1. Horng-Chyi HorngStatistics II127 Summary Table of Influence Procedures for a Single Sample (I) &4-8 (&8-6)

2. Horng-Chyi HorngStatistics II128 Summary Table of Influence Procedures for a Single Sample (II)

3. Horng-Chyi HorngStatistics II129 Testing for Goodness of Fit n In general, we do not know the underlying distribution of the population, and we wish to test the hypothesis that a particular distribution will be satisfactory as a population model. n Probability Plotting can only be used for examining whether a population is normal distributed. n Histogram Plotting and others can only be used to guess the possible underlying distribution type. &4-9 (&8-7)

4. Horng-Chyi HorngStatistics II130 Goodness-of-Fit Test (I) n A random sample of size n from a population whose probability distribution is unknown. n These n observations are arranged in a frequency histogram, having k bins or class intervals. n Let O i be the observed frequency in the ith class interval, and E i be the expected frequency in the ith class interval from the hypothesized probability distribution, the test statistics is

5. Horng-Chyi HorngStatistics II131 Goodness-of-Fit Test (II) n If the population follows the hypothesized distribution, X 0 2 has approximately a chi-square distribution with k-p-1 d.f., where p represents the number of parameters of the hypothesized distribution estimated by sample statistics. n That is, n Reject the hypothesis if

6. Horng-Chyi HorngStatistics II132 Goodness-of-Fit Test (III) n Class intervals are not required to be equal width. n The minimum value of expected frequency can not be to small. 3, 4, and 5 are ideal minimum values. n When the minimum value of expected frequency is too small, we can combine this class interval with its neighborhood class intervals. In this case, k would be reduced by one.

7. Horng-Chyi HorngStatistics II133 Example 8-18 The number of defects in printed circuit boards is hypothesized to follow a Poisson distribution. A random sample of size 60 printed boards has been collected, and the number of defects observed as the table below: The only parameter in Poisson distribution is, can be estimated by the sample mean = {0(32) + 1(15) + 2(19) + 3(4)}/60 = 0.75. Therefore, the expected frequency is: The only parameter in Poisson distribution is, can be estimated by the sample mean = {0(32) + 1(15) + 2(19) + 3(4)}/60 = 0.75. Therefore, the expected frequency is:

8. Horng-Chyi HorngStatistics II134 Example 8-18 (Cont.) n Since the expected frequency in the last cell is less than 3, we combine the last two cells:

9. Horng-Chyi HorngStatistics II135 Example 8-18 (Cont.) 1.The variable of interest is the form of distribution of defects in printed circuit boards. 2.H 0 : The form of distribution of defects is Poisson H 1 : The form of distribution of defects is not Poisson 3.k = 3, p = 1, k-p-1 = 1 d.f. 4. At  = 0.05, we reject H 0 if X 2 0 > X 2 0.05, 1 = 3.84 5.The test statistics is: 6.Since X 2 0 = 2.94 < X 2 0.05, 1 = 3.84, we are unable to reject the null hypothesis that the distribution of defects in printed circuit boards is Poisson.

10. Horng-Chyi HorngStatistics II136 Contingency Table Tests n Example 8-20 A company has to choose among three pension plans. Management wishes to know whether the preference for plans is independent of job classification and wants to use  = 0.05. The opinions of a random sample of 500 employees are shown in Table 8-4. (&8-8)

11. Horng-Chyi HorngStatistics II137 Contingency Table Test - The Problem Formulation (I) n There are two classifications, one has r levels and the other has c levels. (3 pension plans and 2 type of workers) n Want to know whether two methods of classification are statistically independent. (whether the preference of pension plans is independent of job classification) n The table:

12. Horng-Chyi HorngStatistics II138 Contingency Table Test - The Problem Formulation (II) n Let p ij be the probability that a random selected element falls in the ij th cell, given that the two classifications are independent. Then p ij = u i v j, where the estimator for u i and v j are n Therefore, the expected frequency of each cell is n Then, for large n, the statistic has an approximate chi-square distribution with (r-1)(c-1) d.f.

13. Horng-Chyi HorngStatistics II139 Example 8-20

14. Horng-Chyi HorngStatistics II140