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Which of the following best describes the condition necessary to justify using a pooled estimator of the population variance?
1- Assuming this represents a random sample from the population, the sample mean is an unbiased estimator of the population mean. 2-Because they are robust, t procedures are justified in this case. 3- We would use z procedures here, since we are interested in the population mean.
What does n-1 indicate in a calculation for variance?
The n-1 indicates that the calculation is being expanded from a sample of a population to the entire population. Bessel's correction(the use of n − 1 instead of n in the formula) is where n is the number of observations in a sample: it corrects the bias in the estimation of the population variance, and some (but not all) of the bias in the estimation of the population standard deviation. That is, when estimating the population variance and standard deviation from a sample when the population mean is unknown, the sample variance is a biased estimator of the population variance, and systematically underestimates it.
How do you prove that the sample variance is equal to the population variance?
You cannot prove it because it is not true.The expected value of the sample variance is the population variance but that is not the same as the two measures being the same.
How large would your sample have to be for appropriate estimation of the whole population?
First you have chose an estimator for what you want to know about the population. In general the level of variability in the result that any estimator provides will depend on the variability in the population. Therefore, the greater the variability in the population the larger your sample size must be. You will also need to decide how much precision is required in your estimate. The more precision you require the greater your sample size will have to be.
What does it means if the standard deviation is large?
that you have a large variance in the population and/or your sample size is too small
Is sample variance unbiased estimator of population variance?
No, it is biased.
Show that in simple random sampling the sample variance is an unbiased estimator of population variance?
It is a biased estimator. S.R.S leads to a biased sample variance but i.i.d random sampling leads to a unbiased sample variance.
Why is the sample mean an unbiased estimator of the population mean?
The sample mean is an unbiased estimator of the population mean because the average of all the possible sample means of size n is equal to the population mean.
What is the best estimator of population mean?
The best estimator of the population mean is the sample mean. It is unbiased and efficient, making it a reliable estimator when looking to estimate the population mean from a sample.
What does it mean to say that the sample variance provides an unbiased estimate of the population variance?
It means you can take a measure of the variance of the sample and expect that result to be consistent for the entire population, and the sample is a valid representation for/of the population and does not influence that measure of the population.
Which of the following best describes the condition necessary to justify using a pooled estimator of the population variance?
1- Assuming this represents a random sample from the population, the sample mean is an unbiased estimator of the population mean. 2-Because they are robust, t procedures are justified in this case. 3- We would use z procedures here, since we are interested in the population mean.
Differentiate estimate and estimator?
It can get a bit confusing! The estimate is the value obtained from a sample. The estimator, as used in statistics, is the method used. There's one more, the estimand, which is the population parameter. If we have an unbiased estimator, then after sampling many times, or with a large sample, we should have an estimate which is close to the estimand. I will give you an example. I have a sample of 5 numbers and I take the average. The estimator is taking the average of the sample. It is the estimator of the mean of the population. The average = 4 (for example), this is my estmate.
What does n-1 indicate in a calculation for variance?
The n-1 indicates that the calculation is being expanded from a sample of a population to the entire population. Bessel's correction(the use of n − 1 instead of n in the formula) is where n is the number of observations in a sample: it corrects the bias in the estimation of the population variance, and some (but not all) of the bias in the estimation of the population standard deviation. That is, when estimating the population variance and standard deviation from a sample when the population mean is unknown, the sample variance is a biased estimator of the population variance, and systematically underestimates it.
What biased estimator will have a reduced bias based on an increased sample size?
The standard deviation. There are many, and it's easy to construct one. The mean of a sample from a normal population is an unbiased estimator of the population mean. Let me call the sample mean xbar. If the sample size is n then n * xbar / ( n + 1 ) is a biased estimator of the mean with the property that its bias becomes smaller as the sample size rises.
Is sample standard deviation the same as standard deviation?
No, the standard deviation is a measure of the entire population. The sample standard deviation is an unbiased estimator of the population. It is different in notation and is written as 's' as opposed to the greek letter sigma. Mathematically the difference is a factor of n/(n-1) in the variance of the sample. As you can see the value is greater than 1 so it will increase the value you get for your sample mean. Essentially, this covers for the fact that you are unlikely to obtain the full population variation when you sample.
When do you use a z-value rejection region when testing a claim about a population mean?
If you know the population variance or if you have a very large sample then you could reliably use a Z test. Otherwise you should use a t test and use s^2 as an estimator for the population variance.
How is it that a random samples gives a fairly accurate representation of public opinion?
The main point here is that the Sample Mean can be used to estimate the Population Mean. What I mean by that is that on average, the Sample Mean is a good estimator of the Population Mean. There are two reasons for this, the first is that the Bias of the estimator, in this case the Sample Mean, is zero. A Bias other than zero overestimates or underestimates the Population Mean depending on its value. Bias = Expected value of estimator - mean. This can be expressed as EX(pheta) - mu (pheta) As the Sample Mean has an expected value (what value it expects to take on average) of itself then the greek letter mu which stands for the Sample Mean will provide a Bias of 0. Bias = mu - mu = 0 Secondly as the Variance of the the Sample Mean is mu/(n-1) this leads us to believe that the Variance will fall as we increase the sample size. Variance is a measure of the dispersion of values collected from the centre of the data. Where the centre of the data is a fixed value equal to the median. Put Bias and Variance together and you get the Mean Squared Error which is the error associated with using an estimator of the Population Mean. The formula for Mean Squared Error = Bias^2 + Variance With our estimator we can see that as the Bias = zero, it has no relevance to the error and as the variance falls as the sample size increases then we can conclude that the error associated with using the sample mean will fall as the sample size increases. Conclusions: The Random Sample of public opinon will on average lead to a true representation of the Population Mean and therefore the random samle you have will represnt the public opinion to a fairly high degree of accuracy. Finally, this degree of accuracy will rise incredibly quickly as the sample size rises thus leading to a very accurate representation (on average)
