1. Maximum Likelihood Estimator of Proportion Let {s 1,s 2,…,s n } be a set of independent outcomes from a Bernoulli experiment with unknown probability. The likelihood function of probability p is

3. Unbiased Estimator Let G （ X 1,X 2,…,X n ） be an estimator of a parameter r in a distribution. If E[G （ X 1,X 2,…,X n ） ]=r, then G （ X 1,X 2,…,X n ） is called an unbiased estimator of r. For example, is an unbiased estimator of the probability in a Bernoulli distribution, since

4. Estimation of the Expected value of a Normal Distribution is a unbiased estimator of μ, since

5. Estimation of the Variance of a Normal Distribution

7. Maximum Likelihood Estimators of the Normal Distributions Let x 1, x 2,…, x n be random samples from a normal distribution with p.d.f The likelihood function is

9. The Central Limit Theorem Let X 1,X 2, …,X n be random samples from a distribution with mean µ and variance σ 2. Then, Similarly, we can say

10. A Formal Representation of the Central Limit Theorem

12. Approximations for Discrete Distribution The beauty of the central limit theorem is that it holds regardless of the type of the underlining distribution. For example, X 1,X 2,… X n are random samples from a Bernoulli distribution with μ= p and σ 2 = p (1-p). Then, by the central limit theorem,

13. Confidence Intervals for a Statistical Estimator We know that is an unbiased estimator of the probability of a Bernoulli distribution. But, how good is it? According to the central limit theorem, if n is sufficiently large, then approaches N(0,1).

14. One Sided Confidence Interval for Proportions z

15. One Sided Confidence Intervals for Proportion

16. Two Sided Confidence Interval for Proportions

18. Important Observations about Confidence Intervals As n increases, we have a smaller confidence interval. If we require a higher confidence level, then the corresponding confidence interval is larger.

19. An example of Confidence Intervals for Proportions If a poll shows that 700 people out of 1000 favors a public policy, then how confident are we when we say that 70% of the population favor the policy?

20. According to the standard normal distribution table, Therefore, we have 95% of confidence that p  0.6756 and 90% of confidence that p  0.6811. Similarly, we have 95% of confidence that p  0.7233. ~continues

21. Test of Statistical Hypotheses Assume that a product engineer is requested to evaluate whether the defect rate of a manufacturing process is lower than 6%. If the engineer observes 5 out of 200 products checked have defects, can the engineer claim that the quality of the manufacturing process is acceptable?

22. Statistical Hypotheses In this example, we have two hypotheses concerning the defect rate of the manufacturing process r: H 0 : r=0.06; H 1 : r<0.06.

23. H 0 is called the null hypothesis and H 1 is called the alternative hypothesis. There are two possible types of errors: Type I: Rejecting H 0 and accepting H 1, when H 0 is true; Type II: Accepting H 0, when H 0 is false. The probability that Type I error occurs is normally denoted by and is called the significance level of the test. The probability that Type II error occurs is normally denoted by.

24. Design of a Statistical Test Since we do not know exactly what the defect rate of the manufacturing process is, we start with assuming H 0 is true, i.e. r=0.06.

25. Then, if the observed defect ratio r’ is lower than the lower bound of the confidence interval of r, then we say that it is highly unlikely that H 0 is true. In other words, we have a high confidence to reject H 0. In this example, the designer tends to reject H 0 and accept H 1.

26. Let X 1, X 2,…, X 200 correspond to whether each of the 200 random samples is defective or not. According to the central limit theorem, approaches N(0.06,0.000282). According to the central limit theorem,

27. Since 5/200 = 0.025 < 0.0324, we have over 95% confidence, 98% actually, to reject H 0 and accept H 1. In this example, the significance level of the test,, is 0.02.

28. Assume that the engineer wants to have a confidence level of 99%. Then, the criterion for rejecting H 0 must satisfy In practice, the engineer can reject H 0, if the number of defective products is less or equal to 4 in 200 random samples.

29. Chi-Square Test of Independence for 2 x2 Contingence Tables Assume a doctor wants to determine whether a new treatment can further improve the condition of liver cancer patients. Following is the data the doctor has collected after a certain period of clinical trials.

30. ~continues The improvement rate when the new treatment is applied is 0.85 and the rate is 0.77 when the conventional treatment is not applied. So, we observe difference. However, is the difference statistically significant? To conduct the statistical test, we set the following hypothesis H 0 :“ The effectiveness of the new treatment is the same as that of the conventional treatment.”

31. Contingence Table with m Rows and n Columns The chi-square statistic has degree of freedom (m-1)(n-1).

32. ~continues Applying the data in our example to the chi-square statistic equation, we get Therefore, we have over 97.5% confidence that the new treatment is more effective than the conventional treatment.

33. ~continues On the other hand, if the number of patients that have been involved in the clinical trials is reduced by one half, then we get Therefore, we have less than 90% confidence when claiming that the new treatment is more effective than the conventional treatment.

34. A Remark on the Chi-Square Test of independence The chi-square test of independence only tell us whether two factors are dependent or not. It does not tell us whether they are positively correlated or negatively correlated. For example, the following data set gives us exactly identical chi-square value as our previous example.