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Q: Is sample size going to be equal to standard deviation?

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Short answer, complex. I presume you're in a basic stats class so your dealing with something like a normal distribution however (or something else very standard). You can think of it this way... A confidence interval re-scales margin of likely error into a range. This allows you to say something along the lines, "I can say with 95% confidence that the mean/variance/whatever lies within whatever and whatever" because you're taking into account the likely error in your prediction (as long as the distribution is what you think it is and all stats are what you think they are). This is because, if you know all of the things I listed with absolute certainty, you are able to accurately predict how erroneous your prediction will be. It's because central limit theory allow you to assume statistically relevance of the sample, even given an infinite population of data. The main idea of a confidence interval is to create and interval which is likely to include a population parameter within that interval. Sample data is the source of the confidence interval. You will use your best point estimate which may be the sample mean or the sample proportion, depending on what the problems asks for. Then, you add or subtract the margin of error to get the actual interval. To compute the margin of error, you will always use or calculate a standard deviation. An example is the confidence interval for the mean. The best point estimate for the population mean is the sample mean according to the central limit theorem. So you add and subtract the margin of error from that. Now the margin of error in the case of confidence intervals for the mean is za/2 x Sigma/ Square root of n where a is 1- confidence level. For example, confidence level is 95%, a=1-.95=.05 and a/2 is .025. So we use the z score the corresponds to .025 in each tail of the standard normal distribution. This will be. z=1.96. So if Sigma is the population standard deviation, than Sigma/square root of n is called the standard error of the mean. It is the standard deviation of the sampling distribution of all the means for every possible sample of size n take from your population ( Central limit theorem again). So our confidence interval is the sample mean + or - 1.96 ( Population Standard deviation/ square root of sample size. If we don't know the population standard deviation, we use the sample one but then we must use a t distribution instead of a z one. So we replace the z score with an appropriate t score. In the case of confidence interval for a proportion, we compute and use the standard deviation of the distribution of all the proportions. Once again, the central limit theorem tells us to do this. I will post a link for that theorem. It is the key to really understanding what is going on here!

5/32" is as close as you can get which is 3.969 mm. An 11/64 is more than likely going to be too large as it is equal to 4.366.

to introduce what i am going to do

Counting votes for a sample of how an election is going.

letter for dental lab ,going to vacation.

standard error for proportion is calculated as: SE = sqrt [(p)(1-p) / n ] so let us say that "p" is going to represent the decimal proportion of respondents who said YES.... so... p = 20/25 = 4/5 = 0.8 And... we then are going to say that the complement of "p" which is "1-p" is going to represent the decimal proportion of respondents who said NO ... so... 1-p = 1 - 0.8 = 0.2 Lastly, the "n" in the formula for standard error is equal to 25 because "n" represents the sample size.... So now all you have to do is plug the values you found for "p" and for "1-p"... (remember "p = 0.8" and "1-p = 0.2")... and "n=25".... Standard Error (SE) = sqrt [(p)(1-p) / n ] ............................ = sqrt [(0.8)(1-0.8) / 25 ] ............................ = sqrt [(0.8)(0.2) / 25 ] ............................ = sqrt [0.16 / 25] ............................ = sqrt (0.0064) ............................ = +/- 0.08

Yes, you can process it on a TI-83 or higher, your calculator can compute it for you. To do it your problem must include a list of data points. To enter the data on your calculator, push STAT. Then go to EDIT. In edit there is a chart. The top row says L1, L2, etc. If there are already values in your lists, you can clear them by going up and highlighting L1, press clear, and then the down arrow. That will clear the list. To enter data, move to the rectangular box under L1. Type in your first value and press enter. Continue to do so until all values are entered under list one. Next, press STAT. Move over to CALC and select 1-Var Stats. Press enter. A buch of data will pop up. The symbol on your calculator for standard deviation is Sx, and there will be a value listed. That is the standard deviation for your data set. Some other things that are good to know with the info we have produced: _ X = mean/average of data n = number of data points minX = minimum x value Q1= quartile 1 Med = median Q3 = quartile 3 maxX = maximum x value Sx = Standard Deviation And a Q looking thing and a X is (Population Standard deviation) "Qx"

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An example of voluntary response sample would be when a radio talk-show decides to run a call-in survey on a controversial topic, such as gun control. The most likely callers are going to be people with the strongest opinions, thus the sample that results from the survey is going to over-represent those very people.

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