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95 LESSON 11: SAMPLING DISTRIBUTION OF THE SAMPLE MEAN LET'S DEVELOP In a class of 50 students, a research randomly selects 20 students to be the members of a sample. The researcher computes the mean weight of the sample and compared it with the mean weight of the whole class. Is the sample mean equal to the population mean? The researcher randomly selects 20 students to be the members of another sample different from the previous one. He computed the mean weight of the second sample. Are the two sample means equal to each other? The sample means are not equal to the population mean. However, we expect their values to be close to each other. Similarly, the means of the two sample sets are not equal to each other, but we expect their values to be close to each other too. But, how do we detemilne this proximity? How are repeated sample statistics distributed? Explore To answer the questions, let us consider the concept of sampling distribution. The sampling distribution of a sample statistics Is the distribution of values for a sample statistics obtained from repeated samples of the same size drawn from the same population. Recall the inferential statistics is concerned with making Inferences about the population from the samples. The critical part of inferential statistics is that it Involves determining how the sample means vary from the population mean. In this case, the sample statistics are the sample means, while the population parameter is the population mean. These concepts are based from sampling distribution. Example: Find the total number of possible samples of size 2 from a population of size 6. Solution: We can use the combination formula: NCn! = (N!)/(N-n)!n! Where: N = Population n=sample size So, Ncn! = (6!)/(6-2)! 2! = (60/4! 2! = 15 Answer: There are 15 possible samples of size 2 from a population of size 6. 11.1 SAMPLING DISTRIBUTION A sampling distribution is created using sampling method, particularly random sampling. There are two ways of creating a sampling distribution. First, draw samples of the same size from a population and compute sample statistics. Use descriptive method to learn more about sampling distribution. The second relies on the rules of probability, the expected values, and the variance to derive the sampling distribution. 11.2 GETTING POPULATION PARAMETERS We can compute the population parameters, such as population mean, population variance, and population standard deviation. The formulas to get these parameters are listed below. Gabuyo. Y. et.al.. (2016). Statistics and Pmbabiiify. Paligsahan, Quezon City. Where: )1: population mean it = random variables P(x) = probability of the random variables 02 = 2(1 - i402\":r ) Where: 02= population variance )1: population mean it = random variable P(x) = probability of the random variable \"/72 Where : a=populaiion standard deviation 02: population variance Example: Let x = 1,2,3,4,5,6. Compute the following: a. population mean b. population variance c. population standard deviation Solution: Let's recall the probability of distribution of random variable (X). Let x = 1,2,3,4.5,6 and the probability of each random variable Is as follows. - mal_ a. solving for population mean it = {Ix-"(JO =1(1/5) + 2(1/6) + 3(1/5) + 40.16) + 5(1/6) + 6(1/6) = 1/5 + 2/5 + 3/6 +4/6 +5/6 +6/6 = 21/6 = 3.5 Answer the population mean is 3.5. b. solving for the population variance 02 = 2(1 - @2190r ) = (1-3.5)"2 (11'6) + (2-3.5)"2 (116) + (3-3.5)\"2 (116) + (4-3.5)\"2 (1l6) +(5-3.5)"2 (116) + (6-3.5)"2 (1i6) = (-2.5)"2 (1/6) + (-1.5)"'2 (116) + (-0.5)"2 (1/6) + (0.5)\"2 (1l6) +(1.5)"2 (116) + (2.5)\"2 (116) = (6.25) (U6) + (2.25) (U6) + (0.25) (1]6) + (0.5) (lf6) + (1.5) (116) + (2.5) (1f6) = 2.92 Answer: The variance is 2.92. c. Solving for the population standard deviation .=/.z Gahuyo.Y. u.a1.(2016).mndmry. Paligsahan. QuezonCiy. V2.92 approximately 1.71 The population standard deviation is 1.71. 11.3 SAMPLING DISTRIBUTION OF THE SAMPLE MEAN From the population consisting of 6 members (1,2,3,4,5,6), we found out earlier in this lesson that the total number of possible samples with two members that can be formed is 15. Those 15 samples are listed below together with their corresponding means. Samples (n=2) Sample Mean 1,2 1.5 1,3 2.0 1,4 2.5 1,5 3.0 1,6 3.5 2,3 2.5 2,4 3.0 2,5 3.5 2,6 4.0 3,4 3.5 3,5 4.0 3,6 4.5 4,5 4.5 4,6 5.0 5,6 5.5 Here is the frequency distribution of the sample means from the previous table. Sample Mean 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 1 Summation of f = 15 IINNwNNIIiln The sample means 1.5, 2.0, 5.0, 5.5 occurred once, 2.5, 3.0, 4.0, 4.5 occurred twice and the sample mean 3.5 occurred thrice. The sample mean with the highest frequency has a value closest to the population mean. Take a look at the table below. It shows the sampling distribution of the sample mean. Gabuyn, Y. et.a]_, (2016). Stascs and Pmbabigy. Paligsahan, Quezon City. I9sl 98 Sample Mean P(sample mean) 1.50 1/15 = 0.07 2.00 1/15 = 0.07 2.50 2/15 = 0.13 3.00 2/15 = 0.13 3.50 3/15 = 0.20 4.00 2/15 = 0.13 4.50 2/15 = 0.13 5.00 1/15 = 0.07 5.50 1/15 = 0.07 Summation of Probability of Mean = 15/15 = 1.00 A histogram can describe further the sampling distribution of the sample mean. The horizontal axis represents the sample means of the possible outcomes, while the vertical axis represents the corresponding probabilities of each possible outcome. Take a look at the table below. It shows the sampling distribution of the sample mean. Sample Mean P(sample mean) 1.50 1/15 = 0.07 2.00 1/15 = 0.07 2.50 2/15 = 0.13 3.00 2/15 = 0.13 3.50 3/15 = 0.20 4.00 2/15 = 0.13 4.50 2/15 = 0.13 5.00 1/15 = 0.07 5.50 1/15 = 0.07 Summation of Probability of Mean = 15/15 = 1.00 Let's make a probability histogram of the sampling distribution of the sample mean. The probability histogram depicts that there are more sample means that are proximate to the population mean, which is 3.5 e probability histogram depicts the population mean, which is 3.5. Probability Histogra istogram of the Sampling Distribution of the Sample Mean for " - 2 0.25 0.2 0.15 0.13 0.13 0. 13 0.13 0.07 0.07 0.07 0.07 0.5 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 Mean s and Probability 11.5 MEAN OF THE SAMPLING DISTRIBUTION OF THE SAMPLE MEANS In solving the mean of the sampling distribution of the sample mean, use the formula: Gabuyo, Y. et.al., (2016). Statistics and Probability. Paligsahan, Quezon City. 99\"nu'u. .. mu... \\-u-u}l gnu-nun- n - .u-n-nnl. . anon"u", 'uwuu a..." #2 = 2(3?)P(f) Where: yr: mean of the sample mean Example: Find the mean of the sampling distribution of the sample means of the scores 1.2.3.4.5.6 using n=2 without replacement. Solution: 7? PO?) 12130?) 1.50 1/15 1.5/15 2.00 1/15 2/15 2.50 2/15 5/15 3.00 2/15 6/15 3.50 3/15 10.5/15 4.00 2/15 8/15 4.50 2/15 9/15 5.00 1/15 5/15 5.50 1/15 5.5/15 ,uf= 1.5/15 + 2/15 + 5/15 + 6/15 + 10.5/15 + 8/15 + 9/15 + 5/15 + 5.5/15 = 52.5/15 =3.5 Answer: The mean of the sampling distribution of the sample means is 3.5. To solve the mean of the population. we can use the formula: Solving the mean of the population in the previous example, we have ,u= (1+2+3+4+5+6)/6 = 21/5 = 3.5 We can see that the mean of the sampling distribution of the sample means is equal to the mean of the population. Hence. we can infer that pi = p; 11.6 VARIANCE OF THE SAMPLING DISTRIBUTION OF THE SAMPLE MEANS To solve for the variance of the sampling distribution of the sample means (0%)use the formula: (0%) = 20? - 11.92.00?) Where: pi: population mean f= mean P(f)= probability of the mean Example: Find the variance of the sampling distribution of the sample means of the scores 1.2.3.4.5.6 using n=2 without replacement. Gabuyo. Y. et.al.. (2016). Statistics and Probability. Paligsahan, Quezon City. a (5 - its)2 (1? - #292130?) (1.5 -3.5)n2 = 4 (2.5-3.5.... ms 2/15 2/15 (3.0 -3.5)I~2 = 0.2 (3-5 -3-5)"2 = 0 (4-0 -3-5)*2 = 0-2 (4-5 -3-5)"2 = 1 (5-0 -3-5)"2 = 2-25 (5-5 -3-5)*2 = 4 (0'12)= 4/15 + 2.25/15 + 2/15 + 0.5/15 + 0 + 0.5/15 + 2/15 + 2.25/15 + 4/15 = 17.5/15 = 1.17 Answer: The variance of the sampling distribution of the sample means is 1.17. 11.7 STANDARD DEVIATION OF THE SAMPLING DISTRIBUTION OF THE SAMPLE MEANS To solve for the standard deviation of the sampling distribution of the sample means (0%), use the formula: (0%) = MEG? - n'l'l) =a2 Example: Find the standard deviation of the sampling distribution of the sample means of the scores 1,2,3,4,5,6 using n=2 without replacement. Solution: Use the varieties that we obtained in the previous example. If the value of variance is available or given, you can just nd its square root to nd the value of the standard deviation. Thus, (0%) = 402 = 1.17 = 1.03 Answer: The standard deviation of the sampling distribution of the sample means is 1.08. Application 1. Using the numbers 1.2,3.4.5.6 as the demerits of the population, nd the mean of the sample size 3 without replacement. Construct lire sampling distribution of the sample mean and the probability histogram. Compute for the mean, the variance. and the standard deviation of the sampling distribution of the sample means. 2. Divide the class into groups of 10. Measure the height of each memberUust Assume). Find the sampling distribution of the mean height in centimeters of all the possible subgroups consisting of (1) 2 members. (2) 3 members. Gabuyo. Y. el.al.. (1316). Statistics MW. Paligsahan. Qimon City. lOl a. Construct a probability histogram of the sampling distribution of the sample mean. b. Compute for the mean, the variance and the standard deviation of the sampling distribution of the sample means