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Using the formula provided, recall on how to solve sample mean, variance, and standard Lesson 12 Sampling Distribution of the Means deviation in this given example: Mustrative Example 1 Find the sample standard deviation and variance of the following heights (in centimeters) of 10 plants. Background Information 5 12 16 18 19 21 23 32 29 25 In the previous lesson, the concept of random and nonrandom sampling techniques, as well as how to determine the sample size required for a sampling distribution were presented Solution: In this lesson, determining the mean and variance of a sampling distribution of the sample Construct a table by following the steps below. means will be discussed. DEFINITION XI - (x1 - x)2 5 - 20 = -15 -15)3 = 225 The mean, also known as the average, is calculated by dividing the sum of the observations by the 12 12 -20 = -8 (-8) = 64 total number of observations 16 16-20 = -4 (-4) =16 The variance is defined as the average of the squared deviations from the mean. The square root of 18 18 -20 =-2 (-2) =4 this variance is known as the standard deviation. The variance of the sample is denoted as so (read as s squared ) while the population variance is denoted as o (read as sigma squared) 19 19 -20 =-1 -1)2 = 1 21 21 - 20 = 1 12 = 1 23 23 - 20 = 3 31 =9 The formula for solving the sample mean and population mean are as follows: 25 25 - 20 = 5 51 = 2: For the sample mean: 29 29 - 20 =9 92 = 81 - alfx2+x3Jutxn _ 21=1xi 12 - 20 = 12 121 = 144 n 72 where is the sample mean, E FF= 200 I (x1 -x)2 = 570 are the values of observations in the sample; and n is the total number of observations in the samples. STEP 1. Arrange the values in ascending order, as shown in the first column. For the population mean: STEP 2. Calculate the sample mean as follows: _ an Xi_ 200 = 20 10 N N the population mean: STEP 3. Get the individual deviations from the mean, (x - >), as shown in the second column. X'sare the values of observations in the population; and N is the total number of observations in the population. STEP 4. Square the deviation obtained in the second column and write the results in the third column (3 - 2). The formula for the variances of a sample and a population are as follows: STEP 5. Find the sum of the squared deviations as follows: For the variance and sample standard deviation of a sample E " (X, - x) = 225 +64 +16 +4 +1+1+9+25 +81 + 144 = 570 data: (x1-2)2 Divide the sum obtained in step (5) by n - 1 to get the sample variance, =1 n-1 or s = V21=1 n-1 STEP 6. as shown below. where s' is the variance of a sample; 8 = 2 s is the sample standard deviation; is the sample mean $70 are the values of observations in the sample; and n is the total number of observations in the samples. s' = 63. 33 For the variance and sample standard deviation of a population: The sample variance of the heights of 10 plants is 63.33 cm. N (X 1-1)2 0-2 1= N-1 or 0 = V21=1 N-1 Thus, the sample standard deviation is: where of is the variance of a population; is the population standard deviation; s = VE/=1 n-1 a is the population mean; Xiare the values of observations in the population; and Wis the total number of observations in the population. = V63.33 & 7. 96When more observations are included in a data set and wanted to minimize the use of Finding the Mean and Standard Deviation of a set of data using a Scientific Calculator The sample standard deviation of the heights of 10 plants is 7.96 cm. deviation, the formula below can be used for solving sample variance and sample standard deviation. V2 x) - (2x) Illustrative Example 3-1 and S = V. The table shows the number of enrollees for incoming Grade 11 students in 5 days. Find (n-1) (n-1) the mean and standard deviation of the given data using a scientific calculator. where s' is the variance of a sample; Day Number of Applicants s is the sample standard deviation; Monday 80 is composed of the values of observations in the sample; and Tuesday 85 n is the total number of values in observation. Wednesday 92 Illustrative Example 2 Thursday 98 Find the sample standard deviation and variance of the following heights (in centimeters) of Friday 103 10 plants. 5 12 16 18 19 21 23 32 29 25 The calculator to be used is the CASIO fx-82MS Alternative Solution: Here are the steps in finding the mean and variance of the given data using the CASIO fx-82MS. Construct a table by following the steps below. STEP 1. Press (Mode ). Then press 2 STEP 2. Once the letters SD appear on the screen, enter each data. Then press (M+ ) 25 until you enter all the data 144 256 M+ 324 M+ 361 M+ 441 529 M+ 625 M+ 841 1,024 STEP 3. Press ( Shift ). Then press ( 2 ) (S-var). For the mean, press (1 ). Then press (=). (The result must be 91.6) 200 4.570 STEP 1. Arrange the values in increasing order, as seen in the first column. For the sample standard deviation, press ( 3 ). Then press (=). (The result must be 9.34) STEP 2. Compute the sum of the sample as follows E x = 5 + 12 + 16 + 18 + 19 + 21 + 23 + 25 + 29 + 32 = 200 For the population standard deviation, press (2 ). Then press (The result must be 8.36) STEP 3. Square each value, as seen in the second column. Therefore, the mean and standard deviation of the data set are 91.6 and 9.34, respectively. STEP 4. Compute for the total of the squared values (2 5). STEP 5. Solve for the sample variance and sample standard deviation given the To reset the settings of the calculator, press (Shift ). (Mode ). (3 ) and then (= ). formula below Illustrative Example 3-2 For sample variance Given another model of a scientific calculator, which is the CASIO fx-82ES PLUS, the following are the steps in finding the mean and the standard deviation of a set of data. $2 _ "(2x)-(x) 10(4.570)-(200)2 n(n-1) (Substitution of computed values) STEP 1. Press (Mode ), (2 (STAT), and then ( 1 ) (1-VAR). 10 ( 10-1) 45.700-40000 STEP 2. Enter the data. Press (= ) after each data entered. 90 (Simplify) _5,700 STEP 3. Press On Shift ). (1 (STAT), and then 4). 90 $2 # 63.33 For the mean, select (2 )(x). Then press (= For the sample standard deviation, repeat from ( Shift ). (1 ) (STAT). For sample standard deviation After which, press (4 ).(4 ), and then = "( x2) - (2x) = V63.33 * 7.96 n(n-1). To set the calculator back to normal mode, press ( Mode ). Then press (0) The computed sample variance and sample standard deviation of the heights of 10 plants are 63.33 cm and 7.96 cm, respectively.Lesson 13 Mean and Variance of Sampling Distribution of from which the samples were selected. The standard deviation of the sampling distribution of the commonly referred to as the Standard Error of the Mean. This represents the difference the Means between the sample mean (), and the population mean (4). Before proceeding, it is necessary to understand and be able to distinguish between the mean DEFINITION and standard deviation of a population, a sample, and the sampling distribution of the mean. The table The formula for the variance and standard deviation of a sampling distribution of the sample mean are as follows: below shows the symbols for the mean and standard deviation of a population, a sample, and the sampling distribution of the mean. (for variance); and Notation for Mean and Standard Deviation for a Population, a Sample and the Sampling Distribution of the Meal (for standard deviation/ standard error of the mean). Population Sample Sampling Distribution where: is the population variance; of the Mean is the population standard deviation; Portion of data values Means from all possible and samples of size n Description Collection of all data values randomly selected from randomly selected from This theorem is called Central Limit Theorem, which will be formally stated in the next lesson. the population the population Mustrative Example 4 Mean According to the study of TV viewing habits, the average number of hours a teenagerwatches MTV per week is a= 17.9 hours with a standard deviation of o= 3.8 hours. If a sample of 64 teenagers Standard Deviation is randomly selected from the population, then determine the mean and standard deviation of the sampling distribution of the mean. Mustrative Example 1 (Week 5) Solution. Construct a sampling distribution of the mean and a histogram for the set of data below. Given: H = 17.9 86 89 92 95 98 0 =3.8 The population mean of the data is shown below: n = 64 E x 86+89+92+95+98 For the Mean of the Sampling Distribution of the Mean, since up = p, then: H = =92 M-x = 17.9 hours N For the Standard Deviation of the Sampling Distribution of the Mean, Sampling Distribution of the Means (n = 3) VO4 Sample Mean (x) Probability 89 0.1 * 0. 48 hours 90 0.1 91 0.2 THEUREM 92 0.2 The Shape of the Sampling Distribution When Sampling from a Normal Population 93 0.2 If the population being sampled is a normal distribution, then the sampling distribution of the mean is a normal distribution regardless of the sample size, n. 94 J.1 95 0.1 ORIGINAL POPULATION FROM WHICH SAMPLES ARE SELECTED Based on the Sampling Distribution of the Mean of size 3 above, it is shown that there Shape: Normal are ten (10) sample means computed from randomly selecting three (3) values on the problem. The sample means are as follows: 89, 90, 91, 91, 92, 92, 93, 93, 94, 95. To obtain the Mean of the Sampling Distribution of the Mean (up), Me = 21+ 82+8s+.+2n - El=18 Sampling Distribution of the mean for samples of size: 89+90+91+91+92+92+93+93+94+95 (a) (b ) (d) (@) 0 =2 n = 10 n = 20 n = 3 n = 100 MY = 92 Shape Shape: Shape: Shape: Shape: As shown in the above solution, the population mean & of the data is 92 which is norma normal nomal normal normal equal to the mean of the sampling distribution of the mean prwhich is 92 DEFINITION The Mean of the Sampling Distribution of the Mean, px is equal to the mean of the population from which the samples were selected; that is Unlike the mean of the sampling distribution of the mean (#x the standard deviation of FIGURE 5.1 the sampling distribution of the mean is not equal to the standard deviation of the population. It is smaller than the standard deviation of the population; thus, it is less dispersed than the populationActivity Sheet # 6 WRITTEN WORK # 6 A. Multiple Choice Direction: Write the letter corresponding to the correct answer on a separate sheet of paper. Write "e" before the number if your answer is not among the choices. For items 1 -5: Use the set of data below: 12 15 16 21 26 28 30 1. What is the population mean? a. 20.12 C. 24.11 b. 21.14 d. 25.65 2. What is the population variance? a. 5.52 C. 40.12 b. 6.51 d. 42.38 3. What is the population standard deviation? 3.23 C. 6.51 b. 4.58 d. 7.09 4. What is the sample standard deviation if n =4? a. 1.63 C. 6.58 b. 3.26 d. 10.61 5. Which of the following sets is a random sample of the data given? a. {12, 26, 30} C. {12, 14,32 b. {12,29, 30 } d. {14, 27,30} B. Calculate! Direction: Find the mean, sample standard deviation, and variance of the data given below. Provide the complete solution. If a calculator will be used for the solution, provide the model and the steps used in arriving at the answer. 105, 110, 115, 120, 125, 130, 135 PERFORMANCE TASK # 6 Direction: Answer the given problem on a separate sheet of paper. List down the given first before proceeding to the calculation. Show your complete solution. Problem: In Ragma Memorial College, the mean IQ score ofthe studentbody is 107 with a standard deviation of 20. Suppose a random sample of size n is selected from the studentbody, compute the standard error of the mean, of, for samples of size: n = n = c) n =40