Section 6.4. Sampling Distributions and Estimators

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Section 6.4 Sampling Distributions and Estimators

IDEA Ch 5 and part of Ch 6 worked with population. Now we are going to work with statistics. Sample Statistics to estimate population parameters. To make that transition we need the theory in this section.

Sampling Distribution of a Statistics The distribution (organization) of all values of the statistic (mean, standard deviation, variation, variance, median and proportion ) when all possible samples of the same size n are taken from the same population. Assume you take all the possible samples of size n and find the required statistics for each sample If you organize all of those statistics of each different sample into a table, this is a sampling distribution.

Sampling Distribution of a Proportion The distribution of sample proportions, with all samples having the same sample size n taken from the same population. We need to distinguish between a population proportion p and some sample proportion: p = population proportion ˆp = sample proportion (p hat)

Example A -1 Let us assume in the whole world there are only three numbers {1, 2, 5}. Create a sampling distribution of size 2. NOTE Combination is the proper idea but we will use permutation because of the probability. Repetition is allowed because we are dealing with a population.

Example A - 1 Sampling distribution of size 2 Sample n=2 1, 1 1, 2 1, 5 2, 1 2, 2 2, 5 5, 1 5, 2 5, 5

Example A - 2 Determine the probability of each sample.

Example A - 2 Probability of sampling distribution of size 2 Sample n=2 Probability 1, 1 1/9 1, 2 1/9 1, 5 1/9 2, 1 1/9 2, 2 1/9 2, 5 1/9 5, 1 1/9 5, 2 1/9 5, 5 1/9

Example A - 3 Given the whole world of numbers {1, 2, 5} Find the population proportion of odd numbers, p. Find the proportion of odd numbers, ^p for each sample

Example A - 3 Population, p = 2/3 Sample n=2 Probability 1, 1 1/9 2/2 = 1 ^p 1, 2 1/9 ½ = 0.5 1, 5 1/9 2/2 = 1 2, 1 1/9 ½ = 0.5 2, 2 1/9 0/2 = 0 2, 5 1/9 ½ = 0.5 5, 1 1/9 2/2 = 1 5, 2 1/9 ½ = 0.5 5, 5 1/9 2/2 = 1

Example A - 4 Find the average of the proportion of odd numbers, ^p Average = μ= [ ^p p(x)]

Example A - 4 Population, p = 2/3 Sample n=2 Probability, p(x) ^p p(x) * +1, 1 1/9 2/2 = 1 1/9 ^p 1, 2 1/9 ½ = 0.5 1/18 1, 5 1/9 2/2 = 1 1/9 2, 1 1/9 ½ = 0.5 1/18 2, 2 1/9 0/2 = 0 0 2, 5 1/9 ½ = 0.5 1/18 5, 1 1/9 2/2 = 1 1/9 5, 2 1/9 ½ = 0.5 1/18 5, 5 1/9 2/2 = 1 1/9

Example A - 4 Population, p = 2/3 Average of sample = 1/9 + 1/18 + 1/9 + 1/18 +0+ 1/18 + 1/9 + 1/18 + 1/18 What do you notice?

Example A - 4 What do you notice population mean equals the mean of the samples Why does this work? You are taking every single possible sample and you are averaging them.

THEORY What does that mean? In our population we don't have three things we have millions of things so we cant do this all the time, so we need to see the relationship. While every single sample does not give you the population proportion the average of all sample proportions will. Sample proportions target population proportions. They sample we used makes a good estimator

Sampling Distribution of a Mean The distribution of all possible sample means with all samples having the same size n taken from the same population.

Sampling Distribution of a Variance The distribution of sample variance, with all samples having the same sample size n taken from the same population.

Estimators Estimator is a statistic used to infer (estimate) the value of a population parameter. There are two estimators Unbiased Biased

Unbiased Estimator A statistic that targets the value of the population parameter in the sense that the sampling distribution of the statistics has a mean that is equal to the mean of the corresponding parameter. There are three unbiased estimators Mean Variance Proportion

Biased Estimator A statistics that do not target the value of the corresponding population parameter. There are three biased estimators Median Range Standard Deviation (it is a very small bias)

Example 2 Three randomly selected households are surveyed. The numbers of people in the household are 2, 3 and 10. Assume that samples of size n =2 are randomly selected with replacement from the population of s, 3, and 10. Listed below are the nine different samples. 2, 2 2, 3 2, 10 3, 2 3, 3 3, 10 10, 2 10, 3 10, 10

Example 2 - a Find the median of each of the nine samples, then summarize the sampling distribution of the median in the format of a table representing the probability distribution of the distinct median values.

Example 2 - a Sample Median (average of #) 2, 2 2 2, 3 2.5 2, 10 6 3, 2 2.5 3, 3 3 3, 10 6.5 10, 2 6 10, 3 6.5 10, 10 10

Example 2 - a Sample Median Probability 2 1/9 2.5 2/9 3 1/9 6 2/9 6.5 2/9 10 1/9

Example 2 - b Compare the population median to the mean of the samples medians.

Example 2 - b Population Median Median of {2, 3, 10} = 3 Mean of the sample medians Mean of {2, 2.5, 6, 2.5, 3, 6.5, 6, 6.5, 10} Mean = (2+2.5+6+2.5+3+6.5+6+6.5+10) / 9 Mean = 45 / 9 Mean = 5

Example 2 - c Do the sample medians target the value of the population median? In general, do sample medians meet good estimators of population medians? Why or why not?

Example 2 - c No, because the population mean does not equal the mean of the sample medians. In this case we can say that sample medians do not target the population median. Therefore sample medians do not make good estimation of the population medians