4, 5, 6, and 7, the mean will be five. (3 + 4 + 5 + 6 + 7)/5 = 25/5 or 5 If we add 4 to each of those scores we will then have: (7 + 8 + 9 + 10 + 11)/5 = 45/5 = 9 To see that this will work for all sets of scores, think of the formula for the mean: (a1 + a2 + a3 ... + an)/n = original mean Where a represents a single score in the distribution and n = number of scores in the sample. If we were to add 4 to each of the scores, then: [(a1+4) + (a2+4) + (a3+4) ... + (an+4)]/n [(a1 + a2 + a3 ... + an)+4n]/n [(a1 + a2 + a3 ... + an)/n] + 4 original mean + 4
Yes. You could have a biased sample. Its distribution would not necessarily match the distribution of the parent population.
It is 0.0606
Well, sort of. The Chi-square distribution is the sampling distribution of the variance. It is derived based on a random sample. A perfect random sample is where any value in the sample has any relationship to any other value. I would say that if the Chi-square distribution is used, then every effort should be made to make the sample as random as possible. I would also say that if the Chi-square distribution is used and the sample is clearly not a random sample, then improper conclusions may be reached.
The binomial distribution.
Yes. The standard deviation and mean would be less. How much less would depend on the sample size, the distribution that the sample was taken from (parent distribution) and the parameters of the parent distribution. The affect on the sampling distribution of the mean and standard deviation could easily be identified by Monte Carlo simulation.
Sampling distribution is the probability distribution of a given sample statistic. For example, the sample mean. We could take many samples of size k and look at the mean of each of those. The means would form a distribution and that distribution has a mean, a variance and standard deviation. Now the population only has one mean, so we can't do this. Population distribution can refer to how some quality of the population is distributed among the population.
The binomial distribution is defined by two parameters so there is not THE SINGLE parameter.
Because t-score isn't as accurate as z-score, you should use 40 as a safety sample size, rather than 30 as you would for a z-score.
The variance of the estimate for the mean would be reduced.
z =0 and P(X< x) = 0.5 Explanation: z = (x-xbar)/sd, where xbar is the estimated mean or average of the sample, sd is the standard deviation, and x is the value of the particular outcome. We change x to z so that we can use the normal distribution or t-tables tables, which are based on a zero mean and 1 standard deviation. For example: What is the probability that the mean value of the distribution is 5 or less, given the sample average is 5 and the sd is 2? The z-score would be (5-5)/2 which is equal to 0. The probability, if we assume the normal or t-distribution, is 0.50. (see normal distribution tables) I hope this makes sense to you. The normal distribution is symmetrical. Per the example, a sample average of 5 tells you there is equal chance of the population mean being above and below 5.
It would be 3*5 = 15.
According to a z-score calculator, it would be in the 96.9 percentile, which means that the test-taker did better than 96.9% of the sample group of test-takers.