Learning Objectives • Illustrate the Central Limit Theorem • Define the sampling distribution of the sample mean using the Central Limit Theorem; and • Solve problems involving sampling distributions of the sample mean. S ummary 1. The mean of the population is equal to the mean of the sampling distribution. đ = đđą 2. The variance and standard deviation of the sampling distribution is smaller than the variance and standard deviation of the population. đ đ đđą < đ đđą < đ S ampling Error - is the difference between the values from our statistic (sample) and parameter (population). đđą n S tandard Error of the Mean ( đđą ) - measures the accuracy of the statistic (sample) for it to become useful in generating information about our parameter (population). A ccuracy đđą Large Sample Size = Small Standard error Good estimate/statistic- Consistent Central Limit Theorem As n becomes larger, the sampling distribution of the mean approaches the “normal distribution”, regardless of the shape of the population distribution. E ffect of Sample Size Sample size must be n≥ đđ. So that, the CLT, will ensure that the sampling distribution of the sample means tends to have a normal distribution 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 nīĄ The Central Limit Theorem takes care of this situation provided that, the sample size is large enough to approach a normal distribution What is it? đĨ−đ đ= đ đĨ - raw data/score đ= đ= ≡ đ−đ đ- sample mean - population mean - population mean - population s.d - population s.d - sample size - sd of samples Note: This is applicable when dealing with individual data. Note: This is applicable when dealing with samples drawn from the population. E xamples The heights of a group of boys are normally distributed with a mean of 54 inches and standard deviation of 2.5 inches. QUESTIONS: 1. What percentage of the population would have heights between 53 inches and 56 inches? 2. If all possible samples of size 25 are drawn from this population, what percentage of them would have means between 53 inches and 55 inches? 3. If a random sample of size 81 is drawn from this population, what is the probability that the mean of this sample is larger than 53.5 inches? S olution: 1 1 STEP Identify the given values. 1 - 54 inches - 2.5 inches đą - 53 & 56 inches. Find: P(53 < X < 56) S olution: 1 1 STEP Compute for the unknown. Convert the given 2 heights into a standard score đ= đ= đđ − đđ đ= đ. đ −đ đ= đ. đ đđ − đđ đ= đ. đ đ đ= đ. đ đ = −đ. đđ đ = đ. đđ S olution: 1 1 STEP Use the z-table and look the equivalent value of 3 the z-score you calculated in STEP 2. đđ đĢđđŠđĢđđŦđđ§đđŦ đ = −đ. đđ đ. đđđđ đđ đĢđđŠđĢđđŦđđ§đđŦ đ = đ. đđ đ. đđđđ S olution: 1 1 STEP 4 Draw a graph and plot the zscore and its corresponding area. Then, shade the part that you’re looking for: P(53 < X < 56) -0.40 0.80 S olution: 1 STEP 5 Compute for the probabilities or P(53 < X < 56) = 0.7881 - 0.3446 percentage. From the shaded P(53 < X < 56) = 0.4435 or normal distribution, we are going 44.35% to “SUBTRACT” the values. Find: P(53 < X < 56) 53= -0.40 = 0.3446 56= 0.80 = 0.7881 Interpretation: The probability or percentage that the population would have heights between 53 and 56 inches is 44.35% S olution: 2 1 STEP Identify the given values. 1 - 54 inches - 2.5 inches - 53 & 55 inches n - 25 Find: P(53 < < 55) S olution: 2 1 STEP Compute for the unknown. Convert the given 2 heights into a standard score đ= đ= đđ − đđ đ= đ. đ đđ đđ − đđ đ= đ. đ đđ đ đ = −đ đ đ. đ đ = −đ đ đ=đđ đ. đ đ=đ S olution: 2 1 STEP Use the z-table and look the equivalent value of 3 the z-score you calculated in STEP 2. đđ đĢđđŠđĢđđŦđđ§đđŦ đ = −đ đđ đĢđđŠđĢđđŦđđ§đđŦ đ= đ đ. đđđđ đ. đđđđ S olution: 2 1 STEP 4 Draw a graph and plot the zscore and its corresponding area. Then, shade the part that you’re looking for: P(53 < < 55) -2.00 2.00 S olution: 2 STEP 5 Compute for the probabilities or percentage. From the shaded normal distribution, we are going to “SUBTRACT” the values. Find: P(53 < < 55) 53= -2.00 = 0.0228 55= 2.00 = 0.9772 P(53 < < 55) = 0.9772 - 0.0228 P(53 < < 55) = 0.9544 or 95.44% Interpretation: The probability or percentage that all possible samples of size 25 drawn from this population would have means between 53 and 55 inches is 95.44% S olution: 3 1 STEP Identify the given values. 1 - 54 inches - 2.5 inches - 53.5 inches n - 81 Find: P( > 53.5) Note: We assume that the population is infinite. S olution: 3 1 STEP Compute for the unknown. Convert the given 2 heights into a standard score đ= đđ. đ − đđ đ= đ. đ đđ đ đ = −đ. đ đ đ. đ đ = −đ. đ S olution: 3 1 STEP Use the z-table and look the equivalent value of 3 the z-score you calculated in STEP 2. đđ. đ đĢđđŠđĢđđŦđđ§đđŦ đ = −đ. đ đ. đđđđ S olution: 3 1 STEP 4 Draw a graph and plot the zscore and its corresponding area. Then, shade the part that you’re looking for: P( z > -1.80) 0.4641 -1.80 0.5000 S olution: 3 STEP 5 Compute for the probabilities or percentage. From the shaded normal distribution, we are going to “SUBTRACT” the values. Find: P(z > -1.80) 53.5= -1.80 = 0.0359 P(z > -1.80) = 1 – 0.0359 P(z > 1.80) = 0.9641 or 96.41% Interpretation: The probability or percentage that all possible samples of size 81 drawn from this population would have means larger than 53.5 inches is 96.41% T ry these! 1-2 1. A population has a mean of 75 and a standard deviation of 17.4. A sample of 35 items is randomly selected from this population. What is the probability that the mean of this sample lies between 72 and 79? 2. An electric company in Laguna manufactures resistors that have a mean resistance of 120 ohms and a standard deviation of 13 ohms. Find the probability that a random sample of 40 resistors will have an average resistance greater than 114 ohms.
