Statistical Estimation of sampling Theory
Today is Wednesday, May 4, 2011.
Please welcome our reporter Ma’am M. Pilapil. She will be discussing with us the Sample Distribution.
For sampling distribution we can compute a mean, variance, standard deviation is called the standard error.
Any quantity obtained for a sample for the purpose of estimating a population parameter is called a SAMPLE STATISTICS OR STATISTICS.
Kinds of sampling
- Sample Random Sample- N elements is selected from a population of N elements using a sampling plan which each of the possible samples has the same chance of selection.
- Stratified Random Sampling- when the population of interest consists that each subpopulations, or strata, a sampling plan that ensures that each population sample is represented by each sample.
- A Cluster Sample is a simple random sample of clusters from the available cluster in the population
- The kth systematic random sample involves the random selection of one of the first k elements in an ordered population and then systematic selection of every kth element.
Examples:
- Assume that the weights of 3000 female students of UST are normally distributed with mean 45 kg and standard deviation of 2kg. If 80 samples consisting of 25 students each are obtained, what would be the expected mean and standard deviation of the resulting sampling distribution of means if sampling were done a) with replacement, b) without replacement?
Solution:
a) with replacement
and
b) without replacement
and
- In how many example of example1 would you expect to find the mean between:
a) 44.7 and 46.1?
b) Less than 46 kg?
44.7 - 45
a. 44.7 kg in standard units =-----------------
0.4
= -0.75
46.1 - 45
46.1 in standard units =----------------------
0.4
= 2.75
Proportion of samples with mean between44.7 and46.1
=( area under the normal curve between z = - 0.75 and the z= 2.75)
=0.2734 + 0.4970 = 0.7704
Then the expected number of samples= (80)(0.7704) or 62
46-45
b) 46 kg in standard units ---------------- =2.5
0.4
Proportion of samples with means less than 46 kg
=(area under the normal curve less than z=2.5)
= 0.5 + 0.4938 = 0.9938
Then the expected number of samples = (980)(0.9938) 0r 80.
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