Sampling Distribution of the Sample Means PDF

IMPORTANT NOTE: This material is for your own consumption only. Please respect intellectual property rights of the author including the references of this learning material. DO NOT SHARE OR POST this material to other platforms or social media accounts.  CHAPTER 3 Sampling Methods and...

IMPORTANT NOTE: This material is for your own consumption only. Please respect intellectual property rights of the author including the references of this learning material. DO NOT SHARE OR POST this material to other platforms or social media accounts.  CHAPTER 3 Sampling Methods and Sampling Distribution Prepared by: Grace Aquino Pasion October 2024 Module 2 Sampling Distribution of the Sample Means Reference: Francisco, J., Laviña, C., Pasion, G., (2023). Statistics and Probability (A Modern Approach for the New Generation). Johnny and Hansel Publications, Quezon City Module 2 Sampling Distribution of the Sample Means Module Topics Sampling Distribution Population Distribution Sampling Distribution of the Means Standard Error Area under the normal curve Problem Solving Module 2 Sampling Distribution of the Sample Means Each sample may be described by determining the mean, variance, standard deviation, and its own values. By selecting and measuring more samples and observed the sample means, variances and standard deviations, these can be classified as random variables with corresponding distribution which is called sampling distribution. Definition of Sampling Distribution of the Sample Means is a distribution of all possible means of a given sample size drawn from a specific population. Module 2 Sampling Distribution of the Sample Means Suppose SHS students were asked if they are in favor of the DepEd Order on “No periodical test will be given this school year to prevent distance cheating”. Let the response be, 1 for in favor and 0 for not in favor. For possible sample size of 3, the results are shown in the following table: Sample (0,0,0) (1,1,0) (1,0,1) (0,1,1) (1,0,0) (0,1,0) (0,0,1) (1,1,1) Mean 0 2/3 2/3 2/3 1/3 1/3 1/3 1 Getting more responses from three SHS students, what happens is that the mean becomes a random variable, and the sample means 0, 2/3, 1/3… comprise a sampling distribution of the sample means. Module 2 Sampling Distribution of the Sample Means Table 1 Probability Distribution: Sampling Distribution of Sample Means Mean, (ഥ 𝒙) Probability, 𝑷(ഥ 𝒙) 0 1/8 = 0.125 1/3 3/8 = 0.375 2/3 3/8 = 0.375 1 1/8 = 0.125 Module 2 Sampling Distribution of the Sample Means Figure 1 Probability Histogram: Sampling Distribution of Sample Means From the figure above, probability histogram shows approximately normal distribution, which indicates that repeated sampling forms a sampling distribution for all possible sample sizes(n). Module 2 Sampling Distribution of the Sample Means Properties of the Distribution of Sample Means: Population mean is equal to the mean of the sample means. That is: 𝜇𝑥ҧ = 𝜇 note: this is 𝑛 → 𝑁 (Central Limit Theorem, where 𝑛 is large) Population standard deviation leads to a smaller value, and it will be equal to the population standard deviation divided by the square root of the sample size (n). That is: 𝜎 𝜎𝑥ҧ = , also called standard error (SE) 𝑛 Module 2 Sampling Distribution of the Sample Means So, the mean value of observations even if it is small will be normally distributed when samples are drawn from a normally distributed population. In this case, the formula in finding the probability for sampling distribution of the sample means can be applied. That is: 𝑥ҧ − 𝜇𝑥ҧ 𝑥ҧ − 𝜇𝑥ҧ 𝑧= = 𝜎 𝜎𝑥ҧ 𝑛 𝜎 𝜎𝑥ҧ = , also called standard error 𝑛 Module 2 Sampling Distribution of the Sample Means Solution: Example 1: Given: Entrance test results for incoming Grade the population mean, 𝜇 = 84 11 students in a certain school has a mean the population standard deviation, 𝜎 = 8.8 of 84 and a standard deviation of 8.8. The the sample size, 𝑛 = 25 test scores follow a normal distribution. the sample mean, 𝑥ҧ = 85 What is the probability that the mean test Find: 𝑃(𝑥ҧ > 85) score of 25 examinees is above 85? a. Compute for the standard error. 𝜎 𝜎𝑥ҧ = 𝑛 8.8 𝜎𝑥ҧ = 25 𝜎𝑥ҧ = 1.76 Module 2 Sampling Distribution of the Sample Means b. The area of the shaded region in the figure below gives the probability we look for.  Module 2 Sampling Distribution of the Sample Means c. For sample mean 𝑥ҧ =85, use the formula, Using SND table or Excel, 𝑥ҧ − 𝜇𝑥ҧ 𝑧= 𝜎𝑥ҧ 𝑃 𝑥ҧ > 85 = 𝑃 𝑧 > 0.57 Where: = 1 − 0.7157 = 0.2843 𝑥ҧ = 85 𝜇𝑥ҧ = 84 = 28.43% 𝜎𝑥ҧ = 1.76 This concludes that the probability of the average test score of the random sample of 25 examinees is greater than 85 is Thus, approximately 0.2843 or 28.43%. ҧ 𝑥ഥ 𝑥−𝜇 85−84 𝑧= = = 0.57 Excel: 𝜎𝑥ഥ 1.76 =1-norm.s.dist(0.57,true) Module 2 Sampling Distribution of the Sample Means Module 2 Sampling Distribution of the Sample Means Module 2 Sampling Distribution of the Sample Means Module 2 Sampling Distribution of the Sample Means Example 2: A study about the level of happiness is done among SHS students. The scores follow a normal distribution of 1 to 5, with 5 being the highest. If the population mean is 3 and the population standard deviation is 1.5, what is the probability that the mean level of happiness of 35 students is lower than 3.5? Module 2 Sampling Distribution of the Sample Means Solution: Example 2: Given: 𝜇=3 A study about the level of happiness is 𝑛 = 35 done among SHS students. The scores 𝜎 = 1.5 follow a normal distribution of 1 to 5, 𝑥ҧ = 3.5 with 5 being the highest. If the population mean is 3 and the population Find: 𝑃(𝑥ҧ < 3.5) standard deviation is 1.5, what is the probability that the mean level of a) Compute for the standard error. 𝜎 happiness of 35 students is lower than 𝜎𝑥ҧ = 𝑛 3.5? 1.5 𝜎𝑥ҧ = 35 𝜎𝑥ҧ = 0.2535 Module 2 Sampling Distribution of the Sample Means b. The area of the shaded region in the figure below gives the probability we look for. Module 2 Sampling Distribution of the Sample Means Using excel or SND table, c. For sample mean 𝑥ҧ =3.5, use the formula, 𝑥ҧ − 𝜇𝑥ҧ 𝑃 𝑥ҧ < 3.5 = 𝑃 𝑧 < 1.97 = 0.9756 = 97.56% 𝑧= 𝜎𝑥ҧ This concludes that the probability of the level of Where: happiness of the random sample of 35 SHS students 𝑥ҧ = 3.5 lower than 3.5 is approximately 0.9756 or 97.56%. 𝜇𝑥ҧ = 3 Excel: 𝜎𝑥ҧ = 0.2535 =norm.s.dist(1.97,true) 𝑃 𝑧 < 1.97 = 0.9756 Thus, OR try skipping z-score, using the syntax: ҧ 𝑥ഥ 𝑥−𝜇 3.5−3 “=norm.dist(𝑥,ҧ 𝜇, 𝑆𝐸, 𝑡𝑟𝑢𝑒)" 𝑧= = = 1.97 𝜎𝑥ഥ 0.2535 =norm.dist(3.5,3,0.2535,true) 𝑃 𝑥ҧ < 3.5 =0.9757  Module 2 Sampling Distribution of the Sample Means Example 3: A food chain restaurant is using a drive-thru speed-of-service timer to determine the time spends by a customer to order a food. The data show that the average normal time is 5.2 minutes with a standard deviation of 1.6 minutes. If a random sample of 36 customers is observed, find the probability that their mean time at the drive-thru window is between 5 to 5.5 minutes.  Module 2 Sampling Distribution of the Sample Means Solution: Example 3: Given: 𝜇 = 5.2 A food chain restaurant is using a drive-thru 𝑛 = 36 speed-of-service timer to determine the 𝜎 = 1.6 time spends by a customer to order a food. 𝑥ҧ = 5; 𝑥ҧ = 5.5 The data show that the average normal time is 5.2 minutes with a standard deviation of Find: 𝑃(5 < 𝑥ҧ < 5.5) 1.6 minutes. If a random sample of 36 customers is observed, find the probability a) Compute for the SE. that their mean time at the drive-thru 𝜎 𝜎𝑥ҧ = window is between 5 to 5.5 minutes. 𝑛 1.6 𝜎𝑥ҧ = 36 𝜎𝑥ҧ = 0.2667  Module 2 Sampling Distribution of the Sample Means c. For sample mean 𝑥ҧ = 5 and 𝑥ҧ = 5.5, Using excel or SND table, use the formula, 𝑃 5 < 𝑥ҧ < 5.5 = 𝑃 −0.75 < 𝑧 < 1.12 𝑥ҧ − 𝜇𝑥ҧ = 0.8686 − 0.2266=0.6420 𝑧= 𝜎𝑥ҧ The probability of the average time to spend in the drive-thru Where: of the random sample of 36 customers between 5 to 5.5 𝑥ҧ = 5; 𝑥ҧ = 5.5 minutes is approximately 0.6420 or 64.2%. 𝜇𝑥ҧ = 5.2 𝜎𝑥ҧ = 0.2667 Excel: Thus, ҧ 𝑥ഥ 𝑥−𝜇 5−5.2 =norm.s.dist(1.12,true)-norm.s.dist(-0.75,true) 𝑧= = = −0.75 𝑃 −0.75 < 𝑧 < 1.12 =0.6420 𝜎𝑥ഥ 0.2667 ҧ 𝑥ഥ 𝑥−𝜇 5.5−5.2 𝑧= = = 1.12 𝜎𝑥ഥ 0.2667 Module 2 Sampling Distribution of the Sample Means (Assignment, write your answers on your blackboard) Exercises Compute for the mean, variance, and standard error of the sampling distribution of the sample means given the sample size (𝑛) taken from the normal population, the population mean (𝜇), and population standard deviation (𝜎). 1. 𝑛 = 16, 𝜇 = 19, 𝜎 = 4 2. 𝑛 = 7, 𝜇 = 23, 𝜎 = 7 3. 𝑛 = 36, 𝜇 = 55, 𝜎 = 10 4. 𝑛 = 19, 𝜇 = 3.20, 𝜎 = 0.70 5. 𝑛 = 40, 𝜇 = 18.12, 𝜎 = 12.70 Module 2 Sampling Distribution of the Sample Means (from your textbook, answer via Cloud Campus) B. Read and answer the following problems with random samples taken from the normal population. Show all the necessary solutions. (textbook: Assignment) 1. The average online teacher’s salary per hour is ₱380.00 with a standard deviation of ₱51.00. For the sample of 75 online teachers, what is the probability that the sample mean is greater than ₱400.00? 2. The national average percentage score of Grade 6 in National Achievement Test (NAT) recently for Math subject is 36.66 with a standard deviation of 3.57. If a random sample of 200 scores were selected, what is the probability that the sample average percentage score is higher than 45%? (Note: scores in NAT are expressed in percentage) 3. The number of hours spent playing mobile legends a day by teenagers has a mean of 8 hours with a standard deviation of 3.5 hours. Find the probability that if you select 50 random teenagers, the mean number of hours a day spent playing games is between 6.5 and 7.5 hours. Module 2 Sampling Distribution of the Sample Means 4. Kyle has a collection of more than 300 songs in his phone that has a mean duration of 2.75 minutes and a standard deviation of 0.6 minutes. Suppose that every week he makes a playlist by taking random sample of 36 of these songs, calculate the sample mean duration of the songs and the standard error of the sampling distribution of the sample means. 5. On the average, basketball players drink 2.3 liters of water with a standard deviation of 0.9 liter during a game. If the team manager asks his assistant to bring 45 liters of water for 15 players, what is the probability that they will run out of water? 6. Suppose that students who finished JHS in a public school get an average of 5.6 hours of sleep each night with a standard deviation of 1.7 hours. What is the probability that a random selected group of 35 public school students get more than 6 hours of sleep, on the average each night? 7. The average life of a smartphone is 7 years with a standard deviation of 1 year. Assuming that the lives of these phones follow approximately a normal distribution, find the probability that the mean life of a random sample of 9 smartphones fall between 6.4 and 7.2 years. Module 2 Sampling Distribution of the Sample Means 8. Given a normal random variable X with a mean of 20, a variance of 9, and a random sample of n taken from the distribution, what sample size (n) is necessary in order that P(19.9 < X < 20.1) = 0.95? 9. Suppose that systolic blood pressures of a certain population are normally distributed with a mean of 120, a standard deviation of 4, and a sample size 28 is taken from this population, what is the probability that a single sample will have a mean blood pressure of less than 125? 10. Study reports that the average person uses 120 liters of water daily. If the standard deviation is 20 liters, find the probability that the mean of randomly selected sample of 15 people will be between 120 and 125 liters. Module 2 Sampling Distribution of the Sample Means From textbook The average online teacher’s salary per hour is ₱380.00 with a standard deviation of ₱51.00. For the sample of 75 online teachers, what is the probability that the sample mean is greater than ₱400.00? Module 2 Sampling Distribution of the Sample Means From Worksheet 2 Suppose that SHS master teachers in the Philippines earn an average of P720,000 per annum with a standard deviation of P14,000. To prove this, a random sample of 64 master teachers was selected from a database for all master teachers in the country. Find the range of earnings within which 80% (middle 80%) of the average earnings of master teachers for a random sample of 64 will fall. DO NOT INCLUDE THE Peso sign and comma in your answer and answer in two decimal places. 717,757.30 to 722,242.70 End of Module 2 Sampling Distribution of the Sample Means Check your cloud campus/blackboard for corresponding worksheets and/or assignments for submission. GAWIN HABANG NAGPAPAHINGA ☺