Analyze and identify what is asked in the following problems. 1. How many samples ol size n = 3 can be selected from a population with the following sizes? 8pts. 3. N = 12 c. N = 23 b. N = 16 d. N = 55 2. A population consists of the live numbers 2. 3, 6, 8, and 11. Consider samples of size 2 that can be drawn from this population. 12pts. a. List all the possible samples and the corresponding mean. b. Construct the sampling distribution of the sample means. c. Draw a histogram of the sampling distribution 0! the means. LESSON NO. 4: SAMPLING AND SAMPLING DISTRIBUTION Definitions: Sampling - A method of selecting samples, like people. organizations, or objects from a population of interest. Population - Group of a total number of people, objects, or reactions that can be described as having a unique or combination ot qualities. - The entire group that is being studied. Each member of the population is called a unit. Sample - It refers to a limited number of objects selected from the population. - Representative of the entire population. TYPES OF SAMPLING TECHNIQUE A. Probability Sampling (Random Sampling) : Any method of sampling that utilizes some form of random selection, it is periormed by selecting a group of subjects (a sample) for study from a larger group (a population). Each individual is chosen entirely by chance. 1. Simple random samplinglLottery Random Sampling - it is the simplest form of random sampling. Each individual is chosen entirely by chance and each member of the population has an equal chance of being included in the sample. - Sampr'e Size Formula. Steven's Formula = N where: n is the sample size 1 + N32 N is the population size e is the margin of error Example: A researcher wants to study the effects of online learning on Grade 11 students in Baguio City High School. He wishes to use the simple random sampling technique in choosing the members of his sample. If there are 1,000 grade 11 students in the school, how many students should there be in his sample? (use 5% margin ct error) Given: N=1000 e= 5% =0.05 Solution: N 1000 1000 1000 1 +~e2 =W=m= = 285-\" = zastuden\" n: 2. Stratified random sampling - The population is partitioned into several subgroups called strata, based on some characteristics like year level, gender, age, ethnicity, etc., then samples are randomly selected separately from each stratum. 3. Systematic random sampling (with a random start) - Taking every id\" unit from an ordered population, the first unit being selected at random. The value of k is calculated by dividing the number of elements in the population by the number of elements in the desired sample. The value of k is the sampling interval. - k = 5 where k=samp|e interval, N=population size, n=sample size fl. 4. Cluster or area sampling - The entire population is broken into small groups or clusters and then, some of the clusters are randomly selected. The data from the randomly selected clusters are the ones that are analyzed. NON-PROBABILITY SAMPLING: Sampling method that does not involve random selection of samples, the population may or may not be represented well, and it will often be difficult to know how well the population has been represented. 1. Accidental or Haphazard or Convenience sampling - Normally biased since the researcher considers hisfher convenience in the collection of the data. 2. Purposive sampling - Sampling is based on certain criteria laid down by the researcher. People who satisfy the criteria are the ones who can be a part of the sample. 3. Quota sampling - Aims to just fill out the prescribed quota satisfying some definite instructions SAMPLING DISTRIBUTION OF SAMPLE MEANS . SAMPLING DISTRIBUTION It is a probability distribution where in all possible sample size n is repeatedly drawn from a population. - PARAMETER measures that describes a population (Population mean, population variance, population standard deviation etc.). o STATISTICS e measure that describes a sample (sample mean, sample variance, sample standard deviation etc.) Steps on Constructing a Sampling Distribution: 1. Calculate the number of possible outcomes of the data set. N! - - . _ ' NCn (WWW), N_populatlon Size, n_sample Size 2. Construct a random sampling distribution depending on the number of elements and compute the sample mean of the random samples 3. Construct a probability distribution of the sample means 4. Construct a histogram for the distribution Example 1 - Sampling Distribution of Sample Means. A population consists of the numbers 2, 4, 9. 10, and 5. Let us list all possible samples of size 3 from this population and compute the mean of each sample. Given: N=5 (There are 5 given numbers, 2, 4, 9, 10, and 5) n=3 Step1: Calculate the number of possible outcomes of the data set. _ Ni . NCn (WWW) (or you can srmply use Your calcu lator) _ 5! II a: ' u '5 ii I! 563 (53):(3n Press 5 , Shift nCr ,then 3 . _ 5i . _ . . 563 (2)! (31] (note. 5 population size, _ 5x4x3x2x1 _ . 5C3 (2n}(3x2x1) 3 sample srze) 503 _ 120 ' (2X6) 120 563 = = 10 12 Therefore, there are 10 possible samples of size 3 from the given population. Sample Frequency Probability mean (1) P5) Fregueng refers to how many times the 5x! sample mean appears or occur in the table in 3.67 1 1 step 2- E 3.67 appears only once so it's frequency is 1. 5.00 1 i 10 5.33 2 i 10 5.67 1 i .. . .. 10 6.00 1 i 10 6.33 1 i 10 Probability is the quotient of the frequency over 7'00 1 i the total frequency. 110 For the sample mean 3.67, its frequency is 1 and 7'67 1 E the total freouencv is 10. therefore. it has a 8.00 1 i 10_ Total n = 10 _ 1 10 ' Step2: Construct a random sampling distribution depending on the number of elements and compute the sample mean of the random samples Sample Mean _Ex _ 2+4+9 _ 15 _ W SampleMeann 3 3 5 __2. 4. 1__ 5.313 . 2, 4, 5 3.67 2- 9. 10 7-00 Step3: Construct a probability distribution of the sample means 2, 9, 5 5.33 2,10, 5 5.67 4, 9, 10 7.67 __4_._a._._ _____ _.._99__h___ 4, 10, 5 6.33 9, 10, 5 8.00 Step4: Construct a histogram for the distribution 0.25 0.2 0.15 0 LD In \"7! m Probability.I D D U1 mrauomrxhoo mu: m no id vi to r~ Sample mean Example2 - Drawing Cards. Samples 01 two cards are drawn at random from a population of 6 cards numbered from 1 to 6. Given: N = 5, and n = 2 Step1: Calculate the number of possible outcomes of the data set. in (Nn)l(n!) 6! (62}l(2!) NCn = 602 = a 662 : (4)1(21) 6x5x+x3x231 662 _ [4x3x2x1)(2x1} 602 _ 720 _ (2431(2) 602 720 15 _ E _ Step2: Construct a random sampling distribution depending on the number of elements and compute the sample mean of the random samples Step3: Construct a probability distribution of the sample means. Freuenc f Probabilit P x 1.15 =D.07 1 11'1 5=0.07 2 2f15=0.13 2 21'1 5=0.13 3 31'15 or 1f50=.2 2 2f15=0.13 4.5 2 2f15=0.13 5 1 1f15=0.07 5.5 1 1f15=0.07 Total 15 151'15 or 1 Step4: Construct a histogram for the distribution U25 02 005 II II || || || || II II II 0 15 2 15 3 35 4 45 S 55 Sample mean Probability O D i| H U1