19:25 f 15.0 KB/S 26 Ill 57 Let Us Discover 1. Sampling Distribution of Sample Mean with Normal Population when the Variance is Known Consider the grades of three Grade 11 students in their Mathematics subject last academic year. Grading Period Learner 1 Learner 2 earner 3 Quarter 1 88 85 90 Quarter 2 87 90 91 Quarter 3 90 86 90 Quarter 4 91 97 Mean (88+87+90+91)/4=89.00 87.25 90.75 Solving for the mean of the samples, Computing for the population mean, Hi = (89.00+87.25+90.75) + 3 H= (88+85+90+87+90+91+90+86+90+91+88+92) + 12 = 89.00 = 89.00 :. Thus, we can conclude that H& = p = 89.00. GSC-CID-LRMS-ESSLM, v.r. 03.00, Effective June 14, 2021 One of the properties of the sampling distribution of sample mean is that the mean of the sample means is equal to the population mean. That is, if a population has a mean of , then the mean of the sampling distribution of the sample mean is also u. Thus, H& = H. Variance is known in a given normal population when it is an infinite population with mean u and a standard deviation o. An infinite population contains, hypothetically, an infinite number of elements. When the sample size is infinite, the variance (? ,) of the sampling distribution of the sample means is computed by the formula ? & = = and the standard deviation ox) is computed as o = = The standard deviation of the sampling distribution is also known as the standard error of the mean. It measures the degree of accuracy of the sample mean (Ux) as an estimate of the population mean (). Example 1: The average number of senior high students who have a laptop in a certain barangay is 58 students with a standard deviation of 7.35. Find the mean, variance, and standard deviation of the sampling distribution of the sample with a size of 50? STEPS SOLUTION 1. Identify the given information Given: H = 58, o = 7.35, and n = 50 Unknown: H&=?, of=? and ox =? 2. Find the mean of the sampling H& = H distribution = 58 3. Solve the variance (oz) and 7.352 = 1.08 50 standard deviation (ox) of the 7.35 sampling distribution of the means of= = 50 = 1.03 Therefore, the sampling distribution of the 4. Interpret the results sample size of 50 has sample variance (o}) = 1.08 and standard deviation (o,) = 1.03 which is less than compared to the population. The variance and standard deviation of the sample means are lesser than the variance or standard deviation of the population. As the sample sizes increase, the variability of each sampling distribution decreases. The standard deviation does decrease also as the sample19:25 94 STEPS SOLUTION Given: [.1 = 58. o = 7.35, and n = 50 1. lden' the iven information My 9 Unknown: |.lx=?, 0%.:2? and "i =? 2. Find the mean of the sampling pa=|J distribution = 58 3. Solve the variance (0%) and a; = _ 73:2 _ 1.08 standard deviation {0;} of the "a 735 = ' = 1.03 m r 5 Sampling distribution of the means Therefore. the sampling distribution of the sample size of 50 has sample variance 91%) = 1.03 and standard deviation {oi} = 1.03 which is less than compared to the population. 4. Interpret the results The variance and standard deviation of the sample means are lesser than the variance or standard deviation of the population. As the sample sizes increase. the variability of each sampling distribution decreases. The standard deviation does decrease also as the sample size increases since the square root of sample size n appears in the denominator. A good estimate of the mean is obtained if the standard error of the mean is small or close to zero, while a poor estimate if the standard error of the mean is large. II. Sampling Distribution of Sample Mean with Normal Population when the Variance is Unknown In a normal population, variance is unknown when the population is finite. A nite population consists of a finite or fixed number of elements. When variance is unknown or the population is finite. the variance (0' a] and standard deviation (0' g} of the sampling distribution of the sample mean is obtained by using the formulas where: N population size Standard Deviation n sample size a N-n g 2 E - DH E nite population correction factor 03 population variance 2 GSC-Ci'D-LRMSESSLM. VJ. 03.00, Effective June 14, 2691 In general, when the population is large and the sample size is small, the correction factor is not used since it will be very close to 1. Computed standard deviation from the population may be different from the standard deviation for sampling distribution because of the sampling error. Sampling error is influenced by two factors- the population variance and the sample size. The smaller the sampling size. the greater the sampling error. Example 2: Consider a nite population of the age consists of 4Ps beneficiaries whose values are 11.13.15.1320. Find the mean, variance and standard deviation of the sampling distribution of the sample means with a size of 3. STEPS SOLUTION 1. Identify the given Given: x =11, 13. 15. 18. 20. N = 5 and n = 3 information Unknown: u='?, \"g=?. 0%? and i:5i =? 2. Find the sample mean of the I1 = E = 11+13+15+13+2 = 15.4 ; sampling distribution by N 5 solving the population mean Thus "7': P' then "i = 15' 4 _Erxui1_ 3. Solve the variance and \"2 N standard deviation of the r11-114?+113-15.4)1+[15-:5.4]2+(1e-15.4121420-15412=10.\" population using the formula o : J? = m = 3.26 4. Solve the variance (3%) and 2 oz . (N n) _ 10.64. (5 - 3 standard deviation (oi) of \" 3 5 1 the sampling distribution of _ =._ _ = 1-33 )=1aa I1 3 6 5 r 19:25 f 2.00 KB/S 56 In general, when the population is large and the sample size is small, the correction factor is not used since it will be very close to 1. Computed standard deviation from the population may be different from the standard deviation for sampling distribution because of the sampling error. Sampling error is influenced by two factors- the population variance and the sample size. The smaller the sampling size, the greater the sampling error. Example 2: Consider a finite population of the age consists of 4Ps beneficiaries whose values are 11,13,15,18,20. Find the mean, variance and standard deviation of the sampling distribution of the sample means with a size of 3. STEPS SOLUTION 1. Identify the given Given: x = 11, 13, 15, 18, 20, N = 5 and n = 3 information Unknown: [=?, H&=?, of=? and of =? 2. Find the sample mean of the H = = Ex 11+13+15+18+20 - = 15.4 ; sampling distribution by solving the population mean Thus Hy= H, then My = 15.4 3. Solve the variance and = (x -1)2 N standard deviation of the (11-15.4)2+(13-15.4)2+(15-15.4)2+(18-15.4)2+(20-15.4)2 - = 10.64 population using the formula 0 = Vo2 = V10.64 = 3.26 4. Solve the variance (oz) and 10.64 standard deviation (ox) of n 3 the sampling distribution of 3.26 Vn N - 1 V3 = 1.33 the means Therefore, the sampling distribution of the sample 5. Interpret the result. size of 50 has sample variance (o;) = 1.77 and standard deviation (o,) = 1.33 which is less than compared to the population. Based on Example 2, we need to compute first the mean, variance, and standard deviation (standard error) of the population before we can solve for the mean, variance and standard deviation of the sampling distribution of the sample mean. As we can observe, the variance and standard deviation of the sampling distribution is less than compared to the population. Ill. Sampling Distribution of Sample Mean Using Central Limit Theorem If random samples of size n, (usually n 2 30), are drawn from a population with mean u and standard deviation o, then as n becomes larger, the sampling distribution of the sample mean approaches the normal distribution. This is known as Central Limit Theorem. In this theorem, the variance of the sampling distribution of the sample mean is computed by using the formula ? & = - and the standard deviation of the sampling distribution is given by ox = 3 GSC-CID-LRMS-ESSLM, v.r. 03.00, Effective June 14, 2021 Example 3: Given a die, it has 6 Example 4: Given a pair of die, it has 6 faces in faces in which each has which each has either dot's of X = 1, 2, 3, 4, either dot's of X = 1, 2, 3, 4, 5, and 6. Given it as the population, consider 5, and 6 when n=1. the sample of size n = 2. 0=219:26 f 10.0 KB/S 56 11 56 compared to the population. Based on Example 2, we need to compute first the mean, variance, and standard deviation (standard error) of the population before we can solve for the mean, variance and standard deviation of the sampling distribution of the sample mean. As we can observe, the variance and standard deviation of the sampling distribution is less than compared to the population. Ill. Sampling Distribution of Sample Mean Using Central Limit Theorem If random samples of size n, (usually n 2 30), are drawn from a population with mean u and standard deviation o, then as n becomes larger, the sampling distribution of the sample mean approaches the normal distribution. This is known as Central Limit Theorem. In this theorem, the variance of the sampling distribution of the sample mean is computed by using the formula ? & = - and the standard deviation of the sampling distribution is given by ox = 3 GSC-CID-LRMS-ESSLM, v.r. 03.00, Effective June 14, 2021 Example 3: Given a die, it has 6 Example 4: Given a pair of die, it has 6 faces in faces in which each has which each has either dot's of X = 1, 2, 3, 4, either dot/s of X = 1, 2, 3, 4, 5, and 6. Given it as the population, consider 5, and 6 when n=1. the sample of size n = 2. n=1 n=2 Sample 2 3 4 5 6 Samples (1.13 (1,1). (1,3). (1,41. (1,5). (1,61. (2,6). (3,6). 14,61 (5,6). (6,6) (2,1) (2,2). 12,31 (2.4). (2.51. (3,5). (4,5). 15.51 (6,5) Sample Means 2 3 4 56 (3,1) 13,21. (3,31. (3.41. (4,41. (5,4). 16.4 (4,1) (4,21. (4,31. (5,3). (6,3) Frequency 1 23 4 5 (5,1) (5.21 (6,2) Sample Means =(2+31/2=1 2.5 2 2.5 4 4.5 5 5.5 6 Probability Frequency Probability 36 0.20 0.20 0.15 0.15 2 0.10 0.10 20.05 D.05 Probability 0.00 0.00 1 2 3 4 5 6 1 15 2 2.5 3 3.5 4 4.5 5 5.5 6 Sample Mean (x) Sample Mean (x) The trend of the graph is a The trend of the graph tends to a normal straight flat horizontally. distribution. Example 5: The average travel time of senior high school students in General Santos City is 12.05 minutes with a standard deviation of 7.12 minutes. If we take 36 samples what would be the mean, variance, and standard deviation of the sampling distribution of the sample mean? STEPS SOLUTION 1. Identify the given information Given: p = 12.05, o = 7.12, and n = 36 Unknown: Ux=?, of=?, and of =? 2. Find the mean of the sampling distribution H& = P = 12.05 3. Solve the variance (of) and standard = 7.123 - = 1.42 deviation (=) of the sampling 36 7.12 distribution of the means V36 = 1.19 Therefore, the sampling distribution of the 4. Interpret the result sample size of 50 has sample variance (}) = 1.42 and standard deviation (6,) = 1.19 which is less than compared to the population. E O19:26 0.52 KB/S 26 111 56 4 GSC-CID-LRMS-ESSLM, v.r. 03.00, Effective June 14, 2021 Let Us Do Activity 2: Fill me! Direction: Solve problems involving sampling distributions of the sample mean. 1. The scores of individual students on a national test have a mean of 18.6 and a standard deviation of 5.9 in Region XII. What are the mean and standard deviation of the sample mean for 25 students? STEPS SOLUTION Given: H =_, 0 = _, and n = 1. Identify the given information Unknown: H&=?, of=? and ox =? 2. Find the mean of the sampling H& = = distribution 3. Solve the variance (of) and standard o' = 19 =19 deviation (o,) of the sampling distribution of the means Ox 4. Interpret the results 2. A teacher gave a 30-point summative test to a small class of 5 students in Purok 14 of Barangay Tinagacan. The results are 18, 20, 21, 23, and 26. Find a. population mean, variance, and standard deviation and b. mean, variance, and standard deviation of the sampling distribution of the samples means with a sample size of 2. STEPS SOLUTION 1. Identify the given information Given: X =_ ,N= and n = Unknown: =?, H&=?, of=? and ox =? 2. Find the sample mean of the H =: sampling distribution by solving the population mean Thus u= u, then us = 3. Solve the variance and 02 _ _(x - H) standard deviation of the population using the formula 0 = Vo = 4. Solve the variance (o;) and - n standard deviation (ox) of the n sampling distribution of the means Vn N-1 5. Interpret the result. 5 GSC-CID-LRMS-ESSLM, v.r. 03.00, Effective June 14, 2021 E O