1. SAMPLING DISTRIBUTION

2. 2 Introduction In real life calculating parameters of populations is usually impossible because populations are very large. Rather than investigating the whole population, we take a sample, calculate a statistic related to the parameter of interest, and make an inference. The sampling distribution of the statistic is the tool that tells us how close is the statistic to the parameter.

3. 3 Sampling Distribution of the Mean An example – A die is thrown infinitely many times. Let X represent the number of spots showing on any throw. – The probability distribution of X is x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6 E(X) = 1(1/6) + 2(1/6) + 3(1/6)+ ………………….= 3.5 V(X) = (1-3.5) 2 (1/6) + (2-3.5) 2 (1/6) + …………. …= 2.92

4. 4 Suppose we want to estimate  from the mean of a sample of size n = 2. What is the distribution of ? Throwing a die twice – sample mean

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6. 6 The distribution of when n = 2 1 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6/36 5/36 4/36 3/36 2/36 1/36 E( ) =1.0(1/36)+ 1.5(2/36)+….=3.5 V(X) = (1.0-3.5) 2 (1/36)+ (1.5-3.5) 2 (2/36)... = 1.46

7. 7 6 Sampling Distribution of the Mean

8. 8 Notice that is smaller than. The larger the sample size the smaller. Therefore, tends to fall closer to , as the sample size increases. Notice that is smaller than  x. The larger the sample size the smaller. Therefore, tends to fall closer to , as the sample size increases.

9. 9 SAMPLING DISTRIBUTION Let X 1, X 2,…,X n be a r.s. of size n from a population and let T(x 1,x 2,…,x n ) be a real (or vector-valued) function whose domain includes the sample space of (X 1, X 2,…,X n ). Then, the r.v. or a random vector Y=T(X 1, X 2,…,X n ) is called a statistic. The probability distribution of a statistic Y is called the sampling distribution of Y.

10. 10 SAMPLING DISTRIBUTION The sample mean is the arithmetic average of the values in a r.s. The sample variance is the statistic defined by The sample standard deviation is the statistic defined by S.

11. 11 SAMPLING FROM THE NORMAL DISTRIBUTION Properties of the Sample Mean and Sample Variance Let X 1, X 2,…,X n be a r.s. of size n from a N( ,  2 ) distribution. Then,

12. 12 SAMPLING FROM THE NORMAL DISTRIBUTION Let X 1, X 2,…,X n be a r.s. of size n from a N( ,  2 ) distribution. Then, Most of the time  is unknown, so we use:

13. 13 SAMPLING FROM THE NORMAL DISTRIBUTION In statistical inference, Student’s t distribution is very important.

14. 14 SAMPLING FROM THE NORMAL DISTRIBUTION Let X 1, X 2,…,X n be a r.s. of size n from a N(  X,  X 2 ) distribution and let Y 1,Y 2,…,Y m be a r.s. of size m from an independent N(  Y,  Y 2 ). If we are interested in comparing the variability of the populations, one quantity of interest would be the ratio

15. 15 SAMPLING FROM THE NORMAL DISTRIBUTION The F distribution allows us to compare these quantities by giving the distribution of If X~F p,q, then 1/X~F q,p. If X~t q, then X 2 ~F 1,q.

16. 16 CENTRAL LIMIT THEOREM If a random sample is drawn from any population, the sampling distribution of the sample mean is approximately normal for a sufficiently large sample size. The larger the sample size, the more closely the sampling distribution of X will resemble a normal distribution. Random Sample (X 1, X 2, X 3, …,X n ) Sample Mean Distribution X Random Variable (Population) Distribution

17. 17 Sampling Distribution of the Sample Mean If X is normal, is normal. If X is non-normal, is approximately normally distributed for sample size greater than or equal to 30.

18. 18 The amount of soda pop in each bottle is normally distributed with a mean of 32.2 ounces and a standard deviation of 0.3 ounces. – Find the probability that a bottle bought by a customer will contain more than 32 ounces. – Solution The random variable X is the amount of soda in a bottle.  = 32.2 0.7486 x = 32 EXAMPLE 1

19. 19  = 32.2 0.7486 x = 32 Find the probability that a carton of four bottles will have a mean of more than 32 ounces of soda per bottle. Solution – Define the random variable as the mean amount of soda per bottle. 0.9082 EXAMPLE 1 (contd.)

20. 20 The estimate of p = The parameter of interest for nominal data is the proportion of times a particular outcome (success) occurs. To estimate the population proportion ‘p’ we use the sample proportion. Sampling Distribution of a Proportion p^= Xn The number of successes

21. 21 Since X is binomial, probabilities about can be calculated from the binomial distribution. Yet, for inference about we prefer to use normal approximation to the binomial whenever it approximation is appropriate. p^ Sampling Distribution of a Proportion p^

22. 22 Approximate Sampling Distribution of a Sample Proportion From the laws of expected value and variance, it can be shown that E( ) = p and V( )=p(1-p)/n If both np ≥ 5 and n(1-p) ≥ 5, then Z is approximately standard normally distributed.

23. 23 EXAMPLE – A state representative received 52% of the votes in the last election. – One year later the representative wanted to study his popularity. – If his popularity has not changed, what is the probability that more than half of a sample of 300 voters would vote for him?

24. 24 EXAMPLE (contd.) Solution The number of respondents who prefer the representative is binomial with n = 300 and p =.52. Thus, np = 300(.52) = 156 and n(1-p) = 300(1-.52) = 144 (both greater than 5)