1. Addition of Independent Normal Random Variables
Theorem 1 :
Let X and Y be two independent random variables with the N(, 2) distribution. Then, is N(2, 22) .
Proof :

3. Subtraction of Independent Normal Random Variables
Theorem 2 :
Let X and Y be two independent random variables with the N(, 2) distribution. Then, is N(0, 22) .
Proof :
Similar to the proof of Theorem 1.

4. Generalization of Theorem 1
Let X1, X2, ……, Xn be n mutually independent normal random variables with means μ1, μ2, …, μn and variances , respectively.
Then, is

5. Chi-Square Distribution with High Degree of Freedom
Assume that X1, X2, ……, Xk are k independent random variables and each Xi is N(0, 1).
Then, the random variable Z defined by is called a chi-square random variable with degree of freedom =k and is denoted by

6. Addition of Chi-Square Distributions

7. Example of Chi-Square Distribution with Degree of Freedom = 2
Let X,Y be two independent standard normal random variables, and Z = X2 + Y2.

8. Example of Chi-Square Distribution with Degree of Freedom = 3
Let X,Y,Z be three independent standard normal random variables, and W = X2 +Y2 + Z2.

9. Distribution Function of is

10. Note that,
The surface area of a sphere in the k-dimensional vector space is

11. Moment-Generating Function of the Distribution
The moment-generating function of a distribution with p.d.f f(x) is defined to be
Note that

12. Theorem: The moment-generating function
Mk(z) of is
Proof: since the p.d.f of is only defined in [0,∞)
we now consider two cases:
(1) k=2h
(2) k=2h+1

13. Case(1):

14. By applying the same technique repetitively we get
we can prove that for both k=2h and k=2h+1 cases,

15. implies that

16. Estimation of the Expected Value of a Normal Distribution
Let X be a normal random variable with unknown μ and σ2.
Assume that we take n random samples of X and want to estimate μ andσ2 of X based on the samples.
Let denote an estimation of μ.
Then, the likelihood function of n samples is

17. ~continues
Therefore, is a maximum likelihood estimator of μ.

18. ~continues
Let X1, X2, … Xn be the random variables corresponding to sampling X n times.
Since is called an unbiased estimator of μ.
Furthermore, since ,the confidence interval of μ approaches 0 , as n →∞, provided that is used as the estimator of μ.
The confidence interval of μ is

19. Estimation of the Variance of a Normal Distribution

22. That is the confidence interval of S2 approaches 0 as n→∞, provided that S2 is used as the estimator of σ2

23. An Important Observation
There are more general situation in which a degree of freedom of chi-square distributions is lost for each parameter estimated.

24. Test of Statistical Hypotheses
The goal is to prove that a hypothesis H0 does not hold in the statistical sense.
We first assume that H0 holds and design a statistical experiment based on the assumption.
The probability that we reject H0 when H0 actually holds is called the significance level of the test. Typically, we use  to denote the significance level.

26. An Example of Statistical Tests
We want to prove that a coin is biased.
We make the following hypothesis “The coin is unbiased”.
We design an experiment that is to toss the coin n times and we claim the coin is biased, i.e. rejecting the hypothesis, if we observe either one side is up k times or more, where k > ½ n.

27. ~continues
Let X be the random variable corresponding the number of times that one particular side is up in n tosses.
Under the hypothesis “The coin is unbiased”, approaches N(0, 1).
The significance level of the test is

29. ~continues
Assume that we want to achieve a significance level of 0.05 and we toss the coin 100 times. Since ,
Assume that we want to achieve the same level of significance and we toss the coin 1000 times. Then,

30. Test of Equality of Several Means
Assume that we conduct k experiments and all the outcomes of the k experiments are normally distributed with a common variance. Our concern now is whether these k normal distributions, N(1,2), N(2,2),…, N(k,2), have a common mean, i.e. 1= 2=…= k.

31. One application of this type of statistical tests is to determine whether the students in several schools have similar academic performance.
The hypothesis of the test is . 1= 2=…= k.

32. Let ni denote the number of smaples that we take from distribution N(i,2).
As a result, we have the following radom variables:
X11, X12,…, X1n1 : samples from N(1,2).
X21, X22,…, X2n2 : samples from N(2,2).
… … … … …
Xk1, Xk2,…, Xknk : samples from N(k,2).

37. Chi-Square Test of Independence for 2x2 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.

38. ~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 H0 :“ The effectiveness of the new treatment is the same as that of the conventional treatment.”

39. ~continues
Under H0, the two rows of the 2x2 contingence table corresponds to two independent binomial experiments, denoted by X1 and X2 , respectively.
Define parameters as follows.

40. Let Y1, Y2, …, Yn1 be n1 samples taken from a normal distribution N( p, p(1-p) ).
Then, is

41. Therefore, the distribution of Z1 approaches that of
Therefore, the distribution of Z1 approaches that of Similarly, let be samples taken from a normal distribution N( p, p(1-p)). Then the distribution of Z2 approaches that of
Since and are samples from a statisctical test of two means,
is , where

42. Since the distributions of Z1 and Z2 approach those of and , respectively,
approaches
is an estimator of the mean of Yi and Wj, which is p. Therefore, if we use as the estimator,
then we have approaches

43. According to our previous observation,
we have
In conclusion,

44. ~continues
Applying the data in our example to the equation that we just desired, we get
Therefore, we have over 97.5% confidence that the new treatment is more effective than the conventional treatment.

45. ~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.

46. 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.

47. Measure of Correlation
Correlation (A,B) = P(A&B) / P(A)P(B) = P(A|B)P(B) = P(B|A)P(A).
Correlation(a,b) = 1 implies that A and B are independent.
Correlation(a,b) > 1 implies that A and B are positively correlated.
Correlation(a,b) < 1 implies that A and B are negatively correlated.

48. Generalizing the Chi-Square Test of Independence
Given a 32 contingency table as follows.
Let Z1, Z2 and Z3 be the three random variables corresponding to the experiments defined by the three rows in the contingency table.

49. Then

50. The Chi-Square Statistic for Multinomial Experiments
Given the 32 contingency table .
We can regard the table as the outcomes of 2 independent multinomial experiments.

51. Therefore, we derive the following fact about a multinomial experiments as follows:

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

53. Determining the Degree of Freedom of a 2-D contingency Table
Once the counts in cells marked by “” are determined. Then, the remaining counts are determined.

54. Determining the Degree of Freedom on 3-D contingency Table
(rs – 1)(t – 1) + (r – 1)(s – 1) = rst – r – s – t + 2
t layers
s columns
r rows : :