AMS7: WEEK 4. CLASS 3

Transcription

1 AMS7: WEEK 4. CLASS 3 Sampling distributions and estimators. Central Limit Theorem Normal Approximation to the Binomial Distribution Friday April 24th, 2015

2 Sampling distributions and estimators REMEMBER: We want to learn about the population from the sample. BUT: Lets consider the sample mean = n is the sample size. If we take two different samples from the same population, we will obtain most probably two different values

3 REMEMBER The sample mean is a statistic If we take many samples of size n, we have different values of the sample mean Sampling Variability We can talk about the probability distribution of the Sample Mean IN GENERAL: We can talk about the probability distribution of a statistic

4 Some definitions Sampling distributions of the mean: Is the probability distribution of the sample means, for all samples of size n Sampling variability: This is the variability of a statistics when its value changes from sample to sample.

5 Example If I take a random sample from this class to estimate the mean heights of the students ( ), the value of the sample mean for one sample of size n, will vary from the value obtained from another sample of size n. REMEMBER: is a Parameter from the Population is a Statistic from the Sample

6 IMPORTANT RESULT WHEN THE SAMPLE SIZE INCREASES, THE SAMPLE DISTRIBUTION OF THIS SAMPLE MEANS TENDS TO BECOME A NORMAL DISTRIBUTION Other Statistics: 1. Sample Mean 2. Sample Median 3. Sample Variance 4. Sample Standard Deviation 5. Sample Proportions

7 Sampling distributions of proportions Is the probability distribution of sample proportions, with all samples having the same sample size n Example: Use a random sample from the population, for example, 1,000 people, to estimate the proportion who will vote for candidate A in the following elections (using Polls) IMPORTANT RESULT: Under certain conditions, the distribution of sample proportions approximates a normal distribution

8 The Central Limit Theorem 1) Take a sample from a population : Population Mean 2 : Population Variance Random Variable X: Person s height X may or may not be normally distributed X has a Mean and a Variance 2 2) For a sample of size n consider the sample mean =

9 Central Limit Theorem As the sample size increases the distribution of approaches a normal distribution with mean and a Variance (standard deviation is ) NOTATION: 1) Mean of the sample mean = 2) Standard deviation of the sample mean =

10 Example of application of the Central Limit Theorem Assume that men s weights are normally distributed with a given mean = 172 lb. and a standard deviation given by =29 lb. 1) If one man is selected at random, find the probability that his weight is less than 167 lb. Random variable X is mean s weight a) Draw a picture: Weight

11 Example (Cont.) b) Calculate z Score = = = c) Area= P(Weight < 167)=P(z<-0.17)= ) If 36 men are randomly selected, find the probability that they have a mean weight less than 167 lb. Now we have a sample of 36 men

12 Example (Cont.) a) Use Central Limit Theorem: has a normal distribution with mean = and a standard deviation = = = = 4.83 b) Find the z Score for : = =. = ) Find the area below the z Score: Area=P(Mean weight < 167)=P(z<-1.04)=0.1492

13 Normal approximation to the binomial When working with a binomial distribution if np 5 and nq 5 the binomial random variable has a probability distribution that can be approximated by a normal distribution, with mean and standard deviation: = n.p =.. We must verify that it is reasonable to approximate the binomial distribution by the normal distribution

14 Normal approximation to the binomial (cont.) EXAMPLE: Suppose that X is a random variable with a binomial distribution (Example: Number of left-handed in a Statistics class) 1) With n=200, p=0.5 q=1-p=0.5 np= =100 (np 5) qn= =100 (nq 5) 2) Then we can calculate: = n.p= =100 =.. =.. = 7.07

15 Normal approximation to the binomial (cont.) c) Find the probability that X is less than 120 (P(X<120). We can use the normal approximation (it is easier!) but we need to have a continuity correction because the binomial distribution is discrete and the normal distribution is continuous. Number 120 will be represented by the interval ( , )= (119.5,120.5) in the normal curve.

16 Normal approximation to the binomial (cont.) 100 ( ) 120

17 Normal approximation to the binomial (cont.) We want the z Score: = =. = P(z<2.76)=P(X<120) P(x 119.5)= P(z 2.76)= (Using the continuity correction)