7-THE-SAMPLING-DISTRIBUTION-OF-SAMPLE-MEANS-CLT.pptx

lesson

1. THE SAMPLING DISTRIBUTION OF THE SAMPLE MEANS

2. OBJECTIVES: Finds the mean and variance of the sampling distribution of the sample mean. M11/12SP-IIId-5 5. D efines the sampling distribution of the sample mean for normal population when the variance is: (a) known (b) unknown M11/12SP-IIIe-1 6. I llustrates the Central Limit Theorem. M11/12SP-IIIe-2 7. D efines the sampling distribution of the sample mean using the Central Limit Theorem. M11/12SP-III-3 8. S olves problems involving sampling distributions of the sample mean. M11SP-IIIe-f-1

3. Suppose that all possible samples with size n are drawn from a population with mean μ and standard deviation σ . Then, the means obtained from all these possible samples will make up a sampling distribution of the sample means exhibiting the following properties: The mean of the sample means is the same as the population mean μ . The standard error of the sample mean σ is equal to the population standard deviation δ divided by the square root of the sample size, thus, σ = .  

4. Mean and Standard Error of the Sampling Distribution of Sample Means  

5. EXAMPLE 1: Random samples with size 4 are drawn from the population containing the values 14, 19, 26, 31, 48, and 53. Find the mean and standard error of the sample means. Solution: μ = μ = μ = 31.83  

6. To get the standard error, compute first the population standard deviation: Thus , the standard error of the mean is

7. EXAMPLE 2: A school has 900 junior high school students. The average height of these students is 68 in with a standard deviation of 6 in. Suppose you draw a random sample of 50 students. Find the mean, standard deviation, and a variance of the distribution of all sample means that can be derived from the samples.

8. Solution: The mean of the distribution of the sample means is the same as the population mean. Thus The standard error (or standard deviation) of the sample means is equal to The variance is the square of the standard deviation.

9. PRACTICE SET: Compute the mean, variance, and standard error of the sampling distribution of the sample means given the sample size n taken from the population and with mean μ and standard deviation σ . n= 15 , μ = 18 , σ = 3 n= 8, μ = 24 , σ = 8 n= 35, μ = 54 , σ = 9 n= 20, μ = 3.25 , σ = 0.72 n= 30, μ = 8.12 , σ = 2.7

10. DISTRIBUTION OF THE SAMPLE MEAN OF A NORMAL VARIABLE

11. Suppose a random variable X is given and the population distribution of X is known to be normal with mean μ and the variance σ 2 . Then, it follows that the sampling distribution of the mean of all samples of size n selected from X is normal with mean μ = and the variance

12. Knowing the normality of the sampling distribution of the sample mean allows you to compute probabilities involving means of samples using the standard normal distribution. Any score from a normal distribution can be converted into its equivalent standard normal score using the formula Now, since in a sampling distribution of the sample mean, the random variable gives the statistic x̅ with mean μ and standard deviation , the formula is derived as

13. EXAMPLE 1: Suppose that the systolic blood pressures of a certain population are normally distributed with mean μ = 125 and σ = 8 and a sample size 36 is taken from this population. What is the probability that a single sample will have a mean blood pressure of less than 122?

14. SOLUTION: Given: μ = 125, σ = 8, x ̅ = 122, n = 36 Substitute to the formula: z = z = z = z = -2.25 In the z-table, P(-2.25) = P(2.25) = 0.4878  

15. Continuation: Thus, P( x ̅ < 122) = 0.5 – 0.4878 = 0.0122 or 1.22% Conclusion: This means that the probability that a randomly chosen sample from the population will have mean systolic blood pressure less than 122 is 1.22%.

16. EXAMPLE 2: The average time that a person spends in a museum follows a normal distribution with μ = 88 minutes and a variance of 16 minutes. These parameters are taken from a survey conducted to the visitors of the museum in a particular day. Suppose a sample consisting of 9 visitors is selected, what is the probability that the average time these visitors spent in the museum is between 87 and 92 minutes?

17. SOLUTION: Given: μ = 88, σ = , x̅ = 87 and 92, n = 9 Substitute to the formula : P(87 < x̅ < 92) = P( < Z < ) = P(-0.75 < Z < 3) = P(0.75) + P(3) = 0.2734 + 0.4987 P(87 < x̅ < 92) = 0.7721 or 77.21% The probability that the mean time of 9 visitors falls between 87 and 92min is 77.21%.  

18. CENTRAL LIMIT THEOREM

19. CENTRAL LIMIT THEOREM Recall that when samples with any size n are taken from a normally distributed population, then the distribution of all sample means also follows a normal distribution. But what if the population from which samples are taken is not normally distributed? Can you conclude that the sampling distribution of the sample mean will be normal? The answer to this question is “YES” provided that the sample size is large enough. This can be explained using the CENTRAL LIMIT THEOREM.

20. CENTRAL LIMIT THEOREM Given a random variable X with mean μ and variance σ² , then, regardless whether the population distribution of X is normally distributed or not, as the sample size n gets larger, the shape of the distribution of the sample means taken from the population approaches a normal distribution, with mean μ and standard deviation  

21. CENTRAL LIMIT THEOREM Note that the emphasis of the central limit theorem is on the phrase “ as the sample size gets larger ”. You can only assume the sampling distribution to be normal when the sample is large enough. Mostly, n ≥ 30 is sufficient to approximate the normality of the sampling distribution of the sample mean.

22. EXAMPLE 1: The average age of teachers in a certain town is 34 with a standard deviation of 4. If the principal of ABC College employs 100 teachers, find the probability that the average age of these teachers is less than 35.

23. SOLUTION: It is not given that the population is normally distributed but since n > 30, then you can assume that the sampling distribution of the mean ages of 100 teachers is normal according to the CLT. z = z = z = z = 2.5  

24. SOLUTION: Then, P(X < 35) = P(Z < 2.5) P(X < 35) = 0.5 + P(2.5) P(X < 35) = 0.5 + 0.4938 P(X < 35) = 0.9938 or 99.38% Therefore, the probability that the average age of the teachers is less than 35 is 99.38%.

25. EXAMPLE 2: A random sample with size 16 is selected from a normally distributed population with a mean of 16 and a standard deviation of 3.2. What is the probability that the mean of the sample falls between 14 and 16?

26. SOLUTION: Given: μ = 16, σ = 3.2, x̅ = 14 and 16, n = 16 Substitute to the formula: P(14 < x̅ < 16) = P ( < Z < ) = P (-2.5 < Z < 0 ) = P(2.5) + P(0) = 0.4938 + 0 P(87 < x̅ < 92) = 0.4938 or 49.38% The probability that the mean of the sample falls between 14 and 16 is 49.38%.  

27. QUESTION???

28. PRACTICE SET: A population has a mean age of 34 with a standard deviation of 10. What is the probability that the mean age of the sample with size 9 falls between 32 and 35, considering that the given population follows a normal distribution? The heights of the students in a certain college are normally distributed with an average height of 6.2 ft and a standard deviation of 0.8 ft. If a random sample of 36 students is chosen, what is the probability that their height fall between 6.6ft and 6.7ft?