How does the sample size affect our estimate and our decisions?
Parameters are the (usually unknown) measures of a population. Often they are represented by Greek letters like mu and sigma.
Statistics are the calculated measures generated from the samples. Statistics are estimators for parameters.
When the average of a statistic is the parameter itself, it is called an unbiased estimator. X-bar is an unbiased estimator for mu, the population mean.
Sampling distributions are the distributions of all of the averages of all of the samples of size n taken from a population.
When the sample size n increases, the variability of the means of the samples decreases--the graph of the sampling distribution is taller and narrower. When the sample size n decreases, the variability of the means of the samples increases.
This holds for sample proportions. The mean of the sample proportions (p-hats) is the true proportion for the population, p. Under special conditions we can use a formula for the standard deviation of the p-hats: SQRT(p*(1-p)/n).
The condition that allows this is that the sample is less than 1/10th of the population (and, of course, we're talking about simple random samples!!)
Also, the really BIG twist is that we can also use an approximation to the normal distribution when the expected numbers of successes and and failures are both 10 or more.
So, about that CLT thing. . . What was the REALLY BIG idea with the Central Limit Theorem???
How do you express the distributions for a binomial X, a geometric X, a uniform X, a normal X, the sampling distribution (X-bar), and the sample proportions (p-hat)?
When can you assume that the sampling distribution is approximately normally distributed?
What do you have to write to support your calculations of mean and standard deviation? your calculations of probabilities?
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7 comments:
Because they want the distribution of the MEAN, you are correct in assuming that you have to find the mu of x-bar and the sigma of x-bar.
ummm i think the answer to number four on the quiz is incorrect... The question asked what would n have to equal in order to have a standard deviation of .1. Therefore you can square .1 to get .01 = .16/n and then after you divide .16/.01 you would get 16... Therefore I am confused to why the correct answer or so on the quiz is 16?
jf-you are correct. That problem is messed up. The answer "16" will give you a std dev = 0.1.
ba-X and X-bar are used when measurements or counts are provided. They will also provide a std dev in the problem.
p and p-hat are used when you have a guess at the proportion of the time that the success happens.
sf-They didn't give you a std dev, so know that you look for p and p-hat. p = 14.1% p-hat = 3500/25,000. Work it as a proportion problem.
Ok I know its a little late, but are there any major formulas we need for the test tommorow. I have gotten some form the book, but i dont want to miss any.
Clarifying # 4 on the quiz:
sd = sqrt(pq/n), so
sd^2 = pq/n
and n = pq/(sd^2).
Now, p = .2, q = .8, and sd = .1, so the right side = .2*.8/.01
which equals 16.
Why doesn't the online quiz have the right answer? Even the experts have technology blips some days. I elect to look at the positive: it keeps you guys honest!
Good work, all!
Formulas? Which ones do you know already? Which ones can you use?
i hope georgia freezes tomorrow
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